WEBVTT physics/ap-physics-c-electricity-magnetism/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I'm Dan Fullerton, and today, we are going to talk about electric potential and electric potential energy.
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Our objectives include determining the electric potential in the vicinity of 1 or more point charges.
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Calculating the electrical work done on a charge.
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Determining the direction and approximate magnitude of the electric field and various positions given a sketch of equal potentials.
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Calculating the electrostatic potential energy of a system of 2 or more point charges.
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Finally, stating a relationship between electric field and electric potential.
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As we get into this one, please understand electric potential is one of the trickier concepts as we are talking about electricity and magnetism.
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It is not one that typically makes a whole lot of intuitive sense, at least the first couple of times you see that.
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Some of this may feel little bit nebulous and cloudy as you go through it the first time but that is not unusual at all.
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Alright, electric potential energy, let us start there.
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When an object is lifted against gravity by applying a force for some distance, we had to do work to give that object gravitational potential energy.
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When the charged object is moved against an electric field by applying a force for some distance
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we also have to do work to give that object electric potential energy.
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The work done per unit charge in moving a charge between 2 points in an electric field is a scalar known as the electric potential.
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More often times, you will hear that referred to, rather informally as voltage.
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The units are known as V or 1 V is a J/ C.
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The work done is equal to the change in the objects electric potential energy which
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we are going to symbolize U for potential energy and e for electrical,
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where the electrical potential energy is the charge × the electric potential d.
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Taking a look at charge from work.
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A potential difference of 10 V exist between 2 points A and B in an electric field,
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what is the magnitude of charge that requires 2 × 10⁻² J of work to move it from A to B?
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We have a potential difference between A and B of 10 V, we do not know charge but
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we do know that the work done in moving it from A to B was 2 × 10⁻² J.
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Work going from A to B = Q × electric potential.
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Therefore, our charged Q is the work done ÷ the electric potential which is 2 × 10⁻² J ÷ 10 V.
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Or charge is equal to 2 × 10⁻³ C, pretty straightforward problem to start us off.
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Let us take a look at electric energy.
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How much electrical energy is required to move a 4 micro C charge to a potential difference of 36 V?
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We are looking for electric potential energy.
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We know our charge is 4 micro C or 4 × 10⁻⁶ C and our potential difference is 36 V.
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U = QV which is 4 × 10⁻⁶ C × 36 V which gives us the electric potential energy of 1.44 × 10⁻⁴ J,
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another straightforward starter sort of problem.
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When we talk about these different types of energy, oftentimes electrical energy into the work done is a very small portion of the Joule.
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So small that talking about Joules does not always make a whole lot of sense.
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A smaller alternate nonstandard unit of energy that is oftentimes much more convenient to use is known as the electron volt, given the abbreviation eV.
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Where 1 eV is the amount of work done in moving an elementary charge through a potential difference of 1 V.
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The charge on 1 electron move to a potential difference of 1 V is 1 eV.
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The conversion ratio 1 eV is 1.6 × 10 ⁻19 J just like 1 elementary charge is 1.6 × 10 ⁻19 C.
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An example with energy eV, we have a proton move through a potential difference of 10 V in an electric field,
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how much work in eV was required to move this charge?
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Our charge is + 1 elementary charge, our potential difference is 10 V.
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Work done on this charge × V which is 1 elementary charge × 10 V or 10 eV.
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This is straightforward when you are dealing with these tiny units and elementary charges.
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It makes it much more convenient.
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Oftentimes, we you think about topographic maps, maps of showing elevation, they show you lines of equal altitude or equal gravitational potential.
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We have the same sort of thing in the electricity world.
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The lines connecting points of equal electric potential are known as equal potential lines.
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Equal potential lines always cross electrical field lines at right angles.
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They are always perpendicular.
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If you move the charge particle in space and stay on equal potential line, you cannot do any work.
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You have to change the equal potential line you are on.
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You have to change your potential in order to do work.
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If you where to start over here for example, and wander through any path and come up back to the same equal potential line, the work done is 0.
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As equal potential lines get closer together, the gradient of the potential increases, equivalent kind of to a steeper slope of potentials.
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An electric field points from high to low potential.
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And because you had this analogy with slopes, they actually even call this the gradient as you get deeper into physics.
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Let us take a look at some equal potential lines.
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If you wanted to draw an equal potential lines on these electric field diagrams,
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we have to remember that they always intersect field lines at right angles.
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They are perpendicular.
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If I were to do it over here, I would probably draw an equal potential line something like that, especially if I had better art skills.
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And you can draw another equal potential line out here but notice it is always crossing these at right angles.
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Those should be perfect circles but not so good at circles.
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The same idea over here, we would draw equal potential lines they would be circles around these charges,
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with the charges at the center always crossing at 90° angles.
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Down here however, an equal potential line would go like that so it is always crossing at right angles.
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Over here, you might have it looking something more like that.
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Down here, same idea, right in between you might have something like a D shape that is going to,
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When it is curling like that, let me have to draw them so they always intersect at right angles.
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My art skills are not so great, but I think you get the general idea.
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The electric potential energy due to a point charged, find the work required to take a point charge Q2 from infinity,
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some long ways away where its electric potential energy is 0 to some point it is a distance r away from the point charge Q1.
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Assuming these are both positive, if we have this charge due to a long ways away, we have got to do some work.
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We got to push it to get all the way over here to point Q2.
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How much work was done in doing that?
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I’m first going to define that we have an electrical force in this direction along the line from Q1 to Q2 we will call that fe, the electrical force.
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The work done is going to be the integral going from some position infinity to r =R of our force.
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If we are going this way and the force is going one way, the force is going the opposite direction,
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they will have to be - fe ⋅ dl from our definition of work in mechanics force with displacement.
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That is going to be equal to,
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I do not like that minus sign so I'm just going to switch these directions here and say
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we are going to go from r = R to infinity of fe ⋅, instead of dl, I’m going to write it in terms of our variable dr,
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which is going to be the integral from R to infinity of,
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the electrical force we know from Coulomb’s law, that is going to be 1/ 4 π ε₀ Q1 Q2 / R² and we also have dr here.
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I’m going to pull my constants out of the integration, 1/ 4 π ε₀ is all a constant in this case and
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Q1 Q2 their charges are not changing in this problem so they can be pulled out.
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This becomes Q1 Q2/4 π ε₀ integral from R to infinity of r⁻² dr.
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The work done we just have to integrate that, work = we still have our Q1 Q2/4 π ε₀ × the integral of r⁻² is just -1/ r.
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-1/ r evaluated from r to infinity which implies then that the work done is going to be Q1 Q2/4 π ε₀ -1/ infinity is 0, - -1/ r is just going to be 1/ R.
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If that was the work that we did, it implies that the potential energy that it now has must be the same, whatever work we did on,
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if it is starting with 0 now must be its potential energy.
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That is going to be 1/ 4 π ε 0 Q1 Q2/ R.
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Note here that this is independent of path because the Coulombic force, the electrical force is a conservative force.
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It does not matter what path you take to get there, you can push it straight there, you can wobble all around.
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But only the starting and ending points are going to matter.
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Let us see if we can find electric force from electric potential energy.
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Using that relationship for conservative force f = - du dl, in our case that is going to be - d/dr
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of our electric potential energy 1/4 π ε₀ Q1 Q2 / R.
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We will pull our constants out of the derivation so that is -1/ 4 π ε₀ Q1 Q2 × the derivative with respect to r of 1/ r.
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Therefore, the force is going to be 1/ 4 π ε₀, the derivative of 1/ r is going to be -1/ r² so we end up with Q1 Q2 / r², Coulomb’s law.
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We can go in the other direction as well which should make sense.
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How about electric potential due to a point charge?
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Electric potential known as voltage is the work per unit charge required to bring a charge from infinity to some point r in an electric field.
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Potential is a work done per unit charge which is going to be 1/4 π ε₀ Q1 Q2/ R our work ÷ our charge q.
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What we are going to get is 1/ 4 π ε₀ q/ r.
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If you happen to have more than 1 charge, just add up the electric potentials due to each of those individual charges.
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Since, electric potential is a scalar not a vector, you do not have to worry about direction.
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It makes it a whole lot simpler.
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Electric potential is the sum over all i for how many charges you happen to have Qi, Q2, Q3, Q4, however many happen you have of 1/ 4 π ε₀ Qi/ ri.
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Since that is a constant, that can come out as 1/ 4 π ε₀ sum /i of their charges ÷ the distance.
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Let us see if we can do a couple samples to go along with these.
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Find the electric potential at point P located 3 m from the -2 C charge.
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What is the electric potential energy of ½ C charge situated at point P?
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Let us find its potential first.
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The potential point P has 1 /4 π ε₀ Q/ r which is 1/4 π ε₀.
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Our charge is -2 C ÷ our distance 3 m which is -6 × 10⁻⁹ V.
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To find the electric potential energy of ½ C charged at that point, that is going to be charge ×
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our potential which is 0.5 C × our potential which we just said was -6 × 10⁻⁹ V,
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for a total of -3 × 10⁻⁹ J.
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How about the potential due to point charges?
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In this problem should look fairly familiar.
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Last time we were using this to find electric field, now we want to find the electric potential at the origin due to the 3 charges shown in the diagram.
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And then finally, if an electron is placed at the origin what electric potential energy does it possess?
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Let start out with the electric potential and we will start out with electric potential from our green charge up here.
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The potential due to that green charge is 1/4 π ε₀ × Q/ r.
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Q is 2 C, our distance from the origin is going to be 8 m or 2.25 × 10⁹ V.
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Doing the same thing for our red charge down here, our potential is 1/4 π ε₀ × -2 C / 8 m or -2.25 × 10⁹ V.
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We have our blue charge, potential due to the blue charge is 1/4 π ε₀ × our charged 1 C/ the distance,
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these are Pythagorean theorem here, 2 and 2 , √2² + 2².
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We have √2² + 2² which is going to give us about 3.18 × 10⁹ V.
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T get the total then, all we have to do is add these up.
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2.25 × 10⁹ -2.25 × 10⁹, that is 0.
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It leaves us with just 3.18 × 10⁹ V, for the second part of the question.
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If we place an electron at the origin, there it is, what electric potential energy does it possess?
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Electric potential energy is charge × voltage, it is going to be -1.6 × 10 ⁻19 C × our voltage, our electric potential 3.18 × 10⁹ V or -5.1 × 10 ⁻10 J.
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We could also do that in eV if we wanted to make our math a little simpler.
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In that case, this would be -1 e, this would be the same, and we would come up with.
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Let me write that in 3.18 × 10⁹ V, we will multiply those to come up with -3.18 × 10⁹ eV.
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A little bit simpler math but note that this is non standard units.
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Finding the electric field from electric potential, we can do that as well.
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If our potential is 1/4 π ε₀ Q/ R, we can take the derivative of both sides to say that the derivative of the potential with respect to R
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must equal the derivative of the right hand side with respect to R which is 1/4 π ε₀ Q / R.
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Or pulling those constants out, Q is not going to change, 4 π ε₀ is not going to change, this becomes Q/4 π ε₀ × the derivative with respect to R of 1/ R.
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Which implies then that dv dr = we still have Q/ 4 π ε₀.
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The derivative of 1/ R is just -1/ R² so that is equal to -Q/4 π ε₀ R².
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If we multiply both sides in this equation, the dv dr and our answer over here by the unit vector and the R direction r ̂ ,
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This implies then the dv dr r ̂ = -2 /4 π ε₀ R² r ̂.
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If you recall our definition for electric field due to that point charge is Q/4 π ε₀ R², the direction of r ̂.
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Therefore, we can write then this is the opposite of the electric field or electric field = - dv dr in the direction of r ̂.
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There is a relationship we can use between the electric field and the electric potential.
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If we go that way, we should be able to find electric potential from electric field and the answer of course is yes.
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Let us do that.
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Let us start with electric potential V is W/ Q which is going to be 1/Q × our definition of the work done which is the integral from r to infinity of fe ⋅ dr.
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We did that previously, which is going to be the integral from r to infinity of the force ÷ the charge × ⋅dr.
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Which implies then given that the electric field is the force per unit charge, that we now have the potential is the integral from r to infinity of E ⋅ dr.
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True, but usually you see this written a little bit differently.
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Let us go ahead to show that.
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Potential difference which is what we are really talking about is the potential at some point B - the potential at some point A,
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potential difference from A to B, is equal to the opposite of the integral from A to B of E ⋅ dl
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which is how you normally going to see that written and probably is a little bit more useful.
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Both are correct but this one is probably the one that is much more common as opposed to going from here
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where you may see this written as - the integral from infinity to r of E ⋅ dr.
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This by the way is the change in electric potential energy per unit charge as well.
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A couple conversions that you go from potential to electric field and potential energy and how to put all of that wonderful stuff together.
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Let us do a couple more samples now.
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Find the electric potential of the origin due to the following charges.
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We have a 2 micro C charge at 3, 0 and -5 micro C charge at 0, 5 and 1 micro C charge at 4, 4.
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We are going to assume that all of these are given as points on the axis where that is 3 m, 5 m, and so on.
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The electric potential is 1/ 4 π ε₀ and then we just have to add up for all i, our charges ÷ their distance.
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It is going to be 1/ 4 π ε₀ × our first point is 2 × 10⁻⁶ C of 2 micro C and it is 3 m away from the origin.
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Then we have a second charge -5 × 10⁻⁶ C and it is 5 m away from the origin.
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And then finally, we have 1 micro C charge that is at 4, 4.
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Its distance from the origin is going to be √ 4² + 4².
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I do not have to worry about direction again since electric potential is a scalar, just plug through the math here and find that you are going to get about -1410 V.
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Another example, get lots of practicing.
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Given an electric potential function, V is a function of x as 5 x² -7 x, find the magnitude and direction of the electric field at x = 3 m.
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Our electric field we just said was –dv dr in the direction of r ̂ so that is going to be - the derivative with respect to x
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we are in 1 dimension here of 5x² -7x in the i ̂ direction is going to be - the derivative is going to be 10x - 7 in the i ̂ direction or that is 7 -10x (i.) ̂
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But we know that x = 3 m so we can solve for it specifically and say that the electric field is going to be 7 -10 × 3 m in the direction of i ̂.
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7 -30 is just going to be -23 V/ m in the i ̂ direction, one way to solve that.
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Let us put this together with a little bit of energy concerns.
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An electron was released from rest in a uniform electric field of 500 N/ C.
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What is its velocity after its traveled 1 m?
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Our change in kinetic energy is going to be equal to the opposite of the change in potential energy from conservation of energy.
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We also know here that our change in potential energy is going to be -Q integral from A to B of E ⋅ dr.
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Then change in kinetic energy = Q integral from A to B of E ⋅ dr.
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Which implies then that our change in kinetic energy is going to be how the electric field is constant,
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it is a uniform electric field, so we can pull that out.
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QE integral from A to B of dr which is just going to be QE × the change in R.
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Our change in kinetic energy, if it starts from rest is just going to be ½ MV², we have ½ MV² = QE Δr.
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Therefore, solving for V we can say that V = √2Q E Δr/ M.
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Substituting in with our known values that is going to be 2 × 1.6 × 10 ⁻19 C × 500 N/ C × 1 m ÷ mass of our electron 9.11 × 10 ⁻31 kg.
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You can find that on your formula sheet table of contents for the exam.
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There is our square root and I come up with about 1.33 × 10⁷ m/ s.
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Let us keep doing some more examples and make sure we really have these down.
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I will get into a couple of practice AP problems.
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The work required to establish a charge system, fairly common type of question.
00:29:07.000 --> 00:29:17.900
2 point charges 1, 5 micro C and one that is 2 micro C are placed ½ m apart, how much work was required to establish this charge system?
00:29:17.900 --> 00:29:25.900
We had these 2 charges to get them ½ m apart and had you have done some work because they do not want to be there beside each other, they repel each other.
00:29:25.900 --> 00:29:28.300
How much work did you have to do?
00:29:28.300 --> 00:29:53.000
That is the potential energy of the system which is 1/4 π ε 0 Q1 Q2/ R which is 1/4 π ε 0 × 5 × 10⁻⁶ × 2 × 10⁻⁶ ÷ ½ m.
00:29:53.000 --> 00:30:00.900
I get about 0.18 J.
00:30:00.900 --> 00:30:05.700
What is the electric potential halfway between the 2 charges?
00:30:05.700 --> 00:30:24.200
To find that, we will go with our formula for electric potential that is 1/4 π ε₀ × Q1 / R + Q2.
00:30:24.200 --> 00:30:30.000
Let us call that R1, R2, even though R1 and R2 are both going to be the same ¼ m.
00:30:30.000 --> 00:30:51.200
That is going to be 1/4 π ε₀ Q1 5 × 10⁻⁶ C/ R1 .25 m + the second charge 2 × 10⁻⁶ C ÷ ¼ m.
00:30:51.200 --> 00:31:03.900
I come up with about 252,000 V or 252 kV.
00:31:03.900 --> 00:31:10.000
Let us get into some of these AP style questions that are pretty challenging type questions and
00:31:10.000 --> 00:31:15.100
quite likely on the AP exam you are going to see, at least in the free response question.
00:31:15.100 --> 00:31:18.500
question that has some electric potential derivation.
00:31:18.500 --> 00:31:23.000
On most years, you are going to see Gauss’s law question and electric potential question.
00:31:23.000 --> 00:31:27.300
Sometimes, they are even involved in the same questions as we are going to see here.
00:31:27.300 --> 00:31:32.800
We will start off with a 2013 APC E and M exam free response number 1.
00:31:32.800 --> 00:31:36.600
It is highly recommend you pause the video, take a minute, download it and print in out,
00:31:36.600 --> 00:31:42.400
look it over for a few minutes before you come back to the video and hit play again.
00:31:42.400 --> 00:31:49.600
Let me pull the question out as well and we are looking at 2013 free response number 1 on the E and M exam,
00:31:49.600 --> 00:31:53.600
where we have got this very long solid none conducting cylinder.
00:31:53.600 --> 00:32:04.700
The first question it asks us, using Gauss’s law derive an expression for the magnitude of the electric field inside the edges of that cylinder.
00:32:04.700 --> 00:32:07.300
Draw an appropriate Gaussian surface on the diagram.
00:32:07.300 --> 00:32:11.800
First thing, let us give us a representation of that diagram here.
00:32:11.800 --> 00:32:23.200
I will put the center line, the axis, here in green.
00:32:23.200 --> 00:32:25.900
We have this cylinder that goes around it.
00:32:25.900 --> 00:32:28.000
I will put that in blue.
00:32:28.000 --> 00:32:39.700
Here is our cylinder, there we go.
00:32:39.700 --> 00:32:43.700
Using Gauss’s law, derive an expression for the electric field inside that.
00:32:43.700 --> 00:32:50.000
If I'm looking for the most symmetric Gaussian surface I can come up with, I think I’m going to draw a cylinder inside that.
00:32:50.000 --> 00:32:53.500
I will put that in here in purple so you can see it.
00:32:53.500 --> 00:33:04.900
Here is our Gaussian surface and now we are trying to find the magnitude of the electric field there.
00:33:04.900 --> 00:33:18.900
For part A, we will start by writing Gauss’s law integral / the close surface of E ⋅ da, it is our total charge enclosed ÷ ε 0.
00:33:18.900 --> 00:33:27.900
The left hand side and we have done this a couple of times already, with Gauss’s law it is just going to be E × the area of our Gaussian surface.
00:33:27.900 --> 00:33:39.000
The right hand side, our charge enclosed is going to be that volume charge density ρ × the volume that is enclosed by ÷ ε₀.
00:33:39.000 --> 00:34:00.000
Our left hand side becomes E and that area 2 π R × some length l, which we will define as the length of our cylinder must equal ρ × our volume.
00:34:00.000 --> 00:34:07.900
We have got π R² × the length ÷ ε₀.
00:34:07.900 --> 00:34:26.700
Solving for the electric field then, E is going to be equal to, we have got ρ × π R² l / 2π Rl ε₀.
00:34:26.700 --> 00:34:29.400
A couple of cancellation simplifications we can make here.
00:34:29.400 --> 00:34:47.400
We got a π, we have got a π, and l and l, and R and R², and I think that I will do it and come up with ρ × r / 2 ε₀.
00:34:47.400 --> 00:34:55.300
The electric field inside that long solid, none conducting cylinder.
00:34:55.300 --> 00:35:04.400
For part B, we are asks to use Gauss’s law to derive an expression for the magnitude of the electric field outside of that.
00:35:04.400 --> 00:35:09.400
At some r that is greater than the radius R of the blue cylinder.
00:35:09.400 --> 00:35:22.100
We will go back to Gauss’s law again, integral/ the close surface of E ⋅ da = our enclosed charge ÷ ε 0,
00:35:22.100 --> 00:35:38.000
which implies then the left hand side Ea = the charge enclose is just going to be the volume charge density ρ × V ÷ ε₀.
00:35:38.000 --> 00:35:55.200
Our left hand side E × our area 2 π R × length must equal the right hand side, ρ × our volume which is now π R² l ÷ ε₀,
00:35:55.200 --> 00:36:02.900
Because the charge stops when you get to the edges of that R cylinder, when you get to that radius.
00:36:02.900 --> 00:36:18.800
Then a little simplification again E = ρ × π R² l/ 2 π Rl ε₀.
00:36:18.800 --> 00:36:27.800
We have got π and π, an R and l, we do not have an R, we still have that R down there.
00:36:27.800 --> 00:36:30.500
Let me redraw him in R.
00:36:30.500 --> 00:36:45.900
That gives us ρ R²/ 2 ε₀ R.
00:36:45.900 --> 00:36:49.000
Next up, move in on to C.
00:36:49.000 --> 00:36:57.500
On the axis sketch a graph of the electric field as a function of radial distance from r=0 to 2R.
00:36:57.500 --> 00:37:13.100
Let us draw our axis in here first.
00:37:13.100 --> 00:37:24.100
Here we have electric field, here we have r, we have got R in here, and we have got 2R.
00:37:24.100 --> 00:37:27.000
We need to know the electric field.
00:37:27.000 --> 00:37:35.300
As we go from 0 to R, we have already found out that that is a ρ R/ 2 ε₀ so
00:37:35.300 --> 00:37:46.900
we can draw our nice linear fit there and after that it looks like we fall off with respect to 1/ R.
00:37:46.900 --> 00:38:01.000
R there is a constant so then this would be sort of this shape and things I would label right there at that point where r = R.
00:38:01.000 --> 00:38:06.500
We got ρ R/ 2 ε₀.
00:38:06.500 --> 00:38:15.800
We know that because we can plug in R up here for little r or we can plug R down here for little r and get the same thing either way.
00:38:15.800 --> 00:38:24.500
And then we get over here to 2r, we also have another important point, that little r =2R.
00:38:24.500 --> 00:38:30.600
We end up with ρ R/ 4 ε₀.
00:38:30.600 --> 00:38:37.000
I would use that as my graph, therefore part C.
00:38:37.000 --> 00:38:41.200
I think that covers C pretty well.
00:38:41.200 --> 00:38:52.400
Let us take a look now at D, derive an expression for the magnitude of the potential difference from r = 0 to R.
00:38:52.400 --> 00:38:59.500
Alright, let us start with E = - dv dr in the direction of r ̂.
00:38:59.500 --> 00:39:05.100
There are other ways you can do this, you got to know your formulas, you can jump in a few steps ahead
00:39:05.100 --> 00:39:09.900
But it is like doing this every now and then starting from the simpler expressions.
00:39:09.900 --> 00:39:28.800
Then dv dr = - E which implies then that V = - the integral of E ⋅ dr which is going to be - the integral from r = 0 to R.
00:39:28.800 --> 00:39:47.100
We already found that our electric field that is ρ r/ 2 ε₀ dr which implies then that our potential V = let us pull other constants ρ, 2, and ε₀.
00:39:47.100 --> 00:39:57.400
We will have - ρ/ 2 ε₀ integral from 0 to R of r dr.
00:39:57.400 --> 00:40:16.500
Or - ρ/ 2 ε₀ × R²/ 2 evaluated from 0 to R, which is going to be - ρ/ 2 ε₀ × R²/ 2 -0,
00:40:16.500 --> 00:40:27.400
which implies then that our potential is going to be - ρ R²/ 4 ε₀.
00:40:27.400 --> 00:40:41.400
Or the magnitude of the potential difference is just going to be ρ R²/ 4 ε₀.
00:40:41.400 --> 00:40:52.800
Part 2 D2, is a potential wire at r =0 or r = R?
00:40:52.800 --> 00:40:58.900
As we went from 0 to r, we obtained a negative potential difference.
00:40:58.900 --> 00:41:06.600
Let VB - VA was – ρ R²/ 4 ε₀.
00:41:06.600 --> 00:41:31.000
If we said VA - VB must be ρ R² / 4 ε₀, A must be bigger or a potential is higher where A is what we are going to call r = 0 and B is r = R.
00:41:31.000 --> 00:41:43.900
I would say then that we are higher at r = 0.
00:41:43.900 --> 00:41:49.800
And finally, they ask us to do another graph here.
00:41:49.800 --> 00:41:55.900
The non conducting cylinder is replaced with a conducting cylinder of the same shape and same linear charge density.
00:41:55.900 --> 00:42:00.100
Sketch the electric field from r = 0 to 2r.
00:42:00.100 --> 00:42:05.100
Alright, let us go to the next page to do that.
00:42:05.100 --> 00:42:29.800
We will draw our axis again and here we have r, here we have our electric field, and we will mark off R and capital 2R.
00:42:29.800 --> 00:42:33.800
If it is a conductor, we know the electric field inside a conductor is 0.
00:42:33.800 --> 00:42:48.600
This piece becomes nice and simple E = 0 inside the conductor and outside it is the same as when we did part C.
00:42:48.600 --> 00:42:52.300
It does not matter once you are outside that, the enclosed charge is the same.
00:42:52.300 --> 00:42:57.100
We have something that looks like that.
00:42:57.100 --> 00:43:07.100
Where again, we have the same key points ρ r / 2 ε₀.
00:43:07.100 --> 00:43:09.800
I guess that is not drawn to scale very well.
00:43:09.800 --> 00:43:15.900
Or/ 4 ε₀, same as part C.
00:43:15.900 --> 00:43:21.500
That should cover that problem, the 2013 free response number 1.
00:43:21.500 --> 00:43:28.600
But let us keep going, let us take a look at the 2006 E and M free response number 1.
00:43:28.600 --> 00:43:33.400
Another interesting question, this one focusing more on the electric potential and
00:43:33.400 --> 00:43:37.600
some of the inner relationships with the electric field as opposed to Gauss’s law.
00:43:37.600 --> 00:43:46.100
Take a moment, print that out, look at it, give it a shot, come back here and we will see what we have got.
00:43:46.100 --> 00:43:53.000
Here we have 4s² with a point P in the middle and 4 charges on those points.
00:43:53.000 --> 00:44:04.100
We have got point P right here and around that we have this square where we have 4 different charges.
00:44:04.100 --> 00:44:12.100
We will call this corner 1, corner 2, corner 3, and corner 4.
00:44:12.100 --> 00:44:17.100
On the diagram, indicate with an arrow the direction of the net electric field at point P.
00:44:17.100 --> 00:44:21.400
Due to charge 4 it is going to be up into the right because they repel.
00:44:21.400 --> 00:44:24.600
Due to Q, it is going to be up in the right because it is attracted.
00:44:24.600 --> 00:44:27.500
And these 2 both negative are going to cancel out.
00:44:27.500 --> 00:44:33.800
Our net electric field at point P got to go that way.
00:44:33.800 --> 00:44:41.500
Moving on to part B, derive an expression for the magnitude of the electric field at point P.
00:44:41.500 --> 00:44:47.500
By symmetry, the electric field at 1 + electric field at .3 has to be equal to 0.
00:44:47.500 --> 00:44:51.000
We are going to worry about 2 and 4.
00:44:51.000 --> 00:44:59.200
The electric field at 0.2 is going to be Q/4 π ε₀ R².
00:44:59.200 --> 00:45:07.700
The electric field at .4 is going to be Q/4 π ε₀ R².
00:45:07.700 --> 00:45:10.100
But what is R² here?
00:45:10.100 --> 00:45:17.700
If this whole distance is a then that would be a/ 2, that is a/ 2.
00:45:17.700 --> 00:45:31.800
If we want to find R², use the Pythagorean theorem that is a/ 2² + a/ 2² which is 2a/ 2².
00:45:31.800 --> 00:45:46.900
And therefore, R² = 2a²/ 4 which is a²/ 2 and then r must equal a/ √2.
00:45:46.900 --> 00:45:55.100
The electric field at point P is the electric field at point P due to 2 + the electric field at point P due to 4
00:45:55.100 --> 00:46:17.500
which is going to be 2Q /4 π ε₀ R² which is 2Q/4 π ε₀ and R² in the denominator is going to give us a² and 2 up in the numerator.
00:46:17.500 --> 00:46:34.500
2 × 2 and a 4 down there, I end up with Q/ π ε₀ a².
00:46:34.500 --> 00:46:39.800
That will work, how about the electric potential at point P?
00:46:39.800 --> 00:46:45.600
Part 2 of this question of B, find the electric potential at point P.
00:46:45.600 --> 00:46:49.100
That should be a little more straightforward.
00:46:49.100 --> 00:47:01.300
Electric potential at point P just 1/4 π ε₀ × the sum of all those little charges of that Q/ R for each of those charges,
00:47:01.300 --> 00:47:07.400
which is going to be 1/4 π ε₀.
00:47:07.400 --> 00:47:20.900
The r are all the same, that is a constant because they are all the same distance from point P which is going to be a/ √2 × we have +Q and 3 –Q.
00:47:20.900 --> 00:47:36.000
-Q -Q -Q gives us -2Q √2/ 4a π ε₀.
00:47:36.000 --> 00:47:55.300
With a little simplification here, that is going to be -Q/ √2 and we have got π ε₀ a.
00:47:55.300 --> 00:48:03.500
How about C, a positive charge is placed at point P it is then moved from P to point R which is at the midpoint of the bottom side of the square.
00:48:03.500 --> 00:48:08.000
It is going to end up over here.
00:48:08.000 --> 00:48:13.100
If the charge is moved, is the work done by the electric field positive, negative, or 0?
00:48:13.100 --> 00:48:16.800
You got to explain it.
00:48:16.800 --> 00:48:22.100
For part C here, say that has to be negative, why?
00:48:22.100 --> 00:48:31.600
We only know the electric field is up into the right, the charge is being moved against that field so
00:48:31.600 --> 00:48:43.000
we are doing the work against the field, the field is not doing the work.
00:48:43.000 --> 00:48:49.700
That the charges moved against the field so the work done is less than 0.
00:48:49.700 --> 00:48:54.000
The work done on it by the electric field is negative because we are doing the work.
00:48:54.000 --> 00:48:59.300
We are giving the system energy.
00:48:59.300 --> 00:49:11.000
Moving on to part D, describe one way to replace a single charge in this configuration
00:49:11.000 --> 00:49:16.000
that would make the electric field at the center of the square equal to 0 and justify your answer.
00:49:16.000 --> 00:49:22.900
Let us draw this again, just so that we do not lose track of what we are doing.
00:49:22.900 --> 00:49:33.200
We have got these here, we have got P in the middle, and I think they call these 1, 2, 3,and 4.
00:49:33.200 --> 00:49:42.200
The first thing it asks us is what would we do in order to make the electric field at the center of the square =0?
00:49:42.200 --> 00:49:48.900
The way I would do that is I would replace the charge here at 2 that is negative with +Q.
00:49:48.900 --> 00:49:57.100
If we did that, we have +Q here, +Q here, -Q here and -Q here.
00:49:57.100 --> 00:49:59.700
The fields all cancel at point P.
00:49:59.700 --> 00:50:10.300
I would say something like replace Q2 with +Q.
00:50:10.300 --> 00:50:19.900
Fields cancel at point P.
00:50:19.900 --> 00:50:30.200
For part 2, describe a way to replace a single charge in this configuration so electric potential of the center of the square is 0 but the electric field is not.
00:50:30.200 --> 00:50:38.100
We start off again with, we put these charges back -Q -Q -Q and + Q.
00:50:38.100 --> 00:50:48.500
If we replace Q1 over here with + Q, let us make that a +.
00:50:48.500 --> 00:50:52.400
When we do that, the electric potential is going to sum to 0.
00:50:52.400 --> 00:50:56.600
Remember it does not care about direction, it is the scalar piece.
00:50:56.600 --> 00:50:59.700
I have a potential of 0 but the field no longer cancels.
00:50:59.700 --> 00:51:02.600
We are going to have a net electric field to the right.
00:51:02.600 --> 00:51:11.900
I would say that we can replace Q1 with +Q charge.
00:51:11.900 --> 00:51:33.300
V which is the sum of Q/ 4 π ε₀ R = 0 but field no longer cancels at point P.
00:51:33.300 --> 00:51:38.400
The electric field at point P instead would point to the right.
00:51:38.400 --> 00:51:42.500
Another AP style question down, good practice.
00:51:42.500 --> 00:51:49.900
Let us do another one, now we are going to go to 2006 E and M.
00:51:49.900 --> 00:51:51.200
I do not think we needed that page.
00:51:51.200 --> 00:51:59.400
Let us go to the 2005 E and M free response number 1 and then you can print out from the same place, take a minute to work through it,
00:51:59.400 --> 00:52:07.600
and come back here and un pause it, once you have a minute to look it over.
00:52:07.600 --> 00:52:15.100
We are given a graph that shows us an xy plot along with field lines on there,
00:52:15.100 --> 00:52:22.200
that says points A, B, and C are all located at y = 0.06 m and you can see that right from the graph.
00:52:22.200 --> 00:52:28.300
A1, what should these 3 points is the magnitude of the electric field the greatest?
00:52:28.300 --> 00:52:43.400
It is got to be the greatest at point C, since the field lines are the densest there.
00:52:43.400 --> 00:52:48.700
Dense field lines, strongest field.
00:52:48.700 --> 00:52:53.800
For A2, at which of these 3 points is the electric potential the greatest?
00:52:53.800 --> 00:52:56.700
That has got to be at point A.
00:52:56.700 --> 00:53:14.400
I would say that it is A because the field points from high to low potential
00:53:14.400 --> 00:53:26.100
which in this case is left to right, A is farthest left.
00:53:26.100 --> 00:53:38.000
Therefore, A is at highest potential.
00:53:38.000 --> 00:53:44.600
Moving on to question B, an electron is released from rest at point B.
00:53:44.600 --> 00:53:49.700
Qualitatively describe the electrons motion in terms of direction, speed, and acceleration.
00:53:49.700 --> 00:53:56.500
If we release an electron at point B, it is a negative charge so it is going to go in the direction opposite the field line.
00:53:56.500 --> 00:54:20.900
First thing is the electronic accelerates toward the left that moves with increasing speed because it is accelerating.
00:54:20.900 --> 00:54:27.500
But it is going to keep accelerating at smaller and smaller rates, the speed increases by smaller and smaller amounts.
00:54:27.500 --> 00:54:42.100
Let us write that.
00:54:42.100 --> 00:55:06.500
Smaller and smaller amount per unit time as it moves because acceleration is decreasing.
00:55:06.500 --> 00:55:12.800
For B2, calculate the electron speed after it is moved through a potential difference of 10 V.
00:55:12.800 --> 00:55:15.500
Another energy question.
00:55:15.500 --> 00:55:23.200
For B2, we would say that the W = QV.
00:55:23.200 --> 00:55:26.900
The work done in this case is going to be the change in kinetic energy.
00:55:26.900 --> 00:55:36.900
½ MV² = Q × B potential, be careful not to mix up v velocity and potential.
00:55:36.900 --> 00:55:51.500
Therefore, velocity² = 2Q × electric potential ÷ mass or V = √2Q B/ M.
00:55:51.500 --> 00:55:56.300
It gives us some values so we can actually solve for an exact speed.
00:55:56.300 --> 00:56:13.500
V= √2 × 1.6 × 10 ⁻19 C × our voltage 10 ÷ our mass 9.11 × 10 ⁻31 kg.
00:56:13.500 --> 00:56:22.000
I come up with about 1.87 × 10⁶ m/ s.
00:56:22.000 --> 00:56:29.700
Taking a look here at part C, points B and C are separated by a potential difference of 20 V.
00:56:29.700 --> 00:56:35.200
Estimate the magnitude of the electric field midway between them and state any assumptions that you make.
00:56:35.200 --> 00:56:42.900
The electric field is roughly the potential ÷ the distance assuming you are roughly uniform electric field
00:56:42.900 --> 00:56:56.400
which should be 20 V / it looks like we are about 1 cm, .01 m apart there which is 2000 V/ m.
00:56:56.400 --> 00:57:14.400
We have to write what we are assuming and say assuming a roughly uniform e field in that region.
00:57:14.400 --> 00:57:26.100
Onto D, on the diagram draw an equal potential line it passes through point B and intersects at least 3 electric field lines.
00:57:26.100 --> 00:57:34.800
I’m just going to sketch this very quickly, I think you can get the idea without too much,
00:57:34.800 --> 00:57:47.000
The graph looks kind of like that and we have some points coming in there, something kind of like that.
00:57:47.000 --> 00:57:49.400
We have got to be down here somewhere.
00:57:49.400 --> 00:57:57.100
What we need to do is we need to draw an equal potential line so that crosses all of these at right angles and goes through that point.
00:57:57.100 --> 00:58:01.300
Something like that, you have to look at your graph to see exactly how to do it.
00:58:01.300 --> 00:58:06.600
Draw these equal potential lines perpendicular to the field lines and make sure it goes through D.
00:58:06.600 --> 00:58:13.300
And that should cover that 2005 question.
00:58:13.300 --> 00:58:18.600
Finally, let us take a look at one last free response question in this unit then I will you a rest.
00:58:18.600 --> 00:58:22.800
Let us take a look at the 2003 E and M free response number 1.
00:58:22.800 --> 00:58:27.000
Another one that say, a little bit of a more unique sort of question.
00:58:27.000 --> 00:58:29.900
You do not see this very often.
00:58:29.900 --> 00:58:37.000
They give us a spherical cloud of charge of radius R with a total charge +Q and a non uniform volume charge density and
00:58:37.000 --> 00:58:43.100
they gave us the equation for where that the charges distributed.
00:58:43.100 --> 00:58:48.700
Determine the following as a function of r for when you are outside that sphere.
00:58:48.700 --> 00:58:55.100
The magnitude E of the electric field that is just a straightforward Gauss’s law question by now.
00:58:55.100 --> 00:58:58.500
Let us make sure we get that one right, still good practice.
00:58:58.500 --> 00:59:11.200
A1 the integral / the close surface of E ⋅da = the enclose charge ÷ ε₀ which implies then that
00:59:11.200 --> 00:59:22.300
the electric field × that area 4 π R² = our total enclose charge is just Q ÷ ε₀
00:59:22.300 --> 00:59:30.300
Or the electric field = Q/4 π ε₀ R².
00:59:30.300 --> 00:59:34.600
By now, quite a few of you could probably just look at that and say what the answer is because we have done it so much.
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It is same as if it was that + Q point charge in the center.
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However, take the time to go through this and get all the points for showing your work.
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How about the electric potential V in part 2?
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There is part 1, part 2, define the electric potential of V the opposite of the integral from infinity to R
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of E ⋅ dl is going to be - the integral from infinity to R of Q/4 π ε₀ R² dr.
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We will pull other constants -Q /4 π ε₀ integral from infinity to R of r⁻² dr which is -Q/4 π ε₀ × -1/ r
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evaluated from infinity to R which is as we have done before is well Q /4 π ε₀ r.
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You might have been able even to say that from a number of times we have done similar problems, go ahead show the work.
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Get every single possible point here.
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For part B, we have got a proton and we are going to place it at point P as shown and release it.
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Describe its motion for a long time after it is released.
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If we put it at P, the proton is going to move away from the sphere to the right and it is going to accelerate,
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it is going to be moving in increasing velocity but with a decreasing acceleration because the force is going to get smaller the further way it gets.
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I would write something along the lines of, the proton will move away from the sphere to the right, probably important to mention.
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An increasing velocity with a decreasing acceleration.
01:01:50.400 --> 01:02:01.000
Part C, an electron of charged magnitude E is now place at point P which is a distance r from the center of the sphere it is released,
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find the kinetic energy of the electron as a function of R as it strikes the cloud.
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This is a conservation of energy question and we have seen things like this before by now as well.
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Where the potential energy that r= to the potential energy that R + the kinetic R implies that the kinetic R = the potential energy of r - the electric potential energy R.
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Thankfully we know the formula for the electric potential energy U is Q1 Q2/4 π ε₀ R.
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Therefore, this implies then the potential energy at R is going to be - EQ/4 π ε₀ r - -E capital Q/4 π ε₀ R.
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Which implies in that the kinetic energy, if we want to simplify this and combine a little bit is just going to be EQ/4 π ε₀ × 1/ R -1/ r.
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Two more parts to the question, the next 2 pieces are little bit more math intensive but hang with it.
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If we look at part D, it says to derive an expression for ρ 0.
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The way we do that is we realize that the total charge in that sphere is + Q, it gives us the total.
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This is not really an electric fields question, so much as it is kind of a common sense to you, understand what the math and algebra means question.
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Our total charge is Q, which is going to be the integral as we go from our radius = 0 to R,
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the radius of that sphere of our volume charge density × the differential volume.
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Adding up all those charges as we go from 0 to the radius R should give us capital Q.
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We have got the formulas, we just have to back out what that ρ 0 is.
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To do this it would be helpful to start off with some knowledge of dv.
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Dv is going to be, as we go from 0 outwards, what we will take is we will take little tiny shells of area 4 π R² × some thickness dr.
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dv will be4 π R² dr and we also know that our volume charge density function ρ =ρ₀ × 1 - r/ R.
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Back to the math, Q = the integral from 0 to R of ρ which we said is ρ₀ × 1 - r/ R × 4 π R² dr.
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Which implies then that Q = let us pull out our constants.
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We have got a 4 π ρ₀, it can all come out.
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Integral from 0 to R of, we got this R² in here, let us distribute that through r² - r³/ R dr.
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Which will be 4 π ρ₀ the integral of r² is going to be r³/ 3 – r⁴/ 4R, remembering R is a constant,
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and those are evaluate from 0 to R.
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Which is going to be equal to 4 π ρ₀ × R³/ 3 – R⁴/ 4R.
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The 0 is just going to 0 out, we will get anything from that piece which is going to be 4 π ρ₀ ×
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let us see if we can get rid of the r⁴ and make that a cube.
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Get rid of that and find any common denominator of 12 here, we can write this as 4 R³/12 -3 R³/ 12,
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which implies then that Q = 4 π ρ 0 R³/ 12.
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To find ρ₀ itself, ρ₀ then must be 12Q /4 π R³ or 12/ 4 gives us just 3Q/ π R³.
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But we did a lot of work for that, let us put that in the 3D box.
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We should be proud of ourselves, 3Q/ π R³.
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One last piece, hang in there.
01:07:56.500 --> 01:08:07.100
For part E, determine the magnitude E of the electric field as a function of r, for r is less than or equal to R.
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When you are inside that sphere, it looks like you are probably going to be doing a bit more math again.
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I'm going to go with Gauss’s law, the integral / the close surface of E ⋅ da = Q enclose/ ε₀ which implies then since A = 4 π R².
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We know that the Q enclosed is going to be the integral from 0 to R of ρ dV.
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We can write then that 4 π R² × the electric field = 1/ ε₀ × the integral from r= 0 to R of ρ dv.
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We also can now fill in dv, just like before dv = 4 π R² dr.
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This implies then that 4 π r² × our electric field is 1/ ε₀ integral from 0 to r of ρ is ρ₀ × 1 - r / R × 4 π r² dr,
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which implies that then 4 π r² E, our left hand side = we will pull out the constants of the right hand side,
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We have got for constants 4 π, we have got a ρ₀, we got an ε₀ out there already.
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That is going to be 4 π₀/ ε₀ × the integral from 0 to r of r² - r³/ R dr.
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We are getting the set up.
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Let us see, we got a 4 π on both sides, we can maybe do some simplifications now.
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We got to cancel 4 π and 4 π, and write that left hand side r² × electric field = ρ₀ / ε₀ integral of r² is r³/ 3.
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We got r³/ 3 – r⁴/ 4R all evaluated from 0 to r, complies that r² capital E = ρ₀ /ε₀ × r²/ 3 – r⁴/ 4R.
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The 0 does not give us any factors here that we need to worry about.
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And we also now know from our part D problem that ρ₀ is equal to 3Q/ π R³.
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Our electric field if I start solving for E all by itself, we are going to have on our right hand side, we have got ρ know which we just said was 3Q/ π R³.
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We have still got the r² from the left hand side, we have got ε₀ there, and we also have this piece r³/ 3 – r⁴/ 4R.
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And now it becomes an exercise in algebra.
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Just trying to simplify this and pretty it up a little bit.
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We will take a minute and clean it up.
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That will be 3Q/ π ε₀ × R³, we will put the r² back in, in order to see what we can clean up there that will give us
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r² in the denominator will give us r/ 3 - r²/ 4R which is also Q/ I ε₀ R³.
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Let us see, if we multiply all of this by 3, factor out that 3 from there, we can get Q/ π ε₀ r³ 3.
01:13:12.300 --> 01:13:23.900
Put the 3 in that will give us r -3r²/ 4r which implies that E = Q.
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We can factor out a single lowercase r/ π ε₀ R³, 1 -3r/ 4R.
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We might be will simplify that further but I think that is plenty.
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There is our electric field as a function of r when you are inside that radius R.
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Alright, hopefully that gets you going on electric potential and electric potential energy and how all of these starts to play together.
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Thank you for watching www.educator.com.
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