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 in this lesson we are going to talk about electric potentials due to continuous charge distributions.
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Our objectives include calculating the electric potential on the axis of a uniformly charged disk.
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Deriving expressions for electric potential as a function of position for some uniformly charged wires,
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some parallel charge plates, coax cylinders and some concentric spheres.
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Let us dive right in and let us start looking at the potential due to a charged ring.
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Find the electric potential on the axis of a uniformly charged ring of radius R
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and total charge Q at point P located at distance Z from the center of the ring.
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First, let us draw a picture of the situation and we have done this before for electric field.
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That is going to look kind of like this.
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There are our axis, we will call these Z, Y, and that would be X.
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We will put our ring on it like that and we are going to look for the potential at some point P over there.
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The first thing I’m going to do again is break this up into some little portions where we have got some Δ Q over here.
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We will find the distance from Δ Q to that point P, there we go.
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We will call that RI.
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Now that we have done that, let us also mention that this radius is R.
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To find the potential of point P, that is going to be the sum over all these little bitty pieces of their potentials
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which is 1/ 4 π ε₀ × the sum over all I of the charge contained in that little bitty piece QI ÷ RI,
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which implies then the potential at point P is going to be 1/ 4 π ε₀.
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As we make all of these little bitty pieces smaller and smaller and add them up, we can integrate.
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Charge is going to be that little bit of charge DQ ÷ R.
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But in this case, our R, as we go around the circle is going to be the same.
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It is constant everywhere, RI = R.
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Since we know that RI = R, we can say that the potential at point P = 1/ 4 π ε₀.
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This is a constant, it can come out and we can replace it with RI integral of DQ.
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Our integration just got pretty easy.
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If we go all the way around the circle, adding up all the little bits of DQ, the integral of Dq is just our total charge Q.
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We also can tell RI pretty easily using the Pythagorean Theorem.
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That RI is going to be the √ Z² + R².
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Therefore, our potential at point P is going to be Q/ 4 π ε₀ × 1/ √ Z² + R², our RI.
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Pretty straightforward derivation for the potential due to that charged ring.
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Taking a look at the uniformly charged disk we are going to follow the same strategy we did before.
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Where we started off with a little diagram, we had our ring before and we are just going to expand that ring until we get a nice big disk.
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We going to start our little ring small, make it bigger and bigger until we fill out the entire disk.
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Where again, the charge of the little ring is going to be our DQ, R will be the radius of our entire disk, r will be the radius of our expanding ring.
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As we do this, we will have some point over here P, as we draw from our little bit of DQ to P.
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Let us do that in red so it stands out a little bit, there we go.
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Our RI will be the √ C² + R² and that looks pretty similar.
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The width of this is going to be DR, just like we did when you are doing the electric fields in a similar fashion.
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Let us start off by taking a look at DQ is going to be 2 π R, the length of that hoop,
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imagine we cut it and pull it out, × its thickness DR, × our surface charge density σ, where σ is going to be charge ÷ area.
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In this case that is going to be Q/ π R².
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Then, we can do our substitution to say that DQ = 2 π DR × σ which is Q ÷ π R².
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In just a little bit of simplification here, I got a π and π there so we can call this 2Q R DR ÷ R².
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To find our potential, the electric potential at point P is the sum of all these little rings I and the potential of them.
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The sum of all the potentials of those little rings I which is 1/ 4 π ε₀, sum of all of our dq/ RI,
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which implies then that the potential of point P is 1/ 4 π ε₀.
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Making those tinier and tinier, and adding them all up using the integration process going from R = 0
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to R of our DQ which we said was 2Q R DR/ RI.
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It is still down there and ÷ R², as part of our DQ.
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We now have something that we can start to work with.
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Pull other constants again, VP =, 2Q can come out and that R² in the denominator can come out.
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I end up with 2Q ÷ 4 π ε₀ R², integral from 0 to R of, we are going to be left with our RI DR.
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RI is √ Z² c+ R² in the denominator that is going to be Z² + R²⁻¹/2 DR, which implies then that our potential at point P will be Q/2.
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Going from 2/ 4 to ½ π ε₀ R².
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In order to do this, if this is our U, we need a D and we needed 2 over here, so we need a 1/2 there.
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We will end up with ½ here, integral from 0 to R of Z² + R²⁻¹/2 × 2 R DR.
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That is a form we can integrate.
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Our R are variable so we get U ^½.
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Our DU is over here.
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Going to the next step, let us give ourselves a little bit more room here.
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Potential at point P = Q/ 4 π ε₀ R² again, incorporating that ½ we had, integral from 0 to R of Z² + R²⁻¹/2 × 2 R DR.
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Just making this a little bit clear, if we say that U is Z² + R² then DU must be 2R DR.
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If we do that then that means that the integral of U⁻¹/2 DU must be 2U ^½ + Z.
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Applying that to our integral VP = 2/ 4 π ε₀ R² × our integral here,
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we are going to end up with 2 × our U Z² + R² ^½ evaluated from 0 to R.
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Because we have a definite integral, we do not need to use the +C.
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Which is Q/ 4 π ε₀ R² × 2 × Z² + R² ^½ -2 × Z² ^½, or VP = Q, let us factor out that 2.
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2 π ε₀ R² ×, that leave us √ Z² + R² – Z.
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The electric potential at point P due to that uniformly charged disk.
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Let us try the potential due to a spherical shell of charge.
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Find the electric potential both inside and outside a uniformly charged shell of radius R and total charge Q.
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Let us start with outside, the potential outside is going to be the opposite of the integral of E ⋅ DL.
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Choose the opposite of the integral from infinity to R and we by now know the electric field equation pretty well, 1/ 4 π ε₀ Q/ R² DR,
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which implies that the potential outside = -Q/ 4 π ε₀ integral from infinity to R of R⁻² DR
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is going to be -Q/ 4 π ε₀ integral of R⁻² is going to be -1/ R evaluated from infinity to R.
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That the potential outside = -Q/ 4 π ε₀, substituting in we have -1/ R - -1/ infinity that is going to be 0.
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We get Q/ 4 π ε₀ R outside the shell.
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If we want to look inside, it is going to be the opposite of the integral of the E ⋅ DL.
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Again, it is the opposite of the integral from infinity to R of 1/ 4 π ε₀ Q/ R² DR,
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- the integral from R all the way to r, wherever we happen to be of 0 DR.
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We are going to do that piece which means anywhere inside, that we do not have to worry about this we just did.
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That is going to be Q/ 4 π ε₀ R.
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Outside it is a function of how far where you are, inside it is going to be a constant.
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If we want to graph that and I really think we do.
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Let us take a look in plot potential versus distance from the center.
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Let us put a little mark here for R the radius of our shell of charge.
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This will be our potential, here we are.
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Inside we have a constant which is Q/ 4 π ε₀ R.
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Outside, however, we have Q/ 4 π ε₀ r.
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It is proportional to 1/ R.
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There is our potential due to a spherical shell of charge.
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How about if we try a uniform solid sphere?
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Find the electric field and electric potential inside a uniformly charged solid insulating sphere of radius R and total charge Q.
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First thing that I'm going to want to do is to find the electric field
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and we will choose a sphere as our Gaussian surface and I'm going to put that Gaussian sphere inside that sphere of charge.
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Use Gauss’s law to find the electric field first.
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Integral/ the close surface of E ⋅ DA = our total enclosed charge ÷ ε₀.
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The left hand side becomes E × the area of our sphere 4 π R²
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and that is equal to the charge enclosed, the volume charge density ρ × the volume ÷ ε₀.
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As we do this, we have done some of the stuff before so we know that ρ is 3Q/ 4 π R³.
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We have solved that previously.
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The volume of the sphere is 4/ 3 π R³.
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The left hand side becomes E 4 π R² equal to, we have got our ρ 3Q/ 4 π R³ × our volume V 4/3 π R³.
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We still have our ε₀ in the denominator.
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We got a couple of simplifications I think we can make here.
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We have got a 3 here, we have got a 3, we got a π, we got a π, and I think that will do it.
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We got a 4 and a 4.
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That is equal to, we have still got a Q, we have got a R³ ÷, R³ in the denominator,
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we have got an ε₀ in the denominator, and we are dividing by the 4 π R² from the left hand side.
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This implies then that our electric field = Q R/ 4 π ε₀ R³.
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As we do this, it occurs to me that you know it is been a little bit since we did this.
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Maybe we ought to talk about it for just a second about where that ρ = 3Q/ 4 π R³ came from.
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If you want to take just a second and pull that out, for ρ = Q/ V, then that means that is charged / 4/ 3 π R³.
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And that means that ρ = 3Q/ 4 π R³, since we know the total charge there.
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If you are wondering where that came from, then we have done that before.
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It is probably worth taking a second and doing that again.
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We have got our electric field, next we are going to integrate to find the electric potential,
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noting that the total integration from infinity to R, our lower r has to be done piece wise
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because the electric field is discontinuous.
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We are going to integrate from infinity to R and then from R to r because they are different functions.
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Let us take a look at how we might do that.
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Our electric potential = the opposite of the integral from infinity to R of E ⋅ DR.
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Since we are doing this piece wise, V is equal to the opposite of the integral from infinity to R of E ⋅ DR –
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the integral from R to r of E ⋅ DR in that region, which implies then
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that the electric potential is the integral opposite of the integral from infinity to R of our function.
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Our electric field there which is 1/ 4 π ε₀ Q/ R² DR -, inside our sphere the integral from R to r of 1/ 4 π ε₀ R³ × QR DR,
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which is equal to, the left hand side is Q/ 4 π ε₀ R -, in the right hand side we can pull out some of our constants
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and come up with Q/ 4 π ε₀ R³, it can all come out, integral from R to R of R DR.
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This is going to be equal to, we still got our Q/ 4 π ε₀ R - Q/ 4 π ε₀ R³ integral of R is R²/ 2 and that is evaluated from R to R.
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All this becomes Q/ 4 π ε₀ R - Q/ 4 π ε₀ R³ × R²/ 2 - R²/ 2, which implies then,
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let us see if we can put some of this together.
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V = Q/ 4 π ε₀ R -, let us do this separately -Q R²/ 8 π ε₀ R³, let us multiplying these through,
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+ R²/, that will be 8 π ε₀ R³.
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All of that will equal, let us see what do we have here.
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That is 2Q/ 8 π ε₀ R.
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We will end up with 3Q because this becomes R²/ R³ we will gets 3Q/ 8 π ε₀ R combining this first and third term,
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-Qr²/ 8 π ε₀ R³, which implies then that the potential is equal to Q/ 8 π ε₀ R × the quantity, we will bring all that out, 3 - r²/ R².
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The electric potential due to a uniform solid sphere.
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Let us do a couple of AP free response problems and then we will move on here.
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Starting off with the 2012 exam free response question number 1.
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I have mentioned before, probably worth taking a few moment, downloading the question, printing I out, trying it, then come back here.
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Pause buttons are wonderful thing.
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We got two thin concentric conducting spherical shells insulated from each other a 3 DI of .1 and .2 m and it gives you a diagram of that.
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Part A, using Gauss’s law, derive an algebraic expression for the electric field for R between 0.1 and 0.2 m, in that region, between those two.
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For part A, we are going to use Gauss’s law.
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The integral over the close surface of E ⋅ DA which is equal to our enclosed charge ÷ ε 0,
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which implies then that the electric field × 4 π R² = QI, that inner Q ÷ ε₀.
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Therefore, the electric field is QI/ 4 π ε₀ R².
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Moving on to part B, determine an algebraic expression for the electric field when you are greater than 0.2 m,
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the algebraic expression for the electric field.
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I'm going to go to Gauss’s law again, integral/ the closed surface of E ⋅ DA = Q enclosed/ ε₀.
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The left hand side becomes E × that area of our Gaussian surface 4 π R²
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which I’m going to pick outside all of that 0.2 m must equal QI + QO/ ε₀.
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Therefore, the electric field strength is QI + QO/ 4 π ε₀ R².
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Excellent, moving on to part C, determine an algebraic expression for the potential when you are out in that same region as part B.
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Since we already did the work to the electric field, let us use that to help solve this.
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It is V = the opposite of the integral of E ⋅ DL is going to be the opposite of the integral
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coming from infinity to some point R our electric field of what we have previously QI + QO/ 4 π ε₀ R² DR.
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We will pull out our constants -QI + QO, those are not planning on changing any time.
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4 π ε₀ is a constant, we can pull that out, - the integral from infinity to R of R⁻² DR
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which implies then that our potential is - QI + QO/ 4 π ε₀ × -1/ R from infinity to R, is equal to - QI + QO/ 4 π ε₀ × -1/ R.
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The infinity piece is going to go to 0 so we end up with QI + QO/ 4 π ε₀ R.
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You may have been able to predict without having gone through all the math, but still good practice for you.
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Part B, using the numerical information given, calculate the value of the total charge QT on the 2 spherical shells which is QI + QO.
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Since we know the potential there, V = our total Q, QI + QO/ 4 π ε₀ R and we know R = 0.2 m,
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when V = 100 v, that implies then that QT = 4 π ε₀ R × our potential.
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Or 4 π ε₀ × 0.2 × our 100 volts or about 2.23 × 10⁻⁹ C, or 2.23 N/C.
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There is a part E here, on the axis, sketch the electric field E as a function of R, that the positive direction be radially outward.
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Alright, let us see if we can sketch these in.
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We have got a graph here and we are going to have another one here in a minute.
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Let us make both at the same time.
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There is our electric field and in a minute we are going to be asked about potential.
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Let us put in our markers for 0.1, 0.2, and 0.3 m.
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Sketch the electric field as a function of R.
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In our first situation there is 0.1, in between 0.1 and 0.2 where QI is less than 0, we got something like that.
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Between 0.2 onward, we are going to be coming down like that.
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For potential in part F, we are asked to do something similar.
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Sketch the electric potential as a function of R.
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Let us put in our marks again here for 0.1, 0.2, and 0.3.
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It is important to note that the negative of the slope and potential is going to give us the value for electric field.
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That will help us plot this.
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As we go through here, the negative slope of that gives us this.
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If we have got a 0 here, the slope down here has to be 0 and we know that we are going to start off here at -100 V.
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Then up here at 0.2, we are going to have to come up to + 100 V.
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It looks like we go from a steep slope to a lower and lower slope.
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This is probably going to look something like that.
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And then from here on, we have got a shape something like that.
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That the opposite of the slope of this graph gives us the value of the electric field graph.
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Alright let us take a look at the second example.
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This one, let us pull from the 2010 APC E and M exam free response question number 1.
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This is one of the more challenging questions that I have seen especially as you are just learning about potential.
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Let us take a look here and it gives us a charge + Q/ 1/4 circle of radius R but a couple of points there.
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It says rank the magnitude of the electric potential of those points from greatest to least and justify those rankings.
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I would say that the potential at B must be greater than the potential at A and the potential at C, which are the same.
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You can see A and C are the same by symmetry.
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The reason being is B is closest to the charge, therefore it must have the highest potential.
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A and C of the same potential by symmetry, and their lower potential in B because they are further away from the charges.
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Explain that somehow in your answer as you justify that.
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For part B, let me find where is B?
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We have adjusted the problem a little bit.
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Determine an expression for the electric potential at point P due to the charge Q.
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The P is right in the center there due to the charge Q.
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B, we can find the potential there that is going to be the sum over all those little tiny bits of 1/ 4 π ε₀ charged from each of those bits,
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÷ the distance from each of those bits which should be constant.
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That is going to be 1/ 4 π ε₀ integral from, let us go with θ = 3 π/ 4.
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That θ = 5 π/ 4 as we divvy up the pieces of that circle.
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Our Q is going to be the linear charge density × the radius × the differential of θ as we go around that.
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I can draw a picture here that might help explain where we would get that from.
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If we have our charges there going to go from 3 π/ 4 to 5 π/ 4.
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If we break this up into little tiny pieces, there is our point P.
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That distance we will call our R and the charge enclosed there, that Δ Q
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is going to be linear charged density λ × the radius × D θ, where D θ is the angle as we are going through there.
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Θ going from 3 π/ 5 π/ 4 to 3 π/ 4.
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Of course, λ is going to be charged / our length which is going to be total charge / ¼ circle
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is going to have a length of 2 π R ÷ 4, which is going to be π R ÷ 2, which is 2 Q/ π R.
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We can continue this by saying that this then is our DQ λ R D θ/ R, which implies then that potential is going to be λ,
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that should be constant, we can pull that out, / 4 π ε₀ integral from 3 π/ 4 to 5 π/ 4 around the circle of the D θ,
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which is going to be λ/ 4 π ε₀ as we go from 3 π/ 4 to 5 π/ 4, that is π/ 2.
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But going back to our definition of λ, we know that λ = 2Q/ π R.
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That is going to imply then that our potential = λ 2Q/ π R.
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We have a π in the numerator and we still have a 4 π ε₀ in the denominator.
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Λ 2Q/ π R, 4 π and 4 π ε₀.
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We got a 2 down here as well which is going to give us, when we do some simplifications Q/, let us see what do we have.
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4 π ε₀ R.
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Not bad, the map on that one once you get it setup.
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Alright let us take a look there now at part C.
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For C, it says we have a positive point charge Q with mass M placed at point B and released from rest.
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Find an expression for the speed of the point charge and it is very far from the origin, so after it is gone a long ways.
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This looks like a great spot again, these conservation of energy where the initial electric potential energy
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is going to be equal to the final kinetic energy when it is a long ways away.
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Therefore, Q × the potential at point B is going to be equal to ½ Mb velocity².
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We already found the potential at P, we said VP is 1/ 4 π ε₀ Q/ R.
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We can then say that we have Q × VP which is Q/ 4 π ε₀ R equal to ½ Mb².
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Rearranging this to get V all by itself, we can say that V² = 2q Q/ 4 π ε₀ MR.
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A little bit of simplification here that becomes a 2.
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Finally, solving for V itself, we get q Q/ 2 π ε₀ MR √.
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The velocity is your long ways away, converting that electric potential energy into kinetic energy.
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We got a part D on this question.
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We are given a ⋅ on a couple of axis that looks like this.
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Indicate the direction of the electric field at point P due to the charge Q.
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The direction of the electric field at point P should be to the right.
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Do not have any vertical component due to symmetry, that should be pretty straightforward
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Finally, part E.
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Let us go to blue here.
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Derive an expression for the magnitude of the electric field at point P.
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We are only worried about the X direction by symmetry.
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The electric field is just the X component of the electric field, that will be the integral of all the little pieces of electric field X
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due to the portions around there, which will be the integral of DE cos θ which is the integral of DQ/ 4 π ε₀ R² cos θ.
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We have got to do a little bit of work to rearrange this to make it look a little bit easier.
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We already know that λ is going to be 2Q/ π R, I think we said.
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DQ was λ R D θ which is going to be 2Q/ π R, there is our λ R D θ.
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Or DQ = 2Q/ π D θ.
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This implies then that our electric field is going to be the integral of DQ which is 2Q/ π.
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We have still a 4 π ε₀ R² in there.
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We have our cos θ term and we still have our D θ from DQ.
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A little bit more work here still.
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E is going to be equal to, let us pull out our constants, 2 and the 4 that is going to be a 2 in the denominator
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so we end up with Q/ 2 ε₀ π² R² integral from -π/ 4 to π/ 4 cos D θ.
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That is a little bit more reasonable.
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Which is Q/ 2 ε₀ π² R² the integral of the cos is the sin.
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That will be the sin of π/ 4 - the sin of -π/ 4, which implies that the electric field
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is going to be Q/ 2 ε₀ π² R² sin of π/ 4 √ 2/ 2 sin - the sin of -π/ 4 is going to be + √ 2/ 2.
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I can factor out that √ 2 and say that we get Q √ 2/, we have 2 ε₀ π² R² or Q/ √ 2 ε₀ π² R².
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It was a little bit of tricky but certainly doable especially when you get to the point where you have just the cos θ D θ for your integral.
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Let us do say, to think we have 2 more of these to get through.
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Let us try the 2009 exam free response number 1.
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In this one we are given a spherically symmetric charged distribution with net positive charge Q0 distributed within some radius R.
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Its electric potential as a function of distance is given by those formulas.
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That is kind of complicated formulas there.
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For the following regions, indicate the direction of the electric field ER and derive an expression for its magnitude.
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For A1, we are looking inside that sphere.
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The way I would start this is if we are given the potential, the electric field is - DV DR
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which is - the derivative with respect to r of R function for V which is going to be Q₀/ 4 π ε₀ R is given to us, × -2 + 3 × r / R².
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Which is, we can pull out our constants -Q0/ 4 π ε₀ RD/ DR -2 + 3 × R/ R²,
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which implies then that our electric field = –Q0/ 4 π ε₀ R ×, our derivative here is going to be 6R/ R².
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Putting all that together, -3Q₀ r/ 2 π ε₀ R³, complies that the magnitude of the electric field would be 3Q₀ R/ 2 π ε₀ R³.
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That is going to be inward since E is less than 0.
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For part A2, we want to do that for r greater than R.
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Using the same strategy, E = - DV DR is going to be -D/ DR of Q₀/ 4 π ε₀ R.
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Pull out our constants, -Q/ 4 π ε₀ × the derivative with respect to r of R⁻¹.
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Q is just going to be Q/ 4 π ε₀ R².
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No surprise there, we can treat it as if it was a point particle at the center of the sphere.
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Of course, that is going to be outward since E is greater than 0.
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Part A, check.
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Alright part B, for the following regions, derive an expression for the enclosed charge that generates the electric field in that region.
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The first one is for inside that sphere r less than R.
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I'm going to go to Gauss’s law here and say that the integral over the close surface of E ⋅ DA is our enclosed charge ÷ ε₀,
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complies then the left hand side E × the area of our Gaussian sphere 4 π R² = Q enclosed/ ε₀,
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which implies then that Q enclosed is going to be ε₀ E × 4 π R².
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Now since r is less than R, we know the electric field, we just solved for that up here in black, E is -3Q₀ R/ 2 π ε₀ R³.
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We can substitute that in to find Q enclosed = ε₀ ×, we have got our -3Q₀ R ÷ 2 π ε₀ R³.
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We still have our 4 π R².
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Just simplifying this, canceling, and clean this up a little bit.
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Q enclosed would be, we have got the 3 there, ε₀, all of that should give us -6 Q₀ R³/ R³.
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I think we are good.
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Let us highlight that.
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For B2, they want us to do this outside that sphere.
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Let us give ourselves more room here.
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The integral/ the close surface of E ⋅ DA is Q enclosed/ ε₀ which implies that,
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we got 4 π R² ε₀ E = Q enclosed, just like we had from the last part of the problem but substituting in our values now.
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Knowing here that we have our electric field equal to Q₀/ 4 π ε₀ r²,
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we can state that our left hand side becomes 4 π R² ε₀ × our electric field portion Q0/ 4 π ε₀ r².
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All of that = Q enclosed, which implies then that all of this cancel out, cancels out the Q enclosed = Q₀,
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which is kind of it has to, the enclosed charge Q₀ are going to have to be the same there.
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For part C, is there any charge on the surface of the sphere?
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Let us answer that and we are going to justify it.
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Let us start by saying, yes there is charge on the surface, there has to be and then we will justify and prove it.
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How do we know that?
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For r greater than R, the charge enclosed that r greater than R is the surface charge + whatever is enclosed at r = R.
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This is at r greater than R, which implies then that the surface charge = the enclosed charge
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- the enclosed charge and we calculate that at r = R.
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There is a difference in those two so the surface charge = Q enclosed for r greater than R,
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- what we had for charge enclosed to an r is equal the R, which we said was Q₀ - -6Q₀ R³/ R³.
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Which is Q₀ - -6Q₀ or 7Q₀.
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It asks us to do some graphing.
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Let us go to a new page here for the graphing and draw our axis here first.
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We are asked to graph a force that would act on a positive test charge in the region r less than r and r greater than r,
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assuming that the force direct radially outward is positive.
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There is r and let us make a mark here for R.
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For r less than r, we know that we are going to have something that is proportional to r.
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But since we are saying they we are calling radially outward positive, it is going to look something like that.
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R is proportional to r.
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Once we are outside of that, we get to our more familiar portion where we are proportional to 1/ R².
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1 or 2 more example problems here.
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Let us take a look at the 2007 APC E and M exam free response question number 2.
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We are given a figure with a non conducting solid sphere of radius A and charge + Q uniformly distributed throughout its volume
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and we also have a non conducting spherical shell with radius 2A and out radius 3A with a charge -Q uniformly distributed through its volume.
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We are asked to start off with Gauss’s law.
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Looking at A1 here, use Gauss’s law to find the electric field within the solid sphere.
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Let us start off with the volume charge density being Q/ V, it is uniformly distributed it said.
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That is going to be Q/ 4/3 π A² or 3Q/ 4 π A³.
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When we apply Gauss’s law, the integral/ the close surface of E ⋅ DA is equal to the enclosed charge ÷ ε₀,
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which implies then that the electric field × our Gaussian sphere 4 π R² area of our Gaussian sphere.
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It is going to be equal to our volume charged density × the volume that we are enclosing in our Gaussian sphere 4/ 3 π r³/ ε₀.
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We can substitute in for ρ where we know that ρ is 3Q/ 4 π A³,
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to state that the electric field then is going to be, we got our ρ here 3Q/ 4 π A³.
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We have our 4/3 π R³, we have got an ε₀ down here and we are also dividing by 4 π R² from the left hand side.
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All of that is going to simplify to, we have got a 3 and 3, we have got a 4 and 4, we have got a π and a π.
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We got and R² and R³, I would say then that E = Qr/ 4 π ε₀ A³.
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For part 2, it wants us to find the electric field between the solid sphere and the shell.
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Between those two, this is probably going to be a little bit simpler as we have gotten in the Gauss’s law.
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A2 integral/ the close surface of E ⋅ DA = the enclosed charged ÷ ε₀, complies that E × 4 π R² = that enclosed charge/ ε₀ Q.
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Therefore, E is just equal to Q/ 4 π ε₀ R².
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Even if you can do that from your head, notice it tells you to use Gauss’s law to derive the expression.
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Take the time to write all of that out.
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Alright A3, let us give ourselves more room to do a new picture.
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Here we have, within the spherical shell 2A to 3A.
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Q enclosed = Q + ρ B × VB, where ρ B is -Q/ VB, which is going to be -Q/ 4/3 π × 3A³ -4/3 π × 2A³,
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which is going to be equal to -Q/ 4/3 π 27 A³ -8 A³.
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Ρ B = -Q/ 4/3 π 19 A³.
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Applying Gauss’s law then, integral over the closed surface of E ⋅ DA = Q enclosed/ ε₀.
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Therefore, the electric field 4 π R² ε₀ equal to, we have got Q + -Q/ 4/3 π 19 A³ multiplied by 4/ 3 π R³ -4/3 π 2A³.
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Which implies then that the electric field, our left hand side × 4 π ε₀ R² = Q × 1 – R³ / 19 A³ + 8 A³/19 A³.
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That is the algebra coming up here looks like.
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Q × 1 – R³/ 19 A³ + 8/ 19, which implies then that the electric field = Q/ 4 π ε₀ R² × 1 – R³/ 19 A³ + 8 A³/ 19 A³, it is equal to Q/, factoring out that 19.
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4 × 19 = 76 × π × ε₀ × R², 19 – R³/ A³ + 8, which implies that the electric field = Q/76 π ε₀ R² × 27 – R³/ A³.
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Quite the problem going on there.
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A4, outside the spherical shell where r is greater than 3A, Q enclosed = 0.
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Therefore, E = 0.
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Alright, that one was not so bad.
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For part B, what is the electric potential of the outer surface of the spherical shell where R = 3 A and explain your reasoning.
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Let us solve, E = - DV DR which implies then that V = - the integral of E DR,
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which is - the integral from infinity to 3A of E ⋅ DR which is - the integral of 0 or 0.
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We will justify it there.
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C, derive an expression for the electric potential difference between points X and Y as shown in the figure.
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For part C, VX - VY which is Δ V is the integral from A to 2A of E ⋅ DR,
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which is the integral from A to 2A of Q/ 4 π ε₀ R² DR.
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Pulling out our constants, Q/ 4 π ε₀ integral from A to 2A of R⁻² DR.
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Or Q/ 4 π ε₀ the integral of R⁻² is going to be -1/ R evaluated from A to 2A.
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It can be Q/ 4 π ε₀ × -1/2 A - -1/ A or we can say then that we have Q/ 8 π A ε₀.