WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone and welcome back to Educator.com.
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This lesson is on electric potential difference, oftentimes called just electric potential or voltage.
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Our objectives are going to be to find and calculate electric potential energy and electric potential difference, to determine the potential difference due to a series of point charges...
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...finding the electric field strength between two charged parallel plates, finding the energy stored in a parallel plate capacitor and finally, calculating the capacitance of the parallel plate capacitor.
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Let us start by talking about electric potential energy.
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When we lifted an object against gravity by applying a force for some distance, work was done to give that object gravitational potential energy.
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At the same token, when you take a charged object and it is moved against an electric field by applying a force for some distance, you have to do work to give that object electric potential energy.
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Imagine we have a positive charge here; it is stuck in space and a long ways away, infinitely far away, I have a little point charge.
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If I want to bring that point charge -- its positive -- close to this big positive charge, I have to do work -- I have to push harder and harder, and harder, and harder, and harder to get it closer and closer.
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Once it is in this position it has a lot of potential energy -- electric potential energy.
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It wants to be repelled because if I let go of it, it is going to flinging off that way; it is going to convert that electric potential energy into kinetic energy.
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So we had to do work in bringing it from a long ways away until it was at this position.
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If we do work on an object, we have given it energy and in this case, we gave electric potential energy.
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Now the work done per unit charge, and moving that charge through that electric field, is a scalar and that is known as the electric potential or electric potential difference if you are talking about the potential between two different areas.
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The units are volts which is equal to a joule per coulomb (J/C) and the work done is equal to the change in the object's electric potential energy.
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So electric potential energy is charge times electric potential difference.
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Let us start off with a problem.
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If we have a potential difference of 10 volts between two points (A) and (B) in an electric field, 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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Well let us start off with our givens -- electric potential difference (V) is 10 volts; our electric potential energy is going to be 2 × 10^-2 J because that is the amount of work we had to do, and we are looking for charge.
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If electric potential energy is Q × V, then that means (Q) must be electric potential energy divided by (V) or 2 × 10^-2 J/10 volts, which gives us a charge of 2 × 10^-3 C.
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It is pretty straightforward.
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Let us take a look at electric energy.
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How much electric energy is required to move a 4 microcoulomb (MC) charge through a potential difference of 36 volts?
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Well our charge now is 4 MC and MC is 4 × 10^-6 C; our potential difference is V = 36 volts, and we are looking for electrical energy.
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Electrical energy is charge times voltage, which is going to be 4 × 10^-6 C × 36 volts (voltage)...
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...which implies that our electric potential energy will be 0.00014 J.
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Oftentimes, when we are dealing with very small charges -- something like a joule -- is not a very convenient form of energy.
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We are talking about things × 10^-15, ^-16, ^-17, ^-18 J -- it is just not very convenient.
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So there is another non-standard unit of energy that is very commonly used, it is called the electron-volt and it is given the symbol (eV) and it is a very small portion of a joule.
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It is the amount of work done in moving an elementary charge through a potential difference of 1 volt.
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So if you were to think about it, if electric potential energy is charge times voltage and your charge is 1 elementary charge and you move it through 1 volt, the electric potential energy is 1 eV.
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If we were to do that in standard units, we would have said that this is the charge of 1e, which is 1.6 × 10^-19 C × 1 volt and that would have given us 1.6 × 10^-19 J.
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These have to be the same, therefore, 1 eV = 1.6 × 10^-19 J.
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It is just another unit of energy that you oftentimes use when you are dealing with very small charges.
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Keep in mind though, if you are going to use these values for energy in other formulas, the SI unit, the standard unit that is going to make all your units work out, is still going to be joules. That is the standard.
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A proton is moved through a potential difference of 10 V in an electric field. How much work in electron-volts was required to move this charge?
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Let us look at how easy this can be if we use electron-volts.
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If our charge of a proton is +1 elementary charge, our potential difference is 10 volts, then the electric potential energy is charge times potential difference or 1e × 10 volts, which is 10 eV -- very straightforward.
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If we wanted to do that in joules, we could have done 1.6 × 10^-19 C × 10 volts = 1.6 × 10^-18 J if we went through all that, or we could have converted electron-volts to joules and we are all done, knowing that 1 eV = 1.6 × 10^-19 J.
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Thankfully, this problem made it nice and easy though in telling us that the answer was going to be in electron-volts.
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All right. Let us talk about equipotential lines for a minute.
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When you talk about topographic maps, if you have gone hiking or you were in some sort of surveying an organization, you have probably seen a topographic map where you have lines that show you areas of equal altitude.
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Those are lines of equal gravitational potential energy.
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We have the same sort of thing in the electrical world -- we have lines of equal electrical potential which we call equipotential lines.
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Now equipotential lines always cross electric field lines at right angles and if you move a charged particle in space, as long as you stay on an equipotential line, you do not do any work.
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As equipotential lines get closer together, the gradient of the potential increases.
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You have a steeper slope of potentials -- kind of like if you have a topographic map and your equal altitude lines get closer together, you are looking at a steeper cliff, a steeper gradient.
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So what I am showing here on the right is a positive charge -- we have the electric field lines which we have done before -- the black lines radiating away from the positive charge -- our equipotential lines must cross them at right angles.
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Everywhere that an electric field line intersects an equipotential line, we have a 90 degree or a right angle, and equipotential lines show lines of constant potential.
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If I were to take a charge -- let us put a charge right here -- and I want to move it over to here, somewhere else on the same equipotential line.
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The net work done is going to be 0 because you end up at the same potential energy, the same potential because you have the same charge.
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Let us take a look at how we could draw some equipotential lines.
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Around positive point charges, they are pretty easy because it always intersects at a right angle, so these must look like circles.
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I will do my best to draw a circle here -- that would be 1 equipotential line, and we could probably draw another equipotential line here -- pretty close -- and in a perfect world, all of those would intersect at 90 degrees.
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Around a negative point charge, we would have the same idea -- crossing all the electric field lines at right angles, so we will get a circular pattern for our equipotential lines.
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Over here on the bottom left where we have our dipole again of a positive and negative charge -- well now our equipotential lines get a little bit more complicated.
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The one right here is pretty easy, if we draw an equipotential line right through the middle -- that is pretty straightforward -- it crosses all of those at 90 degrees.
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But in order to cross all of these at 90 degrees as we move in here, we are going to have to adjust that a little bit and we are going to start to see some curve to it.
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So there is an equipotential line on that side and on the other side, we are going to have the same sort of thing by symmetry.
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I am trying to cross all these as best as I can at 90 degrees and we will get an equipotential line that looks something like that.
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And over here on the right, now again we have two positive charges, so let us see what we get here.
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Again, right down the middle, we are going to have that point where we do not have anything.
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But as we go just a little bit off here, we are going to have to cross this at 90 degrees; we are going to have to get really, really steep here to cross at 90 degrees, and as we do that, it looks like we are going to come back around.
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I am doing my best to draw that at 90 degree intersecting angles, let us try that again over here.
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We will start from this side this time, crossing all of these at about 90 degrees and that must curve pretty steeply to come back there to give us our equipotential lines.
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So the key point is: equipotential lines always cross electric field lines at right angles.
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Let us talk about the potential due to a point charge just like we talked about the electric field due to a point charge.
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To calculate the potential difference due to a point charge, the electric potential -- well if force is KQ1Q2/R² and we found electric field was going to be KQ/R², well potential is just going to be KQ/R.
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The nice thing about potential is that it is a scalar.
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We do not have to worry about direction, we can add them up in scalar form and save us a lot of work.
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And to find the potential difference due to multiple point charges, we just take the sum of the electric potentials due to each individual point charge -- again not worrying about any vector nature.
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Electric potential energy then can be found by multiplying the electric potential by the charge.
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So the electric potential energy due to a point charge is (QV) or Q × KQ/R, therefore electric potential energy is going to be (K) times the product of your two charges divided by the distance between them.
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Let us take a look at a sample problem here.
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Find the electric potential at point (P) which is located 3 m away from a -2 C charge.
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What is the electric potential energy of a half-coulomb charge situated at point (P)?
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Let us start by finding the electric potential at point (P).
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(V) at point (P) is going to be KQ/R where K = 9 × 10^9 N-m²/C², our Q = -2 C, and our distance = 3 m, so that is going to give us about -6 × 10^9 V.
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What is the electric potential energy of a half-coulomb charge situated at that point?
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Well the electric potential energy is charge times voltage -- our potential -- which is going to be 0.5 C × -6 × 10^9 V or -3 × 10^9 J.
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A nice, straightforward applications of those formulas.
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Let us try one that is a little more involved, kind of mirroring what we did with the electric field.
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Let us find the electric potential at the origin due to the three charges shown in the diagram and at the end it says if we place an electron at the origin, what electric potential energy does it possess?
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Now this is awfully similar to what we did when we were finding the net electric field at the origin, but now we are looking for potential.
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So we are going to do it with the same basic strategy -- let us find the potential at the origin due to each of the three individual charges and then we will add them up.
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So the potential due to the green charge, that is going to be KQ/R = 9 × 10^9N-m²/C², our charge is 2 C and our distance from the origin is 8.
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That is going to be just 2.25 × 10^9 V.
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Now let us do the red charge -- V = KQ/R, which is going to be 9 × 10^9 × -2 C/8, which is going to be -2.25 × 10^9 volts.
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And finally, let us find the potential due to our 1 C charge -- V = KQ/R or 9 × 10^9 × 1/R.
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We said last time that if we make a right triangle here, that side is 2, that side is 2, therefore the hypotenuse must be the square root of 2² + 2²...
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...so (R) is going to be square root of 2² + 2² or square root of 8, which is 3.18 × 10^9 volts.
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Now to find the total -- the potential here at the origin -- we have to sum up the potentials to each of those three charges.
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The total is just going to be: 2.25 × 10^9 + -2.25 × 10^9, so we are going to cancel this out, that is going to be 0, and all we are left with is this 3.18 × 10^9 volts.
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That is the potential due to those three point charges when you are here at the origin.
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Finally, we are asked to find what happens to the electric potential energy if we place an electron at the origin; what is its electric potential energy?
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Well, let us do that.
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The electric potential energy due to that electron is just going to be charge times voltage and our charge is -1.6 × 10^-19 C, because it is an electron; it is a negative...
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Our voltage is 3.18 × 10^9 volts, which is going to give us a potential energy of about -5.1 × 10^-10 J.
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Let us point out one other item here.
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If instead we did this in electron-volts, this would have been -1e × 3.18 × 10^9 volts, which would have given us -3.18 × 10^9 eV.
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So you can see where using electron-volts could be a lot more convenient, but since it does not specify which units it wants for our answer, we will circle both of them -- they will both be correct.
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All right. Let us talk a little bit about parallel plate configurations.
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Configurations in which we have two parallel plates of opposite charge that are situated a fixed distance from each other are very common in physics because this is how we make a capacitor, an electrical device used to store charge.
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And the general look of these is, we will take one plate, put another one just a little bit of distance from it, they have some cross-sectional area (A), and we will situate them at some distance from each other (D), and then typically we are going to put some charge like (+Q) up here and (-Q) there -- there is the basics of a capacitor.
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Now the electric field due to two parallel plates -- as long as you are away from the edges of those plates -- is constant.
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And the electric field, as we know, runs from positive to negative, so anywhere between these parallel plates -- as long as we stay a little bit away from the edges -- is constant throughout this entire region; it is given by the voltage, the potential difference across the plates divided by their distance.
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A nice and easy way to calculate a uniform electric field, and that is true -- equals V/D here, here, here, here, here -- anywhere between those plates, it is constant.
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The magnitude of that electric field strength is given by the potential difference divided by the plate separation and the units are going to be Newton's per coulomb (N/C) again for an electric field or volts over distance (V/m).
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So we are proving that we have the same units (N/C) as equivalent to a (V/m)
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All right. A capacitor is an electrical device used to store charge.
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It consists of two conducting plates separated by some sort of insulator.
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That insulator could be air, it could be vacuum, it could be paper, it could be Jell-O; anything that is an insulator you could put in between the plates and you would still have a capacitor.
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Now once the plates are charged, they are disconnected.
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The charges then stuck on the plates until the plates are reconnected.
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So if you put a charge on each plate, you create an electric field between the plates.
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If you disconnect them, then the charge is stuck there, and you have all that energy that is stored in the electric field.
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The amount of charge a capacitor can store for some given amount of potential difference across it is known as the device's capacitance (C).
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Now notice (C) is capacitance, but it is also used as a unit of charge, coulombs, so we have to be careful with our (C)'s.
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The units of capacitance are coulombs per volt also known as a farad (F), and a farad is a very large amount of capacitance.
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So much more often we will be talking about millifarads, microfarads, nanofarads, even picofarads.
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Our key formula for capacitance -- capacitance is equal to the charge divided by the potential difference between the plates or C = Q/V.
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Capacitors store energy -- because the charges are on opposite plates of the capacitor, they exert forces on each other; it becomes an energy storage device where that energy is stored in the electric field.
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You can find the energy stored in a capacitor by the formula: potential energy in a capacitor is 1/2 times the capacitance times the square of the potential difference.
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If we write this as 1/2 CV² -- well we also just learned that C = Q/V, so I could write this as 1/2 and I am going to replace (C) with Q/V and I still have a V².
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I can do a little bit of simplification here -- (V) -- and the squared goes away and we will come up with 1/2 QV, which is also equal to (V), potential energy stored in a charged capacitor.
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Let us do an example with capacitance.
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A capacitor stores 3 microcoulombs of charge -- Q = 3 × 10^-6 C -- with a potential difference of 1.5 volts across the plates, V = 1.5.
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What is its capacitance and how much energy is stored in it?
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Well its capacitance is C = Q/V or 3 × 10^-6 C/1.5volts = 2 × 10^-6 F which is 2 microfarads.
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How much energy is stored in the capacitor?
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We could do this a couple of different ways, but let us start off with 1/2 CV² -- that is going to be 1/2 times...
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...we just found our capacitance, 2 microfarads or 2 × 10^-6 F, and our potential difference, 1.5 volts ², gives us 2.25 × 10^-6 J).
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Or we could have used u = 1/2 QV, which will be 1/2 × 3 × 10^-6 C (charge) × 1.5 volts (potential difference), which amazingly is 2.25 × 10^-6 J.
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And of course those have to be the same; we said the formulas were equivalent.
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All right. Looking at the charge on a capacitor -- How much charge sits on the top plate of a 200 nanofarad capacitor when charged to a potential difference of 6 volts?
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Well let us start there.
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Capacitance is C = 200 nanofarads or 200 × 10^-9 F and potential difference is V = 6 volts.
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So if C = Q/V, then that means our charge (Q) must be (CV) or 200 × 10^-9 F × 6 volts (potential difference), which is going to be about 1.2 × 10^-6 C.
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How much energy is stored in the capacitor when it is fully charged?
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Well when it is fully charged, u = 1/2 CV² which is 1/2 our capacitance, 200 × 10^-9 or 200 nanofarads × 6 volts², which is going to give us about 3.6 × 10^-6 J.
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How much energy is stored in the capacitor when the voltage across its plate is 3 volts?
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Well when we get to 3 volts, u = 1/2 CV² again -- that is going to be 1/2 × 200 × 10^-9 × 3 volts² = 9 × 10^-7 J.
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That will be 1/4 of that value, so if we cut the voltage in half, we have 1/4 the value, and that is because of that V² relationship -- so pretty good at calculating charge on a capacitor and the energy stored in a charged capacitor.
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Let us talk about the design of capacitors.
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What determines how much charge a capacitor can store?
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Well the area of the plates -- as you have bigger plates you can store more charge, the separation of the plates plays a role in the capacitance and the insulating material between them.
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In short, the capacitance is given by this value, ε -- that is called the permittivity and that is a constant that has to do with the material between the plates -- cross-sectional area in square meters and the separation of the plates (D).
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Now if you have an air gap capacitor or a vacuum capacitor, our baseline ε, our baseline permittivity is 8.85 × 10^-12 C²/N-m².
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If you put something other than air or vacuum between the plates, then you have to go from the permittivity of free space and multiply it by what is known as the dielectric constant, where your permittivity is going to be your dielectric constant (K) × ε0 and materials have larger (K)'s.
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For air or vacuum, that is going to be 1, so ε is equal to that.
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If you have something that is a little bit more resistive, a better insulator for example, that is going to be a bigger number for (K), so that will increase your capacitance as you put a different insulator in there.
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Let us see how that works.
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How far apart should the plates of an air gap capacitor be if the area of the top plate is 5 × 10^-4 m² and the capacitor must store 50 mJ of charge and an operating potential difference of 100 volts.
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It should say it must store 50 mJ of energy -- that makes more sense.
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Well if our area is 5 × 10^-4 m² and our potential energy is 0.05 J, we would have a potential of 100 volts.
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Let us see what we can determine here -- u = 1/2 CV², but we also know that C = ε0 A/D and because it is an air gap capacitor, I can leave that ε0, because I do not have any dielectric constant that I have to put in front of it there.
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So that means that u = 1/2 × ε0 A/D × V².
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If I rearrange this a little bit to find how far apart or the distance between them, I would say that the distance then solving for (D) is going to be...
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...ε0 × (A) × V² divided by 2 × 8.85 × 10^-12 (potential energy) × 5 × 10^-4 m² (area)...
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...× 100 volts² (potential)/2 × 0.05 J.
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When I do all that, I come up with a separation distance of about 4.43 × 10^-10 m, so just combining our formulas in order to solve for the unknown.
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Let us do some more capacitance calculations.
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Find the capacitance of two parallel plates of length 1 mm and with 2 mm if they are separated by 3 micrometers of air.
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All right. To begin with, capacitance is ε0 A/D; it is an air gap capacitor so I do not have to worry about a dielectric constant there.
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That will be 8.85 × 10^-12 × our area, which is length × width, or 0.001 m × 0.002 m/3 micrometers or 3 × 10^-6 m (the separation of the plates).
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If I put that in my calculator, and I come up with about 5.9 × 10^-12 F which is also 5.9 picofarads.
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Now what would the device's capacitance be if we replace that air with glass which has a dielectric constant of 3.9?
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Well the only thing we have to do there is, if before, (C) was ε0 A/D -- because we are no longer using air or vacuum because we have a dielectric constant of 3.9 -- 3.9 ε0 becomes my permittivity.
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So now, all I have to do -- all of this is the same -- is multiply my answer by 3.9.
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That is going to be 3.9 × 5.9 × 10^-12 F or 2.3 × 10^-11 F which is 23 picofarads.
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By inserting a dielectric material with a higher dielectric constant, I increased my capacitance because the dielectric constant was 3.9 times larger than the original, so my capacitance is 3.9 times larger than the original.
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All right. Let us go through a couple more problems to make sure we have everything down.
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An electron sits on a equipotential line of 5 volts.
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How much work is required to move the electron to an equipotential line of -25 volts?
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Let us make this one easy, let us do it in electron-volts.
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As we do this, the potential energy is equal to charge times potential difference.
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The charge on our electron is -1 elementary charge, and our potential difference, if we go from 5 to -25 volts -- well ΔV is final minus initial, so that is going to be -25 - 5 = -30 volts or 30 eV -- nice and straightforward.
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Let us take a look at conservation of energy as we talk about these problems.
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A proton is held at a fixed point in space where the electric potential is 500 kilovolts or 500,000 volts; the proton is then released.
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Assuming no energy is lost to non-conservative forces, what is the speed of the electron at a point in space where the electric potential is 100,000 volts?
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This is a conservation of energy problem.
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It starts out initially with electric potential energy and that must be equal to its final electric potential energy.
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Where did any excess energy go? That must be the final kinetic energy of our proton.
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Well we know the initial is going to be (Q) times our initial voltage, so that must be equal to (Q) times our final voltage and our kinetic energy is going to be 1/2 mv².
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So as I start solving to get v² all by itself, I could say then that v² = Q(Vi - Vf).
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I will have the 2 in there, so 2 will factor out the Q (Vi - Vf) divided by (m), must be equal to V².
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Therefore, v² = 2 × 1.6 × 10^-19 C (charge on our proton); Vi - Vf = 500,000 volts - 100,000 volts all divided by the mass.
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If you have to look up the mass of a proton, you will find that it is about 1.67 × 10^-27 kg.
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Plug all that into my calculator and I come up with something around 7.66 × 10^13 m²/s².
00:32:31.000 --> 00:32:50.000
So if I want just velocity, I take the square root of both sides and get that V = 8.75 × 10^6 m/s -- converting electric potential energy into kinetic energy.
00:32:50.000 --> 00:32:53.000
All right. Let us take a look at a problem with two conducting spheres.
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These two conducting spheres, each with charge (Q) are connected by a wire as shown.
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Do any charges flow between the spheres and how do their potentials compare?
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Well the first thing we have to realize is when we connect them by a wire, all of a sudden they must be at equipotential -- anything that is connected by a conductor is going to be at equipotential.
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So once we do that, these have to be at the same voltage.
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That means that if we look over here at Q1, V1 = KQ1/R1.
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Over here on the right-hand side, V2 = KQ2/R2, and because the potentials must be equal then, we could say that KQ1/R1 = KQ2/R2.
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We have a nice simplification we can make there -- divide (K) out of both sides, therefore Q1/R1 = Q2/R2.
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And since we know that R2 is going to be twice R1 -- measuring those lines, we could figure that out -- then Q1/R1 must equal Q2/2R,1...
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...or with a little bit more rearrangement to say that Q2 = R1Q1/R1 or that is going to be just 2Q1.
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If Q2 is equal to twice Q1, then we must have charge flowing from 1 to 2.
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So our charge must be flowing that way -- we will have charge flowing from Q1 to Q2.
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How did their potentials compare? They have to be equal.
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Equipotential lines for a capacitor -- Draw the equipotential lines for the parallel plate capacitor below.
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The first thing that I am going to do is it is probably easier to draw the electric field first, going from positive to negative.
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So I will put in some electric field lines first, in green here -- electric field is constant between those.
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And then we know that equipotential's are always crossing electric field lines perpendicularly, at right angles, so I could draw my equipotential's that way.
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And if this is 0 volts, that will be maybe 1, 2, 3 volts, 4 volts, 5 volts.
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We have a constant electric field between these but we do not have a constant potential.
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Potential is going to have a linear gradient from 5 volts down to 0 volts.
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Hopefully that gets you a good start on electric potential and electric potential difference.
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We talked about electric potential energy and even some capacitors in there.
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Thanks so much for your time, and make it a great day everyone!