Video: Capacitance of a Parallel Plate Capacitor

An empty parallel plate capacitor has a capacitance of 80 πœ‡F. How much charge must leak off its plates before the voltage across them is reduced by 250v?


Video Transcript

An empty parallel plate capacitor has a capacitance of 80 microfarads. How much charge must leak off its plates before the voltage across them is reduced by 250 volts?

If we sketch a small capacitor here, which is part of an electrical circuit, we know that over time as current flows to this circuit, charge will build up on the capacitor plates, positive on one side and negative on the other. Those opposite charges will help to create an electric field between the plates of this capacitor. And that electric field will indicate a change in potential per unit distance.

In other words, given a certain separation distance between these two plates, a certain electric field between them implies a certain voltage across them. Whatever that voltage is initially, we’re curious in this question to understand how much charge must be lost from these two plates in order for that voltage to be reduced by 250 volts.

Basically, we’re looking for a change in charge on these parallel capacitor plates; we’ll call it Ξ”π˜˜. There’s a bit more we can say about Ξ”π˜˜ because it really represents the change between two states of this capacitor. There is the initial state, where it has its initial amount of charge; we can call that amount 𝘘 one. And then, there’s the capacitor’s final state, where it has less charge; we’ll call that 𝘘 two.

And we can say that 𝘘 one minus 𝘘 two is equal to Ξ”π˜˜; that’s what we want to find. there is a mathematical relationship we can use as a start point to help us solve for Ξ”π˜˜. That relationship says that the capacitance of a capacitor is equal to the charge on one of its plates divided by the potential difference across the plates.

For our capacitor, we’ve identified two different charge amounts. And we can call these two different capacitor states. Yet, throughout this process, the capacitance 𝐢 of the capacitor remains constant. This means we can write that the capacitance of this capacitor 𝐢 is equal to 𝘘 one over 𝑉 one, the potential difference initially across the capacitor, which is also equal to 𝘘 two over 𝑉 two.

And we can realize that we can write an equation for 𝑉 two in terms of 𝑉 one. That’s because the problem statement indicates that our voltage is reduced by 250 volts. We can write then that 𝑉 two is equal to 𝑉 one minus 250 volts. And we can now insert this expression into our equation for 𝑉 two.

With that substitution made, let’s focus on the equality on the right-hand side 𝘘 one over 𝑉 one equals 𝘘 two over 𝑉 one minus 250 volts. If we multiply both sides of this equation by the denominators we see on either side, then we find the quantity 𝑉 one minus 250 volts times 𝘘 one is equal to 𝘘 two times 𝑉 one. Then, if we distribute this factor of 𝘘 one to both terms, we find this resulting equation: 𝘘 one 𝑉 one minus 250 volts times 𝘘 one equals 𝘘 two times 𝑉 one.

Noticing several 𝑉 one factors, what if we divide both sides of the equation by that term. In that case, we see 𝑉 one cancel on our leftmost term and on our rightmost term. But interestingly, on our middle term, something else happens. Notice that in this term, we have 𝘘 one divided by 𝑉 one. What’s that?

Looking back over our starting equation, the capacitance is equal to charge over potential, we realize that this ratio 𝘘 one over 𝑉 one is equal simply to 𝐢, the capacitance of our capacitor. With this simplified equation, let’s make a couple last rearrangements.

First, we’ll add 250 volts times 𝐢 to both sides and then we’ll subtract 𝘘 two from both sides. What we find is a left-hand side of our equation, which is 𝘘 one minus 𝘘 two, and a right-hand side, which is 250 volts times 𝐢. But let’s recall what 𝘘 one minus 𝘘 two is. That is Ξ”π˜˜, the charge value we’re looking for.

We can write then that Ξ”π˜˜, the amount of charge that must leak off the plates, is equal to 250 volts times the capacitance of our capacitor 𝐢. And that value we’re given in the problem statement; it’s 80 microfarads. So to finally solve for Ξ”π˜˜, we plug in the value for 𝐢, 80 times 10 to the negative 6 farads, and multiply it by 250 volts.

And we find a charge amount of 0.020 coulombs. This is how much charge would have to leak off our plates in order for the voltage across them to be reduced by 250 volts.

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