### 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,
weβre 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, weβre looking for a
change in charge on these parallel capacitor plates; weβll call it Ξπ. Thereβs a bit more we can say about
Ξπ 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 π one. And then, thereβs the capacitorβs
final state, where it has less charge; weβll call that π two.

And we can say that π one minus π
two is equal to Ξπ; thatβs what we want to find. there is a mathematical
relationship we can use as a start point to help us solve for Ξπ. 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, weβve identified
two different charge amounts. And we can call these two different
capacitor states. Yet, throughout this process, the
capacitance πΆ of the capacitor remains constant. This means we can write that the
capacitance of this capacitor πΆ is equal to π one over π one, the potential
difference initially across the capacitor, which is also equal to π two over π
two.

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

With that substitution made, letβs
focus on the equality on the right-hand side π one over π one equals π two over
π 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 π one
minus 250 volts times π one is equal to π two times π one. Then, if we distribute this factor
of π one to both terms, we find this resulting equation: π one π one minus 250
volts times π one equals π two times π one.

Noticing several π one factors,
what if we divide both sides of the equation by that term. In that case, we see π 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
π one divided by π one. Whatβs that?

Looking back over our starting
equation, the capacitance is equal to charge over potential, we realize that this
ratio π one over π one is equal simply to πΆ, the capacitance of our
capacitor. With this simplified equation,
letβs make a couple last rearrangements.

First, weβll add 250 volts times πΆ
to both sides and then weβll subtract π two from both sides. What we find is a left-hand side of
our equation, which is π one minus π two, and a right-hand side, which is 250
volts times πΆ. But letβs recall what π one minus
π two is. That is Ξπ, the charge value weβre
looking for.

We can write then that Ξπ, the
amount of charge that must leak off the plates, is equal to 250 volts times the
capacitance of our capacitor πΆ. And that value weβre given in the
problem statement; itβs 80 microfarads. So to finally solve for Ξπ, we
plug in the value for πΆ, 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.