### Video Transcript

What are the voltages across the
capacitors after equilibrium has been reached? a) C one, five volts; C two, five
volts; C three, 10 volts. b) C one, 10 volts; C two, 10 volts; C three, 10 volts. c)
C one, 10 volts; C two, five volts; C three, five volts. d) C one, five volts; C
two, 10 volts; C three, five volts. e) C one, 10 volts; C two, 20 volts; C three, 20
volts.

The question refers to the circuit
on the right, which consists of a single-celled battery connected to three
capacitors. And the question asks us for the
voltage across each capacitor after equilibrium has been reached. Equilibrium means that no voltages
or currents are changing. Letβs recall what this means in the
context of a capacitor. A capacitor stores energy by
separating charges. So, letβs start with an initially
uncharged capacitor. Weβve now added a battery of
voltage π across the capacitor. Current now flows around the
circuit. Conventional current flows out of
the positive terminal and into the negative terminal. Recall, though, that conventional
current is the flow of positive charge. So, positive charge flowing into
the top plate of the capacitor result in a net positive charge on the top plate.

Similarly, conventional current
flowing away from the bottom plate leaves a void of positive charge, which is the
same thing as saying it leaves behind a negative charge. The capacitor now has a charge of
+π on the top plate and βπ on the bottom plate. And these charges keep growing as
the conventional current keeps flowing. This separation of charges results
in a voltage across the capacitor. As the charge on each plate of the
capacitor grows, so does the voltage across it. And as this happens, the
conventional current flowing in the circuit begins to slow.

This is because the greater the
voltage across the capacitor, the harder it is to continue increasing the
charge. And eventually, the electromotive
force of the battery is insufficient to push more charge onto the capacitor. This occurs when the voltage across
the capacitor is equal to the voltage of the battery. And when this occurs, the
conventional current stops flowing through the capacitor. This is what it means for the
capacitor to be in equilibrium. No more conventional current is
flowing through it, so no more charges building up, and the voltage across it is
constant.

If there are more than one
capacitor, as there are in this circuit diagram, the voltage across each capacitor
is not necessarily the voltage of the battery. Nevertheless, the voltage at
equilibrium is constant. Weβve labelled these voltages on
the diagram with the symbols π one, π two, and π three for voltage across
capacitor one, voltage across capacitor two, and voltage across capacitor three. To approach finding these values,
weβll use a property of voltage between two points. The voltage between any two points
is the same regardless of the path taken between those two points. Letβs consider the path between
point A and point B that travels directly across capacitor one. To find the total voltage between
the end points of a path, we simply add up the voltages encountered along that
path.

In this case, the only voltage we
encounter is π one, the voltage across capacitor one. Calling the voltage between point A
and point B π AB, we can write that π AB is equal to π one. If we can independently find a
value for π AB, then we will have found the value for π one. To find π AB independent of π
one, we need a path between A and B with a known voltage. Note that we have a battery with a
given voltage. So, if we can find a path from A to
B that passes through this battery, weβll know the voltage between A and B. To find that path, simply trace the
circuit from A to the battery through the battery and onto point B. The only voltage present on this
path is the voltage of the battery. So, the voltage of this path is 10
volts. But this is also a path between A
and B, so it must have the same voltage as every other path between A and B or π
AB. So, we found that π AB is equal to
10 volts.

Note that itβs positive 10 volts
because we traverse the battery from plus to minus side. If we traverse the other direction,
in other words, a path from B to A, we wouldβve found that the voltage is minus 10
volts. Replacing π AB with 10 volts, we
find π one, the voltage across capacitor one is also 10 volts. Since we know the voltage between
points A and B, letβs find a path from A to B that includes capacitor two and
capacitor three. This time, the path includes two
voltages, the voltage across capacitor two and the voltage across capacitor
three. Again, the total voltage is π AB,
since the path goes from A to B. And this is equal to the sum of the
voltages encountered, π two plus π three.

Just like before, we replace π AB
with 10 volts. But the right-hand side contains
two unknown variables. So, to solve for them, weβll need
another equation, some relation between π two and π three. Weβll first present an intuitive
argument for this relationship and then back it up rigorously. C two and C three are the only two
components on this right-hand branch of the circuit. And since they have the same
capacitance, it ought to be that they behave identically. In other words, since there are no
other components in the right-hand side of the circuit and C two and C three are
effectively identical, thereβs no reason for the circuit to prefer one over the
other when it comes to voltage. So, because theyβre the same
component and ought to behave the same, they should have the same voltage across
them. In other words, π two equals π
three.

Substituting in for π three, we
find that 10 volts is equal to two times π two. Dividing both sides by two, two
divided by two is one and 10 divided by two is five. So, π two equals five volts. And since π two equals π three,
π three is also equal to five volts. We now know the voltages across
each of three capacitors. In order, they are 10 volts, five
volts, and five volts. This is answer choice c, C one, 10
volts; C two, five volts; C three, five volts. To get this answer, we made an
intuitive assumption based on the symmetry of C two and C three that π two is equal
to π three. We can take a longer but more
rigorous route to the same conclusion by understanding how the charges build up on
these capacitors.

Recall that before equilibrium,
charge builds up on the capacitor by conventional current flowing into one plate and
out of the other. Recall also that if current is
flowing in a circuit, then any areas without junctions have the same current
everywhere in that area. Before equilibrium, capacitors two
and three are charging up due to current flowing between points A and B along this
path. This path has no junctions because
the junctions are A and B at the beginning and end. Therefore, the current at all
points along this path must be the same. So, the current flowing into
capacitor two, out of capacitor two and into capacitor three, and out of capacitor
three and back to point B must all have the same value.

But recall that current is the rate
of transfer of charge. So, the current into capacitor two
is depositing charge on that first plate. But that same value of current is
also flowing out of capacitor two, leaving behind a charge of βπ on the other
plate. By the same logic, the currents
flowing into and out of capacitor three are also the same as those into capacitor
two, mean they result in the same charge. The last thing that we need is the
relationship between the charge on a capacitor, the voltage across it, and its
capacitance.

That relationship is voltage across
the capacitor is equal to charge on the capacitor divided by the capacitance. Using this relationship, letβs
write the voltage across capacitor two. π two, the voltage across
capacitor two, is equal to π, the charge on it, divided by 10 picofarads, the
capacitance. As we saw before, capacitor three
has the same charge. And it also is given to us that it
has the same capacitance. So, π three is also equal to π
over 10 picofarads. Since both π two and π three are
equal to π over 10 picofarads, it follows that our original assumption that π two
is equal to π three is indeed correct and justified in this problem.