# AP Physics 2 : Capacitor Energy

## Example Questions

### Example Question #1 : Capacitor Energy

Consider the circuit:

If the voltage drop across C2 is 5V, what is the total energy stored in C2 and C3?

Explanation:

In parallel branches of a circuit, the voltage drops are all the same. Therefore, we know that the voltage drop across C3 is also 5V.

We can then use the following equation to calculate the total stored energy:

Since the voltage is the same for both capacitors, we can simply add the two capacitances to do one calculation for energy:

### Example Question #1 : Capacitor Energy

capacitor is connected to a  battery. Once the capacitor is fully charged, how much energy is stored?

None of the other answers is correct

Explanation:

To find the amount of energy stored in a capactior, we use the equation

.

We're given the capacitance (), and the voltage (), so we'll use the third equation.

### Example Question #3 : Capacitor Energy

You have 4 capacitors, , and , arranged as shown in the diagram below.

Their capacitances are as follows:

If you have a 6V battery connected to the circuit, what's the total energy stored in the capacitors?

Explanation:

The equation for energy stored in a capacitor is

We can find the capacitance by adding the capacitors together, and we have the voltage, so we'll use the second equation, .

Capacitors  and  are in series,  and  are in parallel, and  and  are in parallel.

Now that we have the total capacitance, we can use the earlier equation to find the energy.

The total energy stored is 121.5J.

### Example Question #4 : Capacitor Energy

If ,  and , how much energy is stored in ?

Explanation:

In this circuit, the voltage source,  and , and  are all in parallel, meaning they share the same voltage.

To find the energy, we can use the formula

, with  being the energy,  being the capacitance, and being the voltage drop across that capacitor.

To use the formula we need the voltage across .

Another hint we can use is that  and  having the same charge since they're in series. First let's find the equivalent capacitance:

Now, we can use the formula

to calculate charge in the capacitor.

Now that we know a charge of  exists in both capacitors, we can use the formula again to find the voltage in only .

Finally, we plug this  into the first equation to calculate energy.

### Example Question #1 : Capacitor Energy

Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is not connected to a battery. If some external agent pulls the capacitor apart such that D doubles, did the internal energy, U, stored in the capacitor increase, decrease or stay the same?

Stays constant

Increases

We must know the dielectric constant

Decrease

Increases

Explanation:

Equations required:

We see from the first equation the when D is doubled, C will be halved (since  and  are constant). From the third equation, we seed that when C is halved, the potential energy, U will double.

### Example Question #6 : Capacitor Energy

Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the internal energy, U, stored in the capacitor increase, decrease or stay the same?

Increases by a factor of exactly 2

Increases

Stays constant

Decreases

Decreases

Explanation:

Relevant equations:

We must note for this problem that the voltage, V, is kept constant by the battery. Looking at the second equation, if C changes (by changing D) then Q must change when V is held constant. This means that our formula for U must be altered. How can we make claims about U if Q and C are both changing? We need a constant variable in the equation for U so that we can make a direct relationship between D and U. If we plug in the second equation into the third we arrive at:

Now, we know that V is held constant by the battery, so when C decreases (because D doubled) we see that U actually decreases here. So it matters whether or not the capacitor is hooked up to a battery.