Capacitor Energy - AP Physics 2

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Question

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

Their capacitances are as follows:

$\begin{align} A&=1F \ B&=3F \ C&=2F \ D&=4F \end{align}$

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

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Answer

The equation for energy stored in a capacitor is

$W=\frac{Q^2}{2C}=\frac{CV^2}{2}=\frac{QV}{2}$

We can find the capacitance by adding the capacitors together, and we have the voltage, so we'll use the second equation, $W=\frac{1}{2}CV^2$ .

When adding capacitors, remember how to add in series and parallel.

$(Series)\ \frac{1}{C_{tot}}&=\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}$

$(Parallel)\ C_{tot}&=C_1+C_2+...+C_n$

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

$C_{AB}&=(\frac{1}{C_A}+\frac{1}{C_B})^{-1}$

$C_{AB}=(\frac{1}{1}+\frac{1}{3})^{-1}$

$C_{AB}= \frac{3}{4}$

$C_{ABC}&=C_{AB}+C_C$

$C_{ABC}=\frac{3}{4}+2$

$C_{ABC}= \frac{11}{4}$

$C_{tot}&=C_{ABC}+C_{D}$

$C_{total}=\frac{11}{4}+4$

$C_{total}=\frac{27}{4}$

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

$W=\frac{1}{2}CV^2$

$W= \frac{1}{2}(\frac{27}{4})(6)^2$

The total energy stored is 121.5J.

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Question

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?

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Answer

Equations required:

$C=\frac{\varepsilon_{o}A}{D}$

$U=\frac{1}{2}\frac{Q^{2}}{C}$

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

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Question

Consider the circuit:

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

$C1 = 5F,\ C2 = 7F,\ C3 = 1F$

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Answer

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:

$U=\frac{1}{2}CV^2$

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

$U = \frac{1}{2}(8F)(5V)^2 = 100J$

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Question

If , and , how much energy is stored in ?

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Answer

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

$U=\frac{1}2CV^2$ , 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:

$C_{12}= \frac{C_{1}C_{2}}{C_{1}+C_{2}}=\frac{4}{5}F$

Now, we can use the formula

to calculate charge in the capacitor.

$Q=CV=(\frac{4}{5})(5)=4 C$

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

$Q_{1}=C_{1}\cdot V_{1}$

$V_{1}=\frac{4}{4}=1V$

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

$U=\frac{1}{2}CV^2=\frac{1}{2}(4)(1)^2=2J$

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Question

A $15\mu F$ capacitor is connected to a battery. Once the capacitor is fully charged, how much energy is stored?

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Answer

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

$U=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}QV=\frac{1}{2}CV^2$ .

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

$U=\frac{1}{2}(15\mu F)(9V)$

$U= 6.075\cdot 10^{-4}J$

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Question

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?

Tap to reveal answer

Answer

Relevant equations:

$C=\frac{\varepsilon_{o}A}{D}$

$U=\frac{1}{2}\frac{Q^{2}}{C}$

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:

$U=\frac{1}{2}CV^{2}$

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.

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Question

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

Their capacitances are as follows:

$\begin{align} A&=1F \ B&=3F \ C&=2F \ D&=4F \end{align}$

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

Tap to reveal answer

Answer

The equation for energy stored in a capacitor is

$W=\frac{Q^2}{2C}=\frac{CV^2}{2}=\frac{QV}{2}$

We can find the capacitance by adding the capacitors together, and we have the voltage, so we'll use the second equation, $W=\frac{1}{2}CV^2$ .

When adding capacitors, remember how to add in series and parallel.

$(Series)\ \frac{1}{C_{tot}}&=\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}$

$(Parallel)\ C_{tot}&=C_1+C_2+...+C_n$

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

$C_{AB}&=(\frac{1}{C_A}+\frac{1}{C_B})^{-1}$

$C_{AB}=(\frac{1}{1}+\frac{1}{3})^{-1}$

$C_{AB}= \frac{3}{4}$

$C_{ABC}&=C_{AB}+C_C$

$C_{ABC}=\frac{3}{4}+2$

$C_{ABC}= \frac{11}{4}$

$C_{tot}&=C_{ABC}+C_{D}$

$C_{total}=\frac{11}{4}+4$

$C_{total}=\frac{27}{4}$

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

$W=\frac{1}{2}CV^2$

$W= \frac{1}{2}(\frac{27}{4})(6)^2$

The total energy stored is 121.5J.

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Question

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?

Tap to reveal answer

Answer

Equations required:

$C=\frac{\varepsilon_{o}A}{D}$

$U=\frac{1}{2}\frac{Q^{2}}{C}$

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

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Question

Consider the circuit:

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

$C1 = 5F,\ C2 = 7F,\ C3 = 1F$

Tap to reveal answer

Answer

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:

$U=\frac{1}{2}CV^2$

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

$U = \frac{1}{2}(8F)(5V)^2 = 100J$

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Question

If , and , how much energy is stored in ?

Tap to reveal answer

Answer

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

$U=\frac{1}2CV^2$ , 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:

$C_{12}= \frac{C_{1}C_{2}}{C_{1}+C_{2}}=\frac{4}{5}F$

Now, we can use the formula

to calculate charge in the capacitor.

$Q=CV=(\frac{4}{5})(5)=4 C$

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

$Q_{1}=C_{1}\cdot V_{1}$

$V_{1}=\frac{4}{4}=1V$

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

$U=\frac{1}{2}CV^2=\frac{1}{2}(4)(1)^2=2J$

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Question

A $15\mu F$ capacitor is connected to a battery. Once the capacitor is fully charged, how much energy is stored?

Tap to reveal answer

Answer

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

$U=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}QV=\frac{1}{2}CV^2$ .

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

$U=\frac{1}{2}(15\mu F)(9V)$

$U= 6.075\cdot 10^{-4}J$

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Question

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?

Tap to reveal answer

Answer

Relevant equations:

$C=\frac{\varepsilon_{o}A}{D}$

$U=\frac{1}{2}\frac{Q^{2}}{C}$

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:

$U=\frac{1}{2}CV^{2}$

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.

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Question

Consider the circuit:

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

$C1 = 5F,\ C2 = 7F,\ C3 = 1F$

Tap to reveal answer

Answer

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:

$U=\frac{1}{2}CV^2$

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

$U = \frac{1}{2}(8F)(5V)^2 = 100J$

← Didn't Know|Knew It →

Question

If , and , how much energy is stored in ?

Tap to reveal answer

Answer

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

$U=\frac{1}2CV^2$ , 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:

$C_{12}= \frac{C_{1}C_{2}}{C_{1}+C_{2}}=\frac{4}{5}F$

Now, we can use the formula

to calculate charge in the capacitor.

$Q=CV=(\frac{4}{5})(5)=4 C$

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

$Q_{1}=C_{1}\cdot V_{1}$

$V_{1}=\frac{4}{4}=1V$

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

$U=\frac{1}{2}CV^2=\frac{1}{2}(4)(1)^2=2J$

← Didn't Know|Knew It →

Question

A $15\mu F$ capacitor is connected to a battery. Once the capacitor is fully charged, how much energy is stored?

Tap to reveal answer

Answer

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

$U=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}QV=\frac{1}{2}CV^2$ .

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

$U=\frac{1}{2}(15\mu F)(9V)$

$U= 6.075\cdot 10^{-4}J$

← Didn't Know|Knew It →

Question

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

Their capacitances are as follows:

$\begin{align} A&=1F \ B&=3F \ C&=2F \ D&=4F \end{align}$

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

Tap to reveal answer

Answer

The equation for energy stored in a capacitor is

$W=\frac{Q^2}{2C}=\frac{CV^2}{2}=\frac{QV}{2}$

We can find the capacitance by adding the capacitors together, and we have the voltage, so we'll use the second equation, $W=\frac{1}{2}CV^2$ .

When adding capacitors, remember how to add in series and parallel.

$(Series)\ \frac{1}{C_{tot}}&=\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}$

$(Parallel)\ C_{tot}&=C_1+C_2+...+C_n$

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

$C_{AB}&=(\frac{1}{C_A}+\frac{1}{C_B})^{-1}$

$C_{AB}=(\frac{1}{1}+\frac{1}{3})^{-1}$

$C_{AB}= \frac{3}{4}$

$C_{ABC}&=C_{AB}+C_C$

$C_{ABC}=\frac{3}{4}+2$

$C_{ABC}= \frac{11}{4}$

$C_{tot}&=C_{ABC}+C_{D}$

$C_{total}=\frac{11}{4}+4$

$C_{total}=\frac{27}{4}$

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

$W=\frac{1}{2}CV^2$

$W= \frac{1}{2}(\frac{27}{4})(6)^2$

The total energy stored is 121.5J.

← Didn't Know|Knew It →

Question

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?

Tap to reveal answer

Answer

Equations required:

$C=\frac{\varepsilon_{o}A}{D}$

$U=\frac{1}{2}\frac{Q^{2}}{C}$

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

← Didn't Know|Knew It →

Question

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?

Tap to reveal answer

Answer

Relevant equations:

$C=\frac{\varepsilon_{o}A}{D}$

$U=\frac{1}{2}\frac{Q^{2}}{C}$

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:

$U=\frac{1}{2}CV^{2}$

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.

← Didn't Know|Knew It →

Question

Consider the circuit:

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

$C1 = 5F,\ C2 = 7F,\ C3 = 1F$

Tap to reveal answer

Answer

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:

$U=\frac{1}{2}CV^2$

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

$U = \frac{1}{2}(8F)(5V)^2 = 100J$

← Didn't Know|Knew It →

Question

If , and , how much energy is stored in ?

Tap to reveal answer

Answer

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

$U=\frac{1}2CV^2$ , 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:

$C_{12}= \frac{C_{1}C_{2}}{C_{1}+C_{2}}=\frac{4}{5}F$

Now, we can use the formula

to calculate charge in the capacitor.

$Q=CV=(\frac{4}{5})(5)=4 C$

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

$Q_{1}=C_{1}\cdot V_{1}$

$V_{1}=\frac{4}{4}=1V$

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

$U=\frac{1}{2}CV^2=\frac{1}{2}(4)(1)^2=2J$

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