Capacitor Energy - AP Physics 2
Card 1 of 24
Consider the circuit:
If the voltage drop across C2 is 5V, what is the total energy stored in C2 and C3?
Consider the circuit:
If the voltage drop across C2 is 5V, what is the total energy stored in C2 and C3?
Tap to reveal 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:
Since the voltage is the same for both capacitors, we can simply add the two capacitances to do one calculation for energy:
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:
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If 
, 
and 
, how much energy is stored in 
?
If
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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.
In this circuit, the voltage source,
To find the energy, we can use the formula
To use the formula we need the voltage across
Another hint we can use is that
Now, we can use the formula
Now that we know a charge of
Finally, we plug this
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A 
capacitor is connected to a 
battery. Once the capacitor is fully charged, how much energy is stored?
A
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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.
To find the amount of energy stored in a capactior, we use the equation
We're given the capacitance (
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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?
You have 4 capacitors,
Their capacitances are as follows:
If you have a 6V battery connected to the circuit, what's the total energy stored in the capacitors?
Tap to reveal answer
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, 
.
When adding capacitors, remember how to add in series and parallel.
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.
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,
When adding capacitors, remember how to add in series and parallel.
Capacitors
Now that we have the total capacitance, we can use the earlier equation to find the energy.
The total energy stored is 121.5J.
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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?
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
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.
Equations required:
We see from the first equation the when D is doubled, C will be halved (since
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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?
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
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.
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.
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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?
You have 4 capacitors,
Their capacitances are as follows:
If you have a 6V battery connected to the circuit, what's the total energy stored in the capacitors?
Tap to reveal answer
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, .
When adding capacitors, remember how to add in series and parallel.
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.
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,
When adding capacitors, remember how to add in series and parallel.
Capacitors
Now that we have the total capacitance, we can use the earlier equation to find the energy.
The total energy stored is 121.5J.
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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?
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
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.
Equations required:
We see from the first equation the when D is doubled, C will be halved (since
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Consider the circuit:
If the voltage drop across C2 is 5V, what is the total energy stored in C2 and C3?
Consider the circuit:
If the voltage drop across C2 is 5V, what is the total energy stored in C2 and C3?
Tap to reveal 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:
Since the voltage is the same for both capacitors, we can simply add the two capacitances to do one calculation for energy:
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:
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If , and , how much energy is stored in ?
If
Tap to reveal 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
, 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.
In this circuit, the voltage source,
To find the energy, we can use the formula
To use the formula we need the voltage across
Another hint we can use is that
Now, we can use the formula
Now that we know a charge of
Finally, we plug this
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A capacitor is connected to a battery. Once the capacitor is fully charged, how much energy is stored?
A
Tap to reveal answer
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.
To find the amount of energy stored in a capactior, we use the equation
We're given the capacitance (
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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?
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
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.
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.
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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?
You have 4 capacitors,
Their capacitances are as follows:
If you have a 6V battery connected to the circuit, what's the total energy stored in the capacitors?
Tap to reveal answer
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, .
When adding capacitors, remember how to add in series and parallel.
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.
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,
When adding capacitors, remember how to add in series and parallel.
Capacitors
Now that we have the total capacitance, we can use the earlier equation to find the energy.
The total energy stored is 121.5J.
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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?
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
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.
Equations required:
We see from the first equation the when D is doubled, C will be halved (since
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Consider the circuit:
If the voltage drop across C2 is 5V, what is the total energy stored in C2 and C3?
Consider the circuit:
If the voltage drop across C2 is 5V, what is the total energy stored in C2 and C3?
Tap to reveal 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:
Since the voltage is the same for both capacitors, we can simply add the two capacitances to do one calculation for energy:
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:
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If , and , how much energy is stored in ?
If
Tap to reveal 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
, 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.
In this circuit, the voltage source,
To find the energy, we can use the formula
To use the formula we need the voltage across
Another hint we can use is that
Now, we can use the formula
Now that we know a charge of
Finally, we plug this
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A capacitor is connected to a battery. Once the capacitor is fully charged, how much energy is stored?
A
Tap to reveal answer
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.
To find the amount of energy stored in a capactior, we use the equation
We're given the capacitance (
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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?
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
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.
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.
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Consider the circuit:
If the voltage drop across C2 is 5V, what is the total energy stored in C2 and C3?
Consider the circuit:
If the voltage drop across C2 is 5V, what is the total energy stored in C2 and C3?
Tap to reveal 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:
Since the voltage is the same for both capacitors, we can simply add the two capacitances to do one calculation for energy:
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:
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If , and , how much energy is stored in ?
If
Tap to reveal 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
, 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.
In this circuit, the voltage source,
To find the energy, we can use the formula
To use the formula we need the voltage across
Another hint we can use is that
Now, we can use the formula
Now that we know a charge of
Finally, we plug this
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