Using Capacitor Equations - AP Physics C: Electricity and Magnetism
Card 1 of 77
You are hired to make a capacitor out of two parallel metal sheets. If someone wanted you to make a thin capacitor of 
out of those metal sheets, and the sheets needed to be 
apart, what area do the two metal plates need to be?
You are hired to make a capacitor out of two parallel metal sheets. If someone wanted you to make a thin capacitor of
Tap to reveal answer
For parallel plate capacitors, the equation is 
.
Solve for 
.
Now we can plug in our given values.
For parallel plate capacitors, the equation is
Solve for
Now we can plug in our given values.
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The plates of a parallel plate capacitor are 
apart and 
in area. A potential difference of 
is applied across the capacitor. Find the capacitance.
The plates of a parallel plate capacitor are
Tap to reveal answer
Capacitance is related to plate area and distance by the equation 
.
Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant.
Capacitance is related to plate area and distance by the equation
Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant.
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The plates of a parallel plate capacitor are 
apart and 
in area. A potential difference of 
is applied across the capacitor. Compute the charge on each plate.
The plates of a parallel plate capacitor are
Tap to reveal answer
Charge on a capacitor is given by the equation 
. We know that the voltage is 
, but we need to determine the capacitance based off of the area and distance between the plates.
We can plug this value into the equation for charge.
Charge on a capacitor is given by the equation
We can plug this value into the equation for charge.
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The plates of a parallel plate capacitor are 
apart. A potential difference of 
is applied across the capacitor. Compute the magnitude of the generated electric field.
The plates of a parallel plate capacitor are
Tap to reveal answer
The electric field given by a capacitor is given by the formula 
. We do not have these variables, so we will have to adjust the equation.
The capacitance can be determined by the area of the plates and the distance between them.
This equation simplifies to 
, allowing us to solve using the values given in the question.
The electric field given by a capacitor is given by the formula
The capacitance can be determined by the area of the plates and the distance between them.
This equation simplifies to
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If two identical parallel plate capacitors of capacitance 
are connected in series, which is true of the equivalent capacitance, 
?
If two identical parallel plate capacitors of capacitance
Tap to reveal answer
Relevant equations:
Use the series equation, replacing C1 and C2 with the given constant C:
This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.
Relevant equations:
Use the series equation, replacing C1 and C2 with the given constant C:
This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.
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Two parallel conducting plates each have an area of 
and are separated by a distance, 
. One plate has a charge 
evenly distributed across it, and the other has a charge of 
. A proton (charge 
) is initially held near the positive plate, then released such that it accelerates towards the negative plate. How much kinetic energy has the proton gained in this process?
Two parallel conducting plates each have an area of
Tap to reveal answer
Relevant equations:
According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.
1. Find an expression for potential difference, 
, in terms of 
and 
, by setting the two capacitance equations equal to each other:
2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):
Relevant equations:
According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.
1. Find an expression for potential difference,
2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):
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A 
and 
capacitor are connected in series with a 
battery. What is the magnitude of the charge on one plate of the 
capacitor?
A
Tap to reveal answer
Relevant equations:
First, find the equivalent capacitance of the whole circuit:
Use this equivalent capacitance to find the total charge:
Capacitors in series always have the same charge on each unit, so the charge on the 
capacitor is also 
.
Relevant equations:
First, find the equivalent capacitance of the whole circuit:
Use this equivalent capacitance to find the total charge:
Capacitors in series always have the same charge on each unit, so the charge on the
← Didn't Know|Knew It →
You are hired to make a capacitor out of two parallel metal sheets. If someone wanted you to make a thin capacitor of out of those metal sheets, and the sheets needed to be apart, what area do the two metal plates need to be?
You are hired to make a capacitor out of two parallel metal sheets. If someone wanted you to make a thin capacitor of
Tap to reveal answer
For parallel plate capacitors, the equation is .
Solve for .
Now we can plug in our given values.
For parallel plate capacitors, the equation is
Solve for
Now we can plug in our given values.
← Didn't Know|Knew It →
The plates of a parallel plate capacitor are apart and in area. A potential difference of is applied across the capacitor. Find the capacitance.
The plates of a parallel plate capacitor are
Tap to reveal answer
Capacitance is related to plate area and distance by the equation .
Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant.
Capacitance is related to plate area and distance by the equation
Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant.
← Didn't Know|Knew It →
The plates of a parallel plate capacitor are apart and in area. A potential difference of is applied across the capacitor. Compute the charge on each plate.
The plates of a parallel plate capacitor are
Tap to reveal answer
Charge on a capacitor is given by the equation . We know that the voltage is , but we need to determine the capacitance based off of the area and distance between the plates.
We can plug this value into the equation for charge.
Charge on a capacitor is given by the equation
We can plug this value into the equation for charge.
← Didn't Know|Knew It →
The plates of a parallel plate capacitor are apart. A potential difference of is applied across the capacitor. Compute the magnitude of the generated electric field.
The plates of a parallel plate capacitor are
Tap to reveal answer
The electric field given by a capacitor is given by the formula . We do not have these variables, so we will have to adjust the equation.
The capacitance can be determined by the area of the plates and the distance between them.
This equation simplifies to , allowing us to solve using the values given in the question.
The electric field given by a capacitor is given by the formula
The capacitance can be determined by the area of the plates and the distance between them.
This equation simplifies to
← Didn't Know|Knew It →
If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance, ?
If two identical parallel plate capacitors of capacitance
Tap to reveal answer
Relevant equations:
Use the series equation, replacing C1 and C2 with the given constant C:
This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.
Relevant equations:
Use the series equation, replacing C1 and C2 with the given constant C:
This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.
← Didn't Know|Knew It →
Two parallel conducting plates each have an area of and are separated by a distance, . One plate has a charge evenly distributed across it, and the other has a charge of . A proton (charge ) is initially held near the positive plate, then released such that it accelerates towards the negative plate. How much kinetic energy has the proton gained in this process?
Two parallel conducting plates each have an area of
Tap to reveal answer
Relevant equations:
According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.
1. Find an expression for potential difference, , in terms of and , by setting the two capacitance equations equal to each other:
2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):
Relevant equations:
According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.
1. Find an expression for potential difference,
2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):
← Didn't Know|Knew It →
A and capacitor are connected in series with a battery. What is the magnitude of the charge on one plate of the capacitor?
A
Tap to reveal answer
Relevant equations:
First, find the equivalent capacitance of the whole circuit:
Use this equivalent capacitance to find the total charge:
Capacitors in series always have the same charge on each unit, so the charge on the capacitor is also .
Relevant equations:
First, find the equivalent capacitance of the whole circuit:
Use this equivalent capacitance to find the total charge:
Capacitors in series always have the same charge on each unit, so the charge on the
← Didn't Know|Knew It →
You are hired to make a capacitor out of two parallel metal sheets. If someone wanted you to make a thin capacitor of out of those metal sheets, and the sheets needed to be apart, what area do the two metal plates need to be?
You are hired to make a capacitor out of two parallel metal sheets. If someone wanted you to make a thin capacitor of
Tap to reveal answer
For parallel plate capacitors, the equation is .
Solve for .
Now we can plug in our given values.
For parallel plate capacitors, the equation is
Solve for
Now we can plug in our given values.
← Didn't Know|Knew It →
The plates of a parallel plate capacitor are apart and in area. A potential difference of is applied across the capacitor. Find the capacitance.
The plates of a parallel plate capacitor are
Tap to reveal answer
Capacitance is related to plate area and distance by the equation .
Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant.
Capacitance is related to plate area and distance by the equation
Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant.
← Didn't Know|Knew It →
The plates of a parallel plate capacitor are apart and in area. A potential difference of is applied across the capacitor. Compute the charge on each plate.
The plates of a parallel plate capacitor are
Tap to reveal answer
Charge on a capacitor is given by the equation . We know that the voltage is , but we need to determine the capacitance based off of the area and distance between the plates.
We can plug this value into the equation for charge.
Charge on a capacitor is given by the equation
We can plug this value into the equation for charge.
← Didn't Know|Knew It →
The plates of a parallel plate capacitor are apart. A potential difference of is applied across the capacitor. Compute the magnitude of the generated electric field.
The plates of a parallel plate capacitor are
Tap to reveal answer
The electric field given by a capacitor is given by the formula . We do not have these variables, so we will have to adjust the equation.
The capacitance can be determined by the area of the plates and the distance between them.
This equation simplifies to , allowing us to solve using the values given in the question.
The electric field given by a capacitor is given by the formula
The capacitance can be determined by the area of the plates and the distance between them.
This equation simplifies to
← Didn't Know|Knew It →
If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance, ?
If two identical parallel plate capacitors of capacitance
Tap to reveal answer
Relevant equations:
Use the series equation, replacing C1 and C2 with the given constant C:
This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.
Relevant equations:
Use the series equation, replacing C1 and C2 with the given constant C:
This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.
← Didn't Know|Knew It →
Two parallel conducting plates each have an area of and are separated by a distance, . One plate has a charge evenly distributed across it, and the other has a charge of . A proton (charge ) is initially held near the positive plate, then released such that it accelerates towards the negative plate. How much kinetic energy has the proton gained in this process?
Two parallel conducting plates each have an area of
Tap to reveal answer
Relevant equations:
According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.
1. Find an expression for potential difference, , in terms of and , by setting the two capacitance equations equal to each other:
2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):
Relevant equations:
According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.
1. Find an expression for potential difference,
2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):
← Didn't Know|Knew It →