Capacitors - AP Physics C: Electricity and Magnetism
Card 0 of 99
A parallel plate capacitor has a capacitance of 
. If the plates are 
apart, what is the area of the plates?
A parallel plate capacitor has a capacitance of
The relationship between capacitance, distance, and area is 
. We can rearrange this equation to solve for area.
Now, we can use the values given in teh question to solve.
The relationship between capacitance, distance, and area is
Now, we can use the values given in teh question to solve.
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Charge is distributed uniformly over the area of the two plates of a parallel plate capacitor, resulting in a surface area charge density of 
on the plates (the top plate is positive and the bottom is negative, as shown below). Each plate has area 
and are separated by distance 
. A material of dielectric constant 
has been placed between the two plates.
Which of the following would not result in an increase in the measure of electric potential difference between the two plates?
Charge is distributed uniformly over the area of the two plates of a parallel plate capacitor, resulting in a surface area charge density of
Which of the following would not result in an increase in the measure of electric potential difference between the two plates?
The electric potential difference created between the plates of a parallel plate capactor is given by the equation:
The charge can be calculated by using the equation:
The value of the capacitance is related to the dimensions of the capacitor with the equation:
Combining these equations yields:
The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.
The electric potential difference created between the plates of a parallel plate capactor is given by the equation:
The charge can be calculated by using the equation:
The value of the capacitance is related to the dimensions of the capacitor with the equation:
Combining these equations yields:
The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.
Compare your answer with the correct one above
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
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.
Compare your answer with the correct one above
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
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.
Compare your answer with the correct one above
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
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.
Compare your answer with the correct one above
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
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
Compare your answer with the correct one above
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
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.
Compare your answer with the correct one above
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
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
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
Compare your answer with the correct one above
A parallel plate capacitor has a capacitance of . If the plates are apart, what is the area of the plates?
A parallel plate capacitor has a capacitance of
The relationship between capacitance, distance, and area is . We can rearrange this equation to solve for area.
Now, we can use the values given in teh question to solve.
The relationship between capacitance, distance, and area is
Now, we can use the values given in teh question to solve.
Compare your answer with the correct one above
Charge is distributed uniformly over the area of the two plates of a parallel plate capacitor, resulting in a surface area charge density of on the plates (the top plate is positive and the bottom is negative, as shown below). Each plate has area and are separated by distance . A material of dielectric constant has been placed between the two plates.
Which of the following would not result in an increase in the measure of electric potential difference between the two plates?
Charge is distributed uniformly over the area of the two plates of a parallel plate capacitor, resulting in a surface area charge density of
Which of the following would not result in an increase in the measure of electric potential difference between the two plates?
The electric potential difference created between the plates of a parallel plate capactor is given by the equation:
The charge can be calculated by using the equation:
The value of the capacitance is related to the dimensions of the capacitor with the equation:
Combining these equations yields:
The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.
The electric potential difference created between the plates of a parallel plate capactor is given by the equation:
The charge can be calculated by using the equation:
The value of the capacitance is related to the dimensions of the capacitor with the equation:
Combining these equations yields:
The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.
Compare your answer with the correct one above
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
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.
Compare your answer with the correct one above
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
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.
Compare your answer with the correct one above
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
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.
Compare your answer with the correct one above
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
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
Compare your answer with the correct one above
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
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.
Compare your answer with the correct one above
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
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):
Compare your answer with the correct one above
A and capacitor are connected in series with a battery. What is the magnitude of the charge on one plate of the capacitor?
A
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
Compare your answer with the correct one above
A parallel plate capacitor has a capacitance of . If the plates are apart, what is the area of the plates?
A parallel plate capacitor has a capacitance of
The relationship between capacitance, distance, and area is . We can rearrange this equation to solve for area.
Now, we can use the values given in teh question to solve.
The relationship between capacitance, distance, and area is
Now, we can use the values given in teh question to solve.
Compare your answer with the correct one above
Charge is distributed uniformly over the area of the two plates of a parallel plate capacitor, resulting in a surface area charge density of on the plates (the top plate is positive and the bottom is negative, as shown below). Each plate has area and are separated by distance . A material of dielectric constant has been placed between the two plates.
Which of the following would not result in an increase in the measure of electric potential difference between the two plates?
Charge is distributed uniformly over the area of the two plates of a parallel plate capacitor, resulting in a surface area charge density of
Which of the following would not result in an increase in the measure of electric potential difference between the two plates?
The electric potential difference created between the plates of a parallel plate capactor is given by the equation:
The charge can be calculated by using the equation:
The value of the capacitance is related to the dimensions of the capacitor with the equation:
Combining these equations yields:
The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.
The electric potential difference created between the plates of a parallel plate capactor is given by the equation:
The charge can be calculated by using the equation:
The value of the capacitance is related to the dimensions of the capacitor with the equation:
Combining these equations yields:
The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.
Compare your answer with the correct one above