AP Physics 2 : Capacitors and Electric Fields

Example Questions

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Example Question #1 : Capacitors And Electric Fields

parallel-plate capacitor is connected to a constant voltage source. If the distance between the plates of this capacitor is  and the capacitor holds a charge of , what is the value of the electric field between the plates of this capacitor?

An electric field does not exist between the plates of a parallel-plate capacitor

Explanation:

To solve this problem, we first need to solve for the voltage across the capacitor. For this, we'll need the formula for capacitance:

Solving for the voltage:

Now that we have the voltage, we can make use of the following equation to solve for the electric field:

Example Question #1 : Capacitors And Electric Fields

A set of parallel plate capacitors with a surface area of  has a total amount of charge equal to . What is the electric field between the plates?

Explanation:

The equation for the electric field between two parallel plate capacitors is:

Sigma is the charge density of the plates, which is equal to:

We are given the area and total charge, so we use them to find the charge density.

Now that we have the charge density, divide it by the vacuum permittivity to find the electric field.

Example Question #2 : Capacitors And Electric Fields

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 of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the electric field, E, stored in the capacitor increase, decrease or stay the same?

Decreases

Stays constant

We need to know

Increases

Stays constant

Explanation:

Relevant equations:

Considering we are dealing with regions of charge instead of a point charge (or a charge that is at least spherical symmetric), it would be wise to consider using the definition of the electric field here:

We are only considering magnitude so the direction of the electric field is not of concern. Considering a battery is not connected, we can't claim the V is constant. So we use the second equation to substitute for V in the above equation:

Substitute C from the first equation:

We see here that D actually cancels. This would not have been obvious without the previous substitution. Final result:

These 3 quantities are static in this situation so E does not change.

Example Question #1 : Capacitors And Electric Fields

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 electric field, E, stored in the capacitor increase, decrease or stay the same?

Decreases by exactly

Stays constant

Decreases by exactly

Increases by exactly

Decreases by exactly

Explanation:

Relevant equations:

Considering we are dealing with regions of charge instead of a point charge (or a charge that is at least spherical symmetric), it would be wise to consider using the definition of the electric field here:

The battery maintains a constant V so we can directly relate E and D. Since D is doubled, E will be halved.

Example Question #1 : Capacitors And Electric Fields

If , each plate of the capacitor has surface area , and the plates are  apart, determine the electric field between the plates.

Explanation:

The voltage rise through the source must be the same as the drop through the capacitor.

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

Combine these equations and solve for the electric field:

Convert mm to m and plug in values:

Example Question #131 : Circuits

If , each plate of the capacitor has surface area , and the plates are  apart, determine the excess charge on the positive plate.

Explanation:

The voltage rise through the source must be the same as the drop through the capacitor.

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

Combine equations and solve for the electric field:

Convert mm to m and plugging in values:

Use the electric field in a capacitor equation:

Combine equations:

Converting  to  and plug in values:

Example Question #1 : Capacitors And Electric Fields

Consider the given diagram. If , each plate of the capacitor has surface area , and the plates at  apart, determine the electric field between the plates.

Explanation:

The voltage rise through the source must be the same as the drop through the capacitor.

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

Combine equations and solve for the electric field:

Convert mm to m and plug in values:

Example Question #951 : Ap Physics 2

In the given circuit, the capacitor is made of two parallel circular plates of radius that are apart. If is equal to , determine the electric field between the plates.

None of these

Explanation:

Since it is the only element in the circuit besides the source, the voltage drop across the capacitor must be equal to the voltage gain in the capacitor.

Definition of voltage:

Combine equations:

Solve for

Convert to meters and plug in values:

Example Question #1 : Capacitors And Electric Fields

A parallel plate capacitor with a separation of  and surface area  is in series with a  battery. How will the electric field in the capacitor change if the separation is doubled?

Halved

Doubled

Tripled

It will stay the same

Halved

Explanation:

The voltage drop through the capacitor needs to be equal to the voltage of the battery.

The voltage drop of a parallel plate capacitor is equal to the internal electric field times the distance between them.

Combing equations and solving for

From this, it can be seen that doubling the separation will halve the electric field.

Example Question #957 : Ap Physics 2

A parallel plate capacitor with a separation of  and surface area  is in series with a  battery. How will the electric field in the capacitor change if a second  battery is added in series?

No change

Tripled

Doubled

Impossible to determine

Halved

Doubled

Explanation:

The voltage drop through the capacitor needs to be equal to the voltage of the battery.

The voltage drop of a parallel plate capacitor is equal to the internal electric field times the distance between them.

Combing equations and solving for

From this, it can be seen that doubling the voltage of the battery will doubled the electric field inside the capacitor.

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