# Parallel plate capacitor problem

• smithnh
In summary, the conversation discusses a problem with a parallel plate capacitor connected to a battery and fully charged with Q Coulombs and a voltage of V Volts. The conversation then presents various statements about the effects of changing the plate separation and area on the capacitance and stored energy of the capacitor. The relevant equations (Eq (1), Eq (2), and Eq (3)) are also mentioned. After considering these equations and the fact that Q is constant, the correct answers are determined to be A, B, and E.
smithnh
[SOLVED] Capacitor problem

Problem:
A parallel plate capacitor with adjustible plate separation d and adjustible area A is connected to a battery. The capacitor is fully charged to Q Coulombs and a voltage of V Volts. (C is the capacitance and U is the stored energy.) Give all correct answers concerning a parallel-plate capacitor charged by a battery (e.g. B, AC, CDF).

A) After being disconnected from the battery, increasing the area A will increase U.
B) After being disconnected from the battery, increasing d increases U.
C) After being disconnected from the battery, decreasing d increases C.
D) After being disconnected from the battery, increasing the area A will increase V.
E) With the capacitor connected to the battery, increasing the area A will increase C.
F) After being disconnected from the battery, increasing d decreases V.

Relevant equations:

(1) C=(epsilon_0)A/d

(2) V=E*d=(sigma_0)*d/(epsilon_0)

(3) U=(1/2)(epsilon_0)(E_field)^2*A*d

3. The Attempt at a Solution

A) seems to be true because of Eq (3)
B) seems to be true because of Eq (3)
C) seems to be true because of Eq (1)
D) seems to be false because the potential does not depend on the area in Eq (2). However, I also considered that the if the charge stays the same it will decrease the charge density, sigma_0 for increasing area and thus it will decrease V but this also supports the answer being false.
E) seems to be true because of Eq (1)
F) seems to be false because increasing d will increase V by Eq (2)

AB
ABE
ABCE
ABCDE

I cannot seem to figure this one out despite the fact that it seems so simple, where am I going wrong

What I needed to think about was the fact that the battery cannot suplicate more charge thus Q is constant. After realizing that sigma_0 is defined as Q/A that then brought to question how I was thinking about the electric ield and was able to figure the answer.

?

First of all, it is important to note that the given equations (1), (2), and (3) are only applicable for a parallel plate capacitor with a constant electric field and uniform charge distribution. This means that the plates are perfectly parallel and there is no fringing or edge effects.

With that in mind, let's go through each answer and see which ones are correct:

A) This answer is correct. Increasing the area A will increase the stored energy U because it increases the volume between the plates and thus increases the electric field and the potential energy.

B) This answer is also correct. Increasing the plate separation d will also increase the stored energy U, because it increases the distance between the plates and thus increases the electric field and the potential energy.

C) This answer is false. Decreasing the plate separation d will actually decrease the capacitance C, not increase it. This is because capacitance is inversely proportional to the distance between the plates (see equation (1)).

D) This answer is false. As you correctly pointed out, the potential does not depend on the area in equation (2). Therefore, increasing the area A will not increase the potential V.

E) This answer is correct. Increasing the area A will increase the capacitance C, as shown in equation (1).

F) This answer is also correct. Increasing the plate separation d will decrease the potential V, as shown in equation (2). This is because increasing d will decrease the electric field between the plates, thus decreasing the potential difference.

In summary, the correct answers are A, B, E, and F. It is important to carefully consider the equations and understand the assumptions and limitations before choosing an answer.

## 1. What is a parallel plate capacitor?

A parallel plate capacitor is a type of capacitor that consists of two parallel metal plates separated by a dielectric material. It is used to store electrical energy by creating an electric field between the plates.

## 2. How do you calculate the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor can be calculated by dividing the plate area by the distance between the plates, multiplied by the permittivity of the dielectric material. The formula is C = εA/d, where C is the capacitance, ε is the permittivity, A is the plate area, and d is the distance between the plates.

## 3. What factors affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is affected by the plate area, distance between the plates, and the permittivity of the dielectric material. Additionally, the type of dielectric material used and the presence of any other conductors nearby can also affect the capacitance.

## 4. How does the capacitance change if the distance between the plates is increased?

If the distance between the plates of a parallel plate capacitor is increased, the capacitance will decrease. This is because the electric field strength decreases with distance, resulting in a lower amount of charge being stored on the plates.

## 5. What is the significance of the dielectric material in a parallel plate capacitor?

The dielectric material in a parallel plate capacitor plays a crucial role in determining the capacitance. It has a higher permittivity than air, which allows for a greater amount of charge to be stored on the plates and thus increases the capacitance. It also serves to prevent the plates from coming into direct contact, which could cause a short circuit.

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