# Parallel Plate Capacitor having equal charges on plates

• Tanya Sharma
In summary: The potential difference across the condensator is used to determine the charge density on the inner and outer surfaces of the plates.
Tanya Sharma

## Homework Statement

Q .1 The plates of a parallel plate capacitor are given equal positive charges. What will be the potential difference between the plates? What will be the charges on the facing surfaces and on the outer surfaces?

Q.2 Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a battery .Now,
a) The facing surfaces of the capacitor have equal and opposite charges.
b) The two plates of the capacitor have equal and opposite charges.
c) The battery supplies equal and opposite charges to the two plates.
d) The outer surfaces of the plates have equal charges.

## The Attempt at a Solution

The set up of the two problems looks the same, say, +q, charge is given to the two plates . Let us call the two plates be A and B .The +q charge will distribute uniformly on the two surfaces. Both outer and inner surfaces of A will have +q/2 charge .Now applying Gauss Law ,the inner surface of B will have –q/2 charge and the rest +3q/2 on the outer surface .The charges q/2 on inner surface of A and –q/2 on inner surface of B will contribute to the electric field .

Ans 1) V=Ed
E=q/2εA
Thus ,V=qd/2εA
Charges on the surfaces of the plates are (q/2 ,q/2)on A and (-q/2,3q/2) on B

Is the solution correct ?

Ans 2) I am not clear about the role of battery in affecting the charges on the surfaces of the plates.
Option a) ,c) looks right . Option b)is surely wrong .But not sure about option d)

How will the battery affect the various charges on the inner and outer surfaces of the two plates ?

Why do you expect different charge distributions for plates A and B? The problem statement does not distinguish them - you do not even have a unique way to call them "A" and "B". It does not matter in which order you add the charges.

Option a) ,c) looks right . Option b)is surely wrong .
I agree.

But not sure about option d)
The battery should not break symmetry on the outside I think.

Thanks mfb...

Before the battery is applied , charges on outer surfaces of two plates have to be unequal as the inner surfaces have equal and opposite charges .What is the role of battery after that ?

Before the battery is applied , charges on outer surfaces of two plates have to be unequal as the inner surfaces have equal and opposite charges .
And -0 = 0

What is the role of battery after that ?
It fixes the charge density at the inner sides to a value different from 0.

Connecting the capacitor to the battery, the potential difference across the capacitor becomes the same as the voltage of the battery (U). Accordingly, the charge of the capacitor becomes Q=CU. The charge on the capacitor plates are equal in magnitude and of opposite sign.
You can calculate the electric field outside and inside the capacitor, and you know that the surface charge density on the inner and outer surfaces of the plates are equal to ε0E.

The charged plates can be substituted by planes of charge, and the charge of unit area is σ on one plate and -σ on the other one. (If A is the area of a plate, σ=Q/A).

You know that the charge distribution causes the electric field strength Q/(2ε0) at both sides of a charged plane, and the electric field lines emerge from the positive charges and end in the negative ones.

The electric field in the different regions 1, 2 and 3 are obtained by superposition from the fields of the charge distribution of both planes.
The red arrows show the electric field lines from the positive charges on the upper plane and the blue arrows represent the electric field lines from the negative charges on the lower plane.
You can see that the field lines cancel each other outside the capacitor and the electric field is σ/ε0 inside, pointing downward.

ehild

#### Attachments

• plates.JPG
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mfb said:
And -0 = 0

It fixes the charge density at the inner sides to a value different from 0.

I am unable to understand your explanation

Symmetry requires that the charges on the inside are 0 on both plates in setup 1.

The battery gives a fixed potential difference across the condensator, which can be used to determine the charge density there in setup 2 (see ehild's post for details).

Ah sorry I posted in the wrong thread. Can't delete. Edit only. Sorry I'm a newbie.

mfb said:
Symmetry requires that the charges on the inside are 0 on both plates in setup 1.

The battery gives a fixed potential difference across the condensator, which can be used to determine the charge density there in setup 2 (see ehild's post for details).

Thanks mfb . I have realized my mistakes as well as understood the concept.

Thanks ehild for the wonderful explanation .

## 1. What is a parallel plate capacitor?

A parallel plate capacitor is a device that stores electrical energy by creating an electric field between two parallel conducting plates. It consists of two conductive plates separated by an insulating material, known as the dielectric.

## 2. How does a parallel plate capacitor work?

When equal charges are placed on the two plates of a parallel plate capacitor, an electric field is created between the plates. This electric field causes a potential difference, or voltage, between the plates. The capacitor stores energy in the form of this electric field, which can be released when the capacitor is connected to a circuit.

## 3. What happens if the charges on the plates of a parallel plate capacitor are not equal?

If the charges on the plates of a parallel plate capacitor are not equal, the electric field between the plates will be distorted and the capacitor will not be able to store energy efficiently. This can also cause the capacitor to become unstable and potentially lead to a breakdown of the insulating material between the plates.

## 4. How does the distance between the plates affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is directly proportional to the surface area of the plates and inversely proportional to the distance between them. This means that increasing the distance between the plates decreases the capacitance, while decreasing the distance increases the capacitance.

## 5. What is the significance of equal charges on the plates of a parallel plate capacitor?

Having equal charges on the plates of a parallel plate capacitor is important for maintaining a stable electric field and maximizing the capacitance of the capacitor. If the charges are not equal, the electric field will be distorted, reducing the efficiency of the capacitor and potentially causing damage to the insulating material between the plates.

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