Redistribution of Charges for Two Parallel Plates

Let’s talk about the redistribution of charges.

When two charged bodies are placed near each other, charges induced on them due to one another.

Due to the charges induced, the configuration of the charges will change on either body.

This phenomenon is called the redistribution of the charges.

During this phenomenon, some energy is lost in the form of heat.

To analyse this wastage of energy and new charge on bodies, we study charge redistribution.

Let’s see the redistribution of the charges in parallel plates.

Consider two plates, &, placed parallel to each other. Let the area of each plate is .

The plates and are carrying charges and of different magnitude.

Let’s the and be the charges after redistribution on the left and right surface of the plate .

And and be the charges after redistribution on the left and right surface of the plate .

For plate the total charger will be the sum of and .

And for plate the total charge will be the sum of and .

Now consider a point inside the plate .

As the electric field inside the conductor is always zero, electric field at P will be zero.

Let, , , and be the electric fields due to the charges , , and respectively.

So, the electric fields at point P can be represented as

So, the total electric field at point will be given by,

The electric field due to the charged sheet can be expressed as given above.

So, the electric field at point P can be expressed as,

Replace and , in the equation written for the electric field at the point inside the first plate.

After substitution we got,

On simplifying, we will get the charge on the left surface of the first plate.

Therefore, the charge on the right surface of the first plate will be,

Now take a point Q inside the plate .

Similar to the point P, we can represent the directions of the electric field at Q, due to charges & .

Thus, the total electric field at point will be,

On simplifying the equation, we get

On substituting the value of and in the expression, we get

On solving the equation we will get the expression of charge on the left surface of the plate .

Thus, the charge on the right surface of the second plate will be,

Revision

The charge on the outer surfaces of both the plates are equal, and equal to the mean of the charge supplied to the plates.

The charge on the outer surfaces of the plates is,

The charge on the surfaces facing to each other is equal and opposite, and is equal to the,

The charge of the capacitor is equal to the charge on the inner surface on the positive plate.

If both the plates are considered as capacitor, the charge will be the charge of the capacitor.

The End