Three concentric conducting spherical shell carry charges 4Q on inner shell, -2Q on middle shell and 6Q on outer shell.The charge on inner surface of outer shell is?

Question

Answer

Due to the inner sphere the charge on the inner surface of the middle sphere is -4Q. When this charge comes to the outer surface it will become positive and the total charge on the outer surface of the middle sphere will be -2Q + 4Q = +2Q. So, the charge on the inner surface of the outer sphere will be -2Q. Similarly the total charge on the outer surface of the outer sphere will be 2Q + 6Q = 8Q

Answer

Problem:

Three concentric conducting spherical shells carry charges 4Q on the inner shell, -2Q on the middle shell, and 6Q on the outer shell. The charge on the inner surface of the outer shell is?

Solution:

To find the charge on the inner surface of the outer shell, we need to understand the behavior of charges on the conducting shells.

Charge Distribution:

The charges on a conductor reside only on its outer surface.

The charges distribute themselves over the outer surface of the conductor in such a way that the electric field inside the conductor is zero.

The electric field inside a conductor is zero because any electric field present would exert forces on the charges inside the conductor, causing them to move. This would continue until the charges redistribute themselves in such a way as to cancel out the field.

Charge Calculation:

Let's assume that the radius of the innermost shell is 'a' and the radius of the outermost shell is 'c'.

The charge on the outer surface of the outermost shell will induce a charge on the inner surface of the outermost shell.

Let's assume that the charge induced on the inner surface of the outermost shell is 'q'.

The potential difference between the outer and inner surfaces of the outermost shell is:

V = kQ/c - k(4Q - 2Q + q)/a

V = kQ/c - k(2Q + q)/a

The potential difference between the inner and middle shells is:

V = k(4Q)/a - k(-2Q)/b

V = k(6Q)/c - k(4Q)/a

Equating both potential differences, we get:

kQ/c - k(2Q + q)/a = k(6Q)/c - k(4Q)/a

Solving for 'q', we get:

q = 2Q

Final Answer:

Therefore, the charge on the inner surface of the outermost shell is 2Q.

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