A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. (a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. (a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
Given: The inner radius of spherical conducting shell is r1 and the outer radius is r2 .The charge of the spherical shell is Q.
(a) The surface charge density is the total charge per unit surface area of the conducting shell.
A charge of +q is placed at the centre of the shell. Thus, the charge of magnitude −q will be induced to the inner surface of the shell. The total charge on the inner surface is −q.
The inner surface area of spherical shell is given as,
A1=4πr21
The surface charge density of inner surface of the shell is given as,
σ1=−q4πr21
The charge of magnitude +q will be induced to the outer surface of the shell and a charge of Q is placed on the outer surface so, total charge on the outer surface is Q+q.
The inner surface area of spherical shell is given as,
A2=4πr22
The surface charge density of outer surface of the shell is given as,
σ2=(Q+q)4πr22
Therefore, the surface charge density on the inner and outer surface of the shell is −q4πr21 and (Q+q)4πr22
(b) Electric field inside a cavity is zero even if the shell is not spherical and has any irregular shape. Consider a closed loop, a part of which is inside the cavity along the field line and the remaining part is inside the conductor. The total work done by the field in moving a test charge over the closed loop will be zero because the electric field inside the conductor is zero. Thus, the electric field is zero, no matter what the shape is.
Given: The inner radius of spherical conducting shell is r1 and the outer radius is r2 .The charge of the spherical shell is Q.
(a) The surface charge density is the total charge per unit surface area of the conducting shell.
A charge of +q is placed at the centre of the shell. Thus, the charge of magnitude −q will be induced to the inner surface of the shell. The total charge on the inner surface is −q.
The inner surface area of spherical shell is given as,
A1=4πr21
The surface charge density of inner surface of the shell is given as,
σ1=−q4πr21
The charge of magnitude +q will be induced to the outer surface of the shell and a charge of Q is placed on the outer surface so, total charge on the outer surface is Q+q.
The inner surface area of spherical shell is given as,
A2=4πr22
The surface charge density of outer surface of the shell is given as,
σ2=(Q+q)4πr22
Therefore, the surface charge density on the inner and outer surface of the shell is −q4πr21 and (Q+q)4πr22
(b) Electric field inside a cavity is zero even if the shell is not spherical and has any irregular shape. Consider a closed loop, a part of which is inside the cavity along the field line and the remaining part is inside the conductor. The total work done by the field in moving a test charge over the closed loop will be zero because the electric field inside the conductor is zero. Thus, the electric field is zero, no matter what the shape is.
