Question

A hollow, conducting sphere with an outer radius of 0.250 m and inner radius of 0.200 m has a uniform surface charge density of +6.50µC/m². A charge of -5.00 µC is now introduced into the cavity inside the sphere.

1. What is the new charge density on the outside of the sphere?
2. Calculate the strength of the electric field outside the sphere.
3. What is the electric flux through a spherical surface just outside the inner surface of the sphere?

Answer 1

Surface charge density on the outer sphere is + 6.5 C / m².

Radius of the outer sphere = 0.25 m.

Hence , surface area of the outer sphere = 4 x 0.25² m² = 0.785 m².

So, charge on the sphere was initially Q = 6.5 C / m² x 0.785 m² = 5.1 C.

Now, when a - 5 C charge is kept into the cavity of the spherical shell, a + 5 C charge will be induced at the inner sphere, and - 5 C to the outer one.

Hence, net charge on the outer sphere becomes Q - 5 C = 5.1 C - 5 C = + 0.1 C.

Hence, new charge density on the outside sphere is + 0.1 C / 0.785 m² = + 0.13 C / m².

Net charge enclosed by the sphere, q_enc, including the cavity, is now equal to that on the outer sphere only, so, q_enc = + 0.1 C.

Hence, from Gauss law, we get, the magnitude of the electric field, E, outside the sphere is given by :

E x 4 x 0.25² = q_enc / _o

or, E = q_enc / ( 4 x 0.25² x _o )

or, E = { 0.1 x 10^-6 / ( 4 x 3.14 x 0.25² x 8.854 x 10^-12 ) } V / m

Hence, the strength of the electric field outside the sphere is E = 1.44 x 10⁴ V / m.

When a - 5 C charge is kept into the cavity of the spherical shell, a + 5 C charge will be induced at the inner sphere, thus, net charge enclosed by a spherical Gaussian surface, having surface area, say, A, just outside the inner sphere is :

q = - 5 C + 5 C = 0.

Hence, from Gauss law, we get, the electric flux through the Gaussian surface is : q / _o = 0.

Hence, net electric flux just outside the inner sphere is 0 V.m.