Question

A hollow, conducting sphere with an outer radius of 0.254 mm and an inner radius of 0.207 mm has a uniform surface charge density of +6.38×10⁻⁶ C/m2C/m2 . A charge of -0.640 μCμC is now introduced into the cavity inside the sphere.

a)What is the new charge density on the outside of the sphere?

b)Calculate the strength of the electric field just outside the sphere.

c)What is the electric flux through a spherical surface just inside the inner surface of the sphere?

Answer 1

A)

Charge density, s = (6.38 x 10^-6) - [0.64 x 10^-6/(4 pi x 0.254^2)]

s = 5.59 x 10^-6 C/m^2

B)

Electric field, E = s/e0

E = (5.59 x 10^-6)/(8.85 x 10^-12) = 6.32 x 10^5 N/C

C)

Flux, phi = q/e0

Phi = - (0.64 x 10^-6)/(8.85 x 10^-12) = - 7.23 x 10^4 Nm^2/C

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