A metal sphere with radius r a = 1.40 cm is supported on an insulating stand at

Question

A metal sphere with radius ra = 1.40 cm is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rb = 9.90 cm . Charge +q is put on the inner sphere and charge q on the outer spherical shell. The magnitude of q is chosen to make the potential difference between the spheres 400 V , with the inner sphere at higher potential.

Calculate q.

Are the electric field lines and equipotential surfaces mutually perpendicular?

Are the equipotential surfaces closer together when the magnitude of E  is largest?

Are the equipotential surfaces closer together when the magnitude of  is largest?

Explanation / Answer

Given,

ra = 1.4 cm ; rb = 9.9 cm ; V = 400 V

We know that the voltage diffrence for such an arrangement is given by:

delta-V = 1/4 pi e0 [q (1/r1 - 1/r2)]

solving for q we get

q = V 4 pi e0/(1/r1 - 1/r2)

q = 400 x 4 x 3.14 x 8.85 x 10^-12/(1/0.014 - 1/0.099) = 7.25 x 10^-10 C

Hence, q = 7.25 x 10^-10 C

b)Yes, since E is gradient of V.

c)Yes

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