A thin, uniformly charged spherical shell has a potential of 627 V on its surfac

Question

A thin, uniformly charged spherical shell has a potential of 627 V on its surface. Outside the sphere, at a radial distance of 16.0 cm from this surface, the potential is 416 V.

Calculate the radius of the sphere.

Determine the total charge on the sphere.

What is the electric potential inside the sphere at a radius of 2.0 cm?

Calculate the magnitude of the electric field at the surface of the sphere.

If an electron starts from rest at a distance of 16.0 cm from the surface of the sphere, calculate the electron's speed when it reaches the sphere's surface.

Explanation / Answer

1) The electric field is given by:

E = -V/r = -(416 - 627)/0.16 = 1318 N/C

V1 = 627 = E r = = 1318 r, => r = V1/E = 627/1318 = 0.475 m

Hence radius of the sphere is 0.475 m

2) V1 = kQ/r, => Q = rV1/k = 0.475*627/9x10^9 = 33x10^-9 C = 33 nC

3) E = kQ/r² = 9x10^9(33x10^-9)/0.475² = 1316.3 N/C

4) potential inside the sphere is= Vin= 9*109*Q/R= 9*109*33*10-9/0.02

V= 14850 V

5) To find the velocity, the following well known kinematic eqn:

v² = 2ar

where.

acceleration: a = F/m,

and the electrostatic force: F = qE

hence,

v = 2(F/m)r = 2(qE/m)r

v = {2(qE/m)r} = {[2(1.6x10^-19)*1318/(9.11x10^-31)]0.475.


v = 14.8x10^6 m/s

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