A thin, uniformly charged spherical shell has a potential of 645 V on its surfac

Question

A thin, uniformly charged spherical shell has a potential of 645 V on its surface. Outside the sphere, at a radial distance of 21.0 cm from this surface, the potential is 325 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 21.0 cm from the surface of the sphere, calculate the electron's speed when it reaches the sphere's surface.

Explanation / Answer

The electric field is given by:

E = -V/r = -(325 - 645)/0.21 = 1524 N/C

V1 = 645 = E r = = 1524 r, => r = V1/E = 645/1524 = 0.42 m

So, radius of the sphere = 0.42 m.

V1 = kQ/r,

=> Q = rV1/k = 0.42*645/9x10^9 = 30x10^-9 C = 30 nC

So, total charge of the sphere = 30 nC.

E = kQ/r² = 9x10^9(30x10^-9)/0.42² = 1530.6 N/C

So, electric field at the surface of the sphere = 1530.6 N/C.

To find the velocity utilize 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)*1530.6/(9.11x10^-31)]0.42

v = 10^6{4100*1.6*0.42/9.11}

v = 15.03 x 10^6 m/s

So, the speed of electron when it reaches the sphere's surface = 15.03 x 10^6 m/s.

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