Question

A particle with a charge of -60.0 nC is placed at the center of a nonconducting spherical shell of inner radius 20.0 cm and outer radius 36.0 cm. The spherical shell carries charge with a uniform density of -3.37 µC/m³. A proton moves in a circular orbit just outside the spherical shell. Calculate the speed of the proton.

Answer 1

Using Force balance on proton to find it's speed

Fe = Fc

q*|E| = m*ac

ac = centripetal acceleration = V^2/R

q*|E| = m*V^2/R

V = sqrt (q*|E|*R/m)

q = charge on proton = 1.6*10^-19 C

R = radius of spherical outer shell = 36 cm = 0.36 m

m = mass of proton = 1.67*10^-27 kg

|E| = electric field on the surface of spherical shell = k*|Q|/R^2

Q = total charge inside shell = Q1 + Q2

Q1 = charge of particle placed at center = -60 nC = -60*10^-9 C

Q2 = charge due to uniform volume density = rho*V

rho = -3.37*10^-6 C/m^3

V = Volume of shell = (4*pi/3)*(R^3 - r^3)

R = outer radius = 36.0 cm = 0.36 m

r = inner radius = 20.0 cm = 0.20 m

So,

V = (4*pi/3)*(0.36^3 - 0.20^3) = 0.1619 m^3

So,

Q2 = -3.37*10^-6*0.1619 = -545.6*10^-9 C

Which gives

Q = Q1 + Q2 = -60*10^-9 - 545.6*10^-9

Q = -605.6*10^-9 C

So now electric field will be:

|E| = 9*10^9*605.6*10^-9/0.36^2

|E| = 42055.56 N/C

So velocity of proton will be using above values:

V = sqrt (1.6*10^-19*42055.56*0.36/(1.67*10^-27))

V = 1204383.27 m/sec = 1.20*10^6 m/sec = Speed of proton