Question

A particle with a charge of -60.0 nC is placed at the center of a non-conducting spherical shell of inner radius 20.0 cm and outer radius 25.0 cm. The spherical shell carries charge with a uniform volume density of -1.33 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 = 25 cm = 0.25 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 = -1.33*10^-6 C/m^3

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

R = outer radius = 25 cm = 0.25 m

r = inner radius = 20 cm = 0.20 m

So,

V = (4*pi/3)*(0.25^3 - 0.20^3) = 0.03194 m^3

So,

Q2 = -1.33*10^-6*0.03194 = 42.48*10^-9 C

Which gives

Q = Q1 + Q2 = -60*10^-9 - 42.48*10^-9

Q = -102.48*10^-9 C

So now electric field will be:

|E| = 9*10^9*102.48*10^-9/0.25^2

|E| = 14757.12 N/C

So velocity of proton will be using above values:

V = sqrt (1.6*10^-19*14757.12*0.25/(1.67*10^-27))

V = 594528.344 m/sec

V = Speed of proton = 5.95*10^5 m/sec

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