consider a charge of q=30 micro collums, sitting at the center of a hollow condu

Question

consider a charge of q=30 micro collums, sitting at the center of a hollow conducting sphere of outer radius 1m and inner radius of 0.5m. a) what amount of charge is on the inner and outer surface of the sphere? b) what is the electric field at 5m from the center of the sphere? c) what is the electric potential at 5m from the center of the sphere? d) an electron is held at 5m from the center of the sphere. what is the potential energy? e) if the electron is released, how fast will it be moving as it hits the sphere?

Explanation / Answer

As given in the question,

Charge at the center: Q = 30 C = 3*10^-5 C

(a) As we know that the electric field inside a conductor in a static situation is zero.

=> By Gauss’s Law, there can be no net charge inside the conductor

So, Q(inner surface) = - Q = - 3*10^-5 C  = - 30 C

and   Q(outer surface) = Q = 3*10^-5 C = 30 C

(b)   The electric field at 5 m from the center of the sphere,

r = 5m

E  = k*Q / (r)^2  = (8.9*10^9)*(3*10^-5) / (5)^2 = 1.068*10^4 N/C

(c)   The electric potential at 5 m from the center of the sphere,

  V  = k*Q / r  = (8.9*10^9)*(3*10^-5) / (5) = 5.34*10^4 V

(d)   The electric potential energy because of electron at 5 m from the center of the sphere,

Charge on electron: qe  = - 1.6*10^-19 C

  PE = k*Q*qe / r  =   (8.9*10^9)*(3*10^-5)*(- 1.6*10^-19) / (5) = 8.544*10^-15 J

(e)   For the speed of electron as it hits the sphere,

Mass of electron: me = 9.1*10^-31 kg

KE(of electron) = PE

=>   (1/2)*me*(v)^2  = PE

=>   (1/2)*(9.1*10^-31)*(v)^2  = (8.544*10^-15) => v = 1.37*10^8 m/s

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