A hollow metal sphere has a 5 cm inner radius and a 12 cm outer radius. There is

Question

A hollow metal sphere has a 5 cm inner radius and a 12 cm outer radius. There is a particle at the center of the sphere. The surface charge density on the inside surface of the sphere is ?150nC/m2. The surface charge density on the exterior surface is +150nC/m2.

A. What is the charge on the particle at the center of the sphere?
B. Beginning with Gauss's Law, find the electric field at a point 4.00 cm from the center. Remember, electric field is a vector, so this answer must be a vector.
C. What is the electric field at a point 8.00 cm from the center?
D. Beginning with Gauss's Law, find the electric field at a point 20 cm from the center.

Explanation / Answer

A)charge on inner surface, Q_inner = inner surface charge density*area

= -150*10^-9*4*3.14*0.05^2

= -4.71*10^-9 C

we know electric filed inside a metal is zero.

let q is the charge at the center

so, q + Q_inner = 0

q = -Q_inner

= +4.71*10^9 C <<<<<<<<<<Answer for part A

B) E = k*q/r^2 = 9*10^9*4.71*10^-9/0.04^2 = 26493.75 N

direction of E is radially outward

C) E = 0 (inside metal E = 0)

D) Q_outer = 150*10^-9*4*3.14*0.12^2 = 27.1296*10^-9 C


Qinside = q + Q_inner + Q_outer = (4.71-4.71+27.1296)*10^-9 = 27.13*10^-9 C

E = k*Q_inside/r^2 = 9*10^9*27.13*10^-9/0.2^2
= 6104.25 N/c

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