Chapter 23, Problem 049 In the figure a solid sphere of radius a 2.90 cm is conc

Question

Chapter 23, Problem 049 In the figure a solid sphere of radius a 2.90 cm is concentric with a sphencal conducting shell of inner radius b 2.00a and outer radius c shell has a net charge q2--q1. What is the magnitude of the electric field at radial distances (a) r = 0 cm, (b) r = a/2.00, (c) r = a, (d) on the (g) inner and (h) outer surface of the shell? 2.40a. The sphere has a net uniform charge q1 +7.59 fc; the 1.50e, (e) r = 2.30a, and (f) r = 3.50a? What is the net char (a) Number (b) Number (c) Number (d) Number (e) Number (f) Number (g) Number (h) Number Units Units Units Units Units Units Units Units

Explanation / Answer

From gauss law of electrostatics ,net flux through a closed loop is phi = q_inside/eo

for a) construct a gaussina surface having radius r = 0,then no charges resides in that gausian sphere of radius r = 0 ,then

net flux is E*A = q_inside/eo

E*A = 0

E = 0 N/C

b) at a point inside the sphere at r = a/2

E = k*q1*r/R^3 = (k*q1*(a/2))/a^3 = k*q1/(2*a^2) = (9*10^9*7.59*10^-15)/(2*0.029^2) = 40.6*10^-3 N/C

C) at a point on the sphere r = a

then E = k*q1/a^2 = (9*10^9*7.59*10^-15)/(0.029^2)

E = 81.2*10^-3 N/C

D) at r = 1.5a

E = k*q1/r^2 = (9*10^9*7.59*10^-15)/(1.50.029)^2 = 36*10^-3 N/C

E) at r = 2.3 a

then flux through a gausian surface of radius r = 2.3 a,then

flux is E*A = q_inside /eo = (q1-q1)/eo = 0

So E = 0 N/C

f) similarly at r = 3.5a

then E = 0 N/C

g) q_inner = 0 C

h) q_outer = -7.59*10^-9 C

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