B 1 u .mx, x\' A··.HIn. :E· .._ . 1Nornal 1No Spac. Heading 1 -ing 2 1-e The pic

Question

B 1 u .mx, x' A··.HIn. :E· .._ . 1Nornal 1No Spac. Heading 1 -ing 2 1-e The picture below shows a uniform spherical charge distribution = 4.35 mc or radius a 0.182 m. This charge distribution is located at the center of a conducting shell of inner radius b = 0.256 m and outer rad carries an excess charge of -23.5 nC. Find the electric field at of the following distances from the center of the shell. 0.412 m. The conducting r = 0.150 m (7 points) r = 0.225 m (8 points) r = 0.350 m (7 points) r = 0.450 m (8 points) a. d,

Explanation / Answer

part a

r < a

charge inside gaussian sphere of radius r , q = rho*(4/3)*pi*r^3

from gauss law

E*4*pi*r^2 = q/epsilon0

E*4*pi*r^2 = rho*(4/3)*pi*r^3/epsilon0

E = rho*r/(3*epsilon0)

E = 4.35*10^-6*0.15/(3*8.854*10^-12)

E = 2.45*10^4 N/C

=============================

part(b)

a < r < c

charge inside gaussian sphere of radius r , q = rho*(4/3)*pi*a^3

from gauss law

E*4*pi*r^2 = q/epsilon0

E*4*pi*r^2 = rho*(4/3)*pi*r^3/epsilon0

E = rho*a^3/(3*epsilon0*r^2)

E = 4.35*10^-6*0.182^3/(3*8.854*10^-12*0.225^2)

E = 1.95*10^4 N/C

======================================================

part ( c)

b < r < c

r is inside the conductor

Electric field inside the conductor = 0

E = 0

==========================================

part (d)

r > c

Q = -23.5

charge inside gaussian sphere of radius r , q = rho*(4/3)*pi*a^3 + Q

q = 4.35*10^-6*(4/3)*pi*0.182^3 - 23.5*10^-9

q = 8.63*10^-8 C

from gauss law

E*4*pi*r^2 = q/epsilon0

E = q/(4*pi*epsilon0*r^2)

E = 8.63*10^-8/(4*pi*8.854*10^-12*0.45^2)

E = 3.83*10^3 N/C

DONE please check the answer. any doubts feel free to ask

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