Question

5) A solid conducting sphere of radius 2.00 cm has a charge of 8.00uC. A conducting spherical shell of inner radius 4.00 cm and outer radius 5.00 cm is concentric with the solid sphere and has a charge of -4.00uC. Find the electric field at

a) r= 1.00 cm

b) r= 3.00 cm

c) r= 4.50 cm

d) r= 7.00 cm

from the center of this charge configuration.

Answer 1

1.Sol::

Given:

radius of inner solid sphere r1 = 0.02 m
charge on inner solid sphere q1 = 8 x 10^-6 C
inner radius of shell r2 = 0.04 m
outer radius of shell r3 = 0.05 m
charge on shell q2 = -4.00 x 10^-6 C

We have Surface area of a sphere as
A = 4πr^2
Volume of a sphere
V = (4/3)*π*r^3

Let charge density of the sphere be 'P'.
Let the radius of the solid sphere be 'a'
And 'Q' is the charge of object,
'V' is the volume of the object.

a = r1 = 0.02 m

P = Q / V
= Q / (4/3)*π*a^3
= (8 x 10^-6 ) / (4/3)*(3.14)*(0.02)^3
= 0.238 C/m^3

a)
As per given data there is no charge inside of a solid conductor
So

|E| = 0

b)

r = 0.03 m , radius of the Gaussian surface

|E| = Q / 4*π*ε0*r^2

|E| = (8x10^-6)/4*(π)*(8.85x10^-12)*(0.03)^2

= 79.94 x 10^6 N/C

c)
When r = 4.50 cm

As There is no charge inside the solid of a conductor, so there is no electric field.

|E| = 0

d)
He we have 2 different charges coming from 2 concentric spheres.
To get the electric field, we take the sum of the two electric fields from the radius indicated in the problem.

r = 0.07 m , radius of the Gaussian surface
Q_inner = q1 = 8 x 10^-6 C
Q_outer = q2 = -4.00 x 10^-6 C

|E_inner| = Q_inner / 4*π*ε0*r^2
|E_outer| = Q_outer / 4*π*ε0*r^2
|E_net| = |E_inner| + |E_outer|

= [Q_inner/4*π*ε0*r^2] + [Q_outer/4*π*ε0*r^2]
= [Q_inner+Q_outer] / [4*π*ε0*r^2]
= [(8-4)x10^-6]/[4π*(8.85x10^-12)*(0.07)^2]

= 7.341 x 10^6 N/C

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