Question

A solid conducting sphere of radius 1.00 cm has a uniform charge of -5.00 µC. It is surrounded by a concentric spherical shell, with a radius of 2.50 cm, that has a uniform charge of +6.00 µC. Determine the magnitude and direction of the electric field (a) at the center of the sphere (r = 0), (b) at r = 0.500 cm, (c) at r = 2.00 cm, and (d) at r = 3.00 cm.

Answer 1

a)

R = radius of the solid conducting sphere = 1 cm = 0.01 m

Q = charge on the solid conducting sphere = - 5 x 10-6 C

Rs = Radius of concentric spherical shell = 2.50 cm

Q' = charge on concentric spherical shell = 6 x 10-6 C

consider a Gaussian spherical surface of radius "r"

At r = 0

Qenc = Charge enclosed within the Gaussian spherical surface = 0

Using Gauss's law

E (4r2) = Qenc /

E (4r2) = 0/

E = 0

b)

R = radius of the solid conducting sphere = 1 cm = 0.01 m

Q = charge on the solid conducting sphere = - 5 x 10-6 C

Rs = Radius of concentric spherical shell = 2.50 cm

Q' = charge on concentric spherical shell = 6 x 10-6 C

consider a Gaussian spherical surface of radius "r"

At r = 0.5 cm = 0.005 m

Qenc = Charge enclosed within the Gaussian spherical surface = 0

Using Gauss's law

E (4r2) = Qenc /

E (4r2) = 0/​​​​​​​

E = 0

c)

R = radius of the solid conducting sphere = 1 cm = 0.01 m

Q = charge on the solid conducting sphere = - 5 x 10-6 C

Rs = Radius of concentric spherical shell = 2.50 cm

Q' = charge on concentric spherical shell = 6 x 10-6 C

consider a Gaussian spherical surface of radius "r"

At r = 1 cm = 0.01 m

Qenc = Charge enclosed within the Gaussian spherical surface = Q = charge on the solid conducting sphere = - 5 x 10-6 C

Using Gauss's law

E (4r2) = Qenc /

E (4(3.14)(0.01)2) = (5 x 10-6 )/(8.85 x 10-12)​​​​​​​

E = 4.5 x 108 N/C

d)

R = radius of the solid conducting sphere = 1 cm = 0.01 m

Q = charge on the solid conducting sphere = - 5 x 10-6 C

Rs = Radius of concentric spherical shell = 2.50 cm

Q' = charge on concentric spherical shell = 6 x 10-6 C

consider a Gaussian spherical surface of radius "r"

At r = 3 cm = 0.03 m

Qenc = Charge enclosed within the Gaussian spherical surface = - 5 x 10-6 +  6 x 10-6 = 1 x 10-6

Using Gauss's law

E (4r2) = Qenc /

E (4(3.14)(0.03)2) = (1 x 10-6 )/(8.85 x 10-12)​​​​​​​

E = 1 x 107 N/C