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.

Answer 1

Step-by-step explanation:

The surface charge density on the inside surface of the sphere is -150nC/m2

Charge on the inner surface = - 4 π x (.05)² x 150 x 10⁻⁹

= - 4.71 x 10⁻⁹ C

This is the bound charge . This charge will be bound by charge at the centre

equal to the same charge but of opposite nature .

So charge at the centre

Q = + 4.71 x 10⁻⁹ C

B) Imagine a Gaussian surface with radius equal to 4 cm

If E be electric field on the surface

Flux through the surface

4 π x (.04)² x E = Q / ε₀

= 4.71 x10⁻⁹ / 8.85 x 10⁻¹²

= .532 x 10³

E = .532 x 10³ / 4 π x (.04)²

= 26473 N/C

C )Imagine a Gaussian surface with radius equal to 8 cm

If E be electric field on the surface

Flux through the surface

4 π x (.08)² x E = Q / ε₀

= 0 / 8.85 x 10⁻¹² ( Total charge inside will be -Q + Q = 0 )

= 0

D)

Total charge on the outer surface

= 4 π x (.12)² x 150 x 10⁻⁹

= 27.13 x 10⁻⁹ C

Imagine a Gaussian surface with radius equal to 20 cm

If E be electric field on the surface

Flux through the surface

4 π x (.12)² x E = Q / ε₀

= 27.13 x10⁻⁹ / 8.85 x 10⁻¹² ( - Q + Q will cancel each other )

= 3.065 x 10³

E = 3.065 x 10³ / 4 π x (.12)²

= 16946.43 N/C