A uniformly charged ball of radius a and charge? Q is at the center of a hollow metal shell with inner radius b and outer radius c. The hollow sphere has net charge + 2Q.

Part A

Determine the electric field strength in the region r? a. Give your answer as a multiple of Q/?0.

Express your answer in terms of some or all of the variables a, b, c, r, and the constant?.

Park B

Determine the electric field strength in the region a

Express your answer in terms of some or all of the variables a, b, c, r, and the constant?.

Part C

Determine the electric field strength in the region b

Part D

Determine the electric field strength in the region c

Express your answer in terms of some or all of the variables a, b, c, r, and the constant?.

Question

Physics

Baker

30 March, 05:58

A uniformly charged ball of radius a and charge? Q is at the center of a hollow metal shell with inner radius b and outer radius c. The hollow sphere has net charge + 2Q.

Part A

Determine the electric field strength in the region r? a. Give your answer as a multiple of Q/?0.

Express your answer in terms of some or all of the variables a, b, c, r, and the constant?.

Park B

Determine the electric field strength in the region a

Express your answer in terms of some or all of the variables a, b, c, r, and the constant?.

Part C

Determine the electric field strength in the region b

Part D

Determine the electric field strength in the region c

Express your answer in terms of some or all of the variables a, b, c, r, and the constant?.

+5

Answers (1)

Know the Answer?

Milo Wong · Answer 1

A) E = 1/4π ε₀ r/3a³, B) E = 1/4πε₀ Q / r²

, C) E = 0

, D) E = 1/4πε₀ 3Q / r²

Explanation:

For this exercise let's use Gauss's law

Ф = ∫ E. dA = / ε₀

Since the two objects are spherical we will use a sphere as our Gaussian surface

E A = q_{int} / ε₀

E 4π r² = q_{int} / ε₀

Part A

The radius is r
In this case the chage inside of the Gaussian surface can be found using the concept of density

ρ = q / V

q_{int} = ρ 4/3 π r³

We replace

E 4π r² = rho 4/3 π r³ 1/ε₀

E = rho r/3ε₀

E = 1/4π ε₀ r/3a³ r
Part B

In this case they ask for the field between the sphere and the shell

a
In this case the charge inside is the charge of the sphere, we replace

E (4π r²) = Q / ε₀

E = 1 / 4πε₀ Q / r²

Part C

The electric filed in the region inside the spherical shell

b
An electric conductor in equilibrium all the charge is on its surface so the charge inside the conductor is zero

q_{int} = 0

E = 0

Part D

In the region outside ofthe hollow shell

c
The charge inside is

q_{int} = Q + 2Q = 3 Q

The electric field is

E = 1 / 4πε₀ 3Q / r²