Negative charge –Q is distributed uniformly over the surface of a thin spherical insulating

shell with radius R.

Part A Calculate the magnitude of the force that the shell exerts on a positive point charge q

located a distance r>R from the center of the shell (outside the shell).

Express your answer in terms of the variables q, Q, r, R, and constants π and ϵ0.

Part B Find the direction of the force that the shell exerts on a positive point charge q located a

distance r>R from the center of the shell (outside the shell).

Part C Calculate the magnitude of the force that the shell exerts on a positive point charge q

located a distance r
Express your answer in terms of the variables q, Q, r, R, and constants π and ϵ0.

Explanation:

The force of attraction or repulsion is given by

F=q1q2/4πϵ0r²

A. It is given that q1= -Q i.e a negative charge

And q2=q and it is at a distance r from Q.

Since it is a positive and negative charge then it is a force of attraction

So applying the equation,

F=q1q2/4πϵ0r²

F=Qq/4πϵ0r²

B. The direction of the force on q due to the Q is force of attraction and it is moving towards each other.

C. It is given that q1= -Q i.e a negative charge

And q2=q and it is at a distance r from Q.

Since it is a positive and negative charge then it is a force of attraction

So applying the equation,

F=q1q2/4πϵ0r²

F=Qq/4πϵ0r²

But if r is less than R i.e it is inside the sphere, then the F is zero inside the sphere

F=0N

a. spokes of a wheel.

as indicated in the question radial contains rad, also found in radius, so the fracture's would span the radius.