Charge Distribution Inside a Conductive Shell: Solving the Homework Question

[Dell]

a ball with a radius R and a charge of +Q is placed inside a conductive shell with an inner radius of 3R and outer radius of 4R, the shell has a charge of -4Q, after the ball is placed inside what will the charge on each radius be??

i thought (intuition) that if i place a ball with a charge of Q then more electrons will come to the inner radius making the 3R have a charge of -3Q and the 4R a charge of -Q,... (since at the beginning there was a total of -4Q)

by using gauss surface , and taking a spherical surface with a radius of just more than 3R, then EA=0=Q+(xQ), giving me the result that x=-1, therefore the inner radius will have a charge of -Q and the outer will have a charge of -3Q

 

[LowlyPion]

3R have a charge of -3Q and the 4R a charge of -Q

I think you have reversed 3R and 4R, but otherwise it looks like you have grasped the fundamentals.

 

[Hao]

Intuitively, we know that the stable electric field inside the conductor must be zero, otherwise the non-zero electric field keep moving charge until the electric field inside is indeed zero.

Since we know that electric field around a charge distribution is proportional to the charge enclosed, and we know that the electric field inside the conductor is zero, we can guess that the net charge enclosed on the inside sums up to zero.

As the charge on the ball is fixed, this means that the charge on the inner surface is equal in magnitude but opposite in sign.

Then by conservation of charge, the charge on the outer surface can be found.

This is essentially a restatement of Gauss' Law, but I think this is how your line of intuitive thought should go.