Solving Spherical Conductor: Net Charge and Electric Field

[tica86]

A solid spherical conductor of radius 1.50 m has a small spherical cavity of radius 0.50 m with the same center. The net charge of the conductor is zero. An object with a net charge of 4.00 nC is placed inside the cavity in such a way that it is isolated from the conductor. I know the answers I just want to be able to understand the concept.

1)What is the charge on the inner cavity surface?

I know it is -4.00nC but why??


2)What is the electric field magnitude 3.00m from the center of the sphere?
E(4pi*r^2)=Q/E0
THE ANSWER is 3.996 V/m again I don't understand..

 

[Falcons]

For 1) It's induction. The 4 nC charge pulls a minus 4 nC charge towards itself from the conductor. 2) Is Gauss' law, and there really isn't another way to explain it.

 

[SammyS]

You can also explain (1) using Gauss's Law. Consider a sphere of radius 0.60 m (It could anything greater than 0.50 and less than 1.50 m.) with center at the same location as the other two spheres. The electric field is zero, everywhere on the surface of this sphere.

How much flux passes through this closed surface?

What is the net charge enclosed by the surface of this sphere?

 

[gomunkul51]

perfect symmetry -> use Gauss' Law :)

(Gaussian surface inside the conduction outer shell)