# Calculating the Electric Field Inside a Conductor with Two Cavities

For this problem I am stuck on how to calculate the electric field. I know to apply Gauss' Law: $$\oint \overrightarrow{ E} \cdot \overrightarrow {dA} = \frac{q_{enclosed}}{\epsilon_0}$$ $$E(4\pi r^2) = \frac{Q + -Q}{\epsilon_0}$$ I don't know $$r$$, but knowing that the $$Q's$$ are equal magnitude can this be stated that the electric field inside is zero?

• Is the answer E? Sep 21 '20 at 4:22
• That's what I am trying to verify, but I don't know if my approach is valid.
– E__
Sep 21 '20 at 4:25
• I am writing an answer wait Sep 21 '20 at 4:27

Consider a point $$D$$ on in the meat of the conductor as shown.
We know that field inside the meat of a conductor is zero. So, electric field at $$D$$ is zero for all positions of $$D$$. Hence, the distribution on the outer surface of the sphere must be uniform so that it behaves like a shell. Also, the net charge on the surface is equal to the net charge inside a conductor. Hence, you may replace the conductor by an uncharged sphere to analyze Electric field and Potential outside the conductor.
So, your answer is $$0$$.