How does gravity affect the permitivity of a dielectric?

Intuitively, I would not expect gravity to significantly affect electric permitivity. But, Consider a neutral black hole with a plastic sphere outside the event horizon. The protons-electrons of the plastic are polarized with the heavier protons being pulled preferentially closer. A negative charge would be induced on the surface of the sphere. Now, Consider a positively charged relatively mass-less body (positive ion cloud?) surrounded by a plastic sphere. The electrons on the plastic sphere are pulled towards the central charge resulting in a net positive charge measured on the outer surface of the plastic sphere. Measuring the electric field outside the sphere can tell us its permitivity. \$E = Q/4*pi*r^2 * permitivity.

Now consider a positively charged black hole with a plastic sphere outside the event horizon. The protons-electrons of the plastic field are affected oppositely by gravity and the central charge. The measured field $$E = Q/4*pi*r^2*Permitivity$$ would therefore be less outside the plastic sphere.

So for the field to be different outside the plastic sphere , the implication is that the permitivity is different.

is this intuition correct?

• The equivalence principle prohibits that "protons-electrons of the plastic are polarized with the heavier protons being pulled preferentially closer". Jan 12 '20 at 20:11
• the equivalence principle posits the equivalence of acceleration due to motion and the acceleration due to gravity. I dont see how that is relevant to the effect of gravity on charge Jan 18 '20 at 1:44