Derive the Analog of Gauss' law

[KT KIM]

I am stuck with this (d) all the other (a),(b),(c) were just like
(a) E= −∇φ
(b) ∇ · E= −∇2φ
(c) ∇ × E

I know only very basic about Gauss' law, the problem is not from Electrodynamics course.
I don't know what should I do to solve (d)

 

[DuckAmuck]

Gauss' law is:
$$\nabla \cdot E = \rho / \epsilon_0$$
We also know that by definition, potential and electric field are related by
$$E = - \nabla \cdot \phi$$
So we then have
$$-\nabla^2 \phi = \rho/\epsilon_0$$

You know the form of phi, so you can just integrate over r from 0 to R to get the charge contained in the region. Since there's no angular dependence, 4pi comes out, and it's just a simple integral over r.
$$- 4\pi \epsilon_0 \int \limits_0^R (\nabla^2 \phi) r^2 dr = Q$$

I'll let you work out the rest.

 

[KT KIM]

Thankyou!