Applying Gauss Law?

In the diagram. we see that the radius of the sphere is twice the radius of the cavity. Actually, we can prove that the field is uniform regardless of how much bigger the sphere is compared to the cavity. Let $b > a$ denote the radius of the sphere. Without loss of generality, we may assume $\rho > 0$ . Let $\vec E_1$ be the electric field generated by a solid sphere of radius $a$ with uniform charge density $-\rho$ that fills the cavity, and let $\vec E_2$ be the electric field generated by a solid sphere of radius $b$ with uniform charge density $\rho$ and no cavity. We can compute $\vec E$ as the superposition of $\vec E_1$ and $\vec E_2$ .

Suppose $\vec E_2$ is centered at the origin and $\vec E_1$ is centered at $\vec d$ , with $|\vec d| = d \leq a - b$ . Any point inside the cavity can be written as $\vec d + \vec r$ for some $\vec r$ with $|\vec r| = r < a$ . Consider the value of $\vec E_1(\vec d + \vec r)$ . By Gauss's law, $\Phi = \frac {Q} {\epsilon_0},$ where $\Phi$ is the electric flux of $\vec E_1$ through the spherical surface $|\vec v - \vec d| = r$ , $Q$ is the negative charge contained within this surface, and $\epsilon_0$ is the electric constant. Clearly $Q = -4\pi r^3 \rho / 3$ . Since $|\vec E_1|$ is the same at all points on the spherical surface, for some constant field strength $E_r$ we have $\Phi = E_r \cdot 4\pi r^2 = -\frac{ 4 \pi r^3 \rho }{ 3 \epsilon_0 } \implies E_r = -\frac {r\rho}{3 \epsilon_0}.$ By symmetry, $\vec E_1$ is radial about $\vec d$ , so $\vec E_1(\vec d + \vec r) = E_r \cdot \frac {\vec r}{r} = -\frac {\vec r \rho}{3\epsilon_0}.$ Now consider the value of $\vec E_2(\vec d + \vec r)$ . Applying the same reasoning we used to find $\vec E_1(\vec d + \vec r)$ , we find $\vec E_2(\vec d + \vec r) = \frac {(\vec d + \vec r) \rho} {3 \epsilon_0}.$ By superposition, $\vec E(\vec d + \vec r) = \vec E_1(\vec d + \vec r) + \vec E_2(\vec d + \vec r) = -\frac {\vec r \rho}{3\epsilon_0} + \frac {(\vec d + \vec r) \rho}{3\epsilon_0} = \frac {\vec d \rho}{3\epsilon_0}.$ Since the value of $\vec E$ depends only on $\vec d$ , which is fixed, we have shown that $\vec E$ is uniform inside the cavity.