Holy field

A standard "trick" in dealing with a cavity in a uniform (constant density) charge distribution is to treat it as a volume with the same charge density of opposite sign . We thus impose a sphere of radius $\ \frac{R}{4} \$ with charge density $\ -\rho \$ , embedded within a space with charge density $+\rho \$ . (It is not terribly important that this "space" is inside a sphere, since its surface is relatively far away.) This "imagined" negative charge distribution "cancels" the ambient positive charge, effectively producing a region of zero net charge.

Our little sphere is centered at $\ r \ = \frac{R}{4}$ , relative to the center of the large sphere. So the point P is at a distance of $\ \frac{R}{8} \$ from the centers of both spheres. Using "Gauss' Law" (the Divergence Theorem, as applied to electostatic charge distributions), since the densities are constant and the fields of electric charges are "radial" (they point away or toward the charge in every direction, with a specific field strength at any fixed distance from the charge), we may write, describing the field at P ,

$\iint_S \ \vec{E} \ \cdot \ d\vec{A} \ = \ \iiint_V \ \nabla \cdot \vec{E} \ dV \ \ \Rightarrow \ \ \vec{E} \cdot \ 4 \pi \ \left( \frac{R}{8} \right)^2 \ = \ \frac{\rho}{\epsilon_0} \cdot \frac{4 \pi}{3} \left( \frac{R}{8} \right)^3 ,$

the strength of the field $\ \vec{E} \$ over the surface area of the sphere of radius $\ \frac{R}{8} \$ , which extends from either center to point P , being equal to the total charge within that sphere divided by $\ \epsilon_0 \$ . This reduces to

$\vec{E} \ = \ \frac{\rho}{3 \epsilon_0} \left( \frac{R}{8} \right) \ = \ \frac{\rho \ R}{24 \ \epsilon_0}$

for the field due to each charge distribution. For the positive charge distribution of the large sphere, the field points radially outward from its center, and thus toward P . For the effective negative charge distribution, its field points radially inward toward the center of the cavity, and so also toward P . The sum of these fields is then the total field at P ,

$\vec{E_P} \ = \ \frac{\rho \ R}{12 \ \epsilon_0} \$ directed outward from the center of the large sphere.