Gauss' law:

$\overline{)\begin{array}{rcl}\mathbf{\oint}\mathbf{E}\mathbf{\xb7}\mathbf{d}\mathbf{A}& {\mathbf{=}}& \frac{{\mathbf{q}}_{\mathbf{enc}}}{{\mathbf{\epsilon}}_{\mathbf{0}}}\\ \mathbf{E}\mathbf{\xb7}\mathbf{A}& {\mathbf{=}}& \frac{\mathbf{q}}{{\mathbf{\epsilon}}_{\mathbf{0}}}\end{array}}$

The cavity is inside the sphere, the sphere is charged, and the electric field is to be found inside the cavity.

A Gaussian surface completely inside the cavity encloses no charge.

$\begin{array}{rcl}\mathbf{\oint}\mathbf{E}\mathbf{\xb7}\mathbf{dA}& \mathbf{=}& \frac{\mathbf{0}}{{\mathbf{\epsilon}}_{\mathbf{0}}}\\ \mathbf{E}\mathbf{\xb7}\mathbf{A}& \mathbf{=}& \mathbf{0}\end{array}$

Charge in a gaussian surface,

Area of a sphere, A:

$\overline{){\mathbf{A}}{\mathbf{=}}{\mathbf{4}}{\mathbf{\pi}}{{\mathbf{r}}}^{{\mathbf{2}}}}$

An insulating sphere of radius *a*, centered at the origin, has a uniform volume charge density *ρ*.

A spherical cavity is excised from the inside of the sphere. The cavity has radius *a*/4 and is centered at position **h**, where |**h**| < 3*a*/4 , so that the entire cavity is contained within the larger sphere. Find the electric field *inside the cavity*.

Express your answer as a vector in terms of any or all of *ρ*, *ε*_{0}, **r**, and **h**.

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