Question

A sphere of radius $$R$$ has a uniform volume charge density $$\rho$$. A spherical cavity of radius $$b$$ whose centre lies at $$\vec {r} = \vec {a}$$ is removed from the sphere.
a. Find the electric field at any point inside the spherical cavity.
b. Find the electric field outside the cavity.

Answer 1

a. The field within the cavity or outside is the superposition of the field due to the original uncut sphere, plus the field due to a sphere of the size of the cavity but with a uniform negative charge density. The effective charge distribution is composed of a uniformly charged sphere of radius $$r$$, charge density $$\rho$$, superposed on it, a charge density $$-\rho$$ filling the cavity. Electric field $$\vec {E}_{1}$$ caused by the charge distribution $$+\rho$$ at a point $$\vec {r}$$; inside the spherical cavity.
$$\vec {E}_{1} = \dfrac {\rho r}{3\epsilon_{0}} \hat {r} = \dfrac {\rho \vec {r}}{3\epsilon_{0}}$$; where $$\hat {r}$$ is a unit vector in radial direction.
Similarly, the electric field $$\vec {E}_{2}$$ formed by the charge density $$-\rho$$ inside the cavity is
$$\vec {E}_{2} = \dfrac {(-\rho)\vec {s}}{3\epsilon_{0}}; \vec {s} = \vec {r} - \vec {a}$$
$$\vec {s}$$ is the radius vector from cavity centre to the point $$P$$.
$$\therefore \vec {E}_{2} = \dfrac {-\rho (\vec {r} - \vec {a})}{3\epsilon_{0}}$$
The resultant electric field inside the cavity is, therefore, given by the superposition of $$\vec {E}_{1}$$ and $$\vec {E}_{2}$$, i.e.
$$\vec {E} = \vec {E}_{1} + \vec {E}_{2} = \dfrac {\rho \vec {r}}{3\epsilon_{0}} + \left [\dfrac {-\rho (\vec {R} - \vec {a})}{3\epsilon_{0}}\right ]$$
$$= + \dfrac {\rho \vec {a}}{3\epsilon_{0}} = constant$$
$$\Rightarrow \vec {E} = \dfrac {\rho \vec {a}}{3\epsilon_{0}}$$.
b. (i) Electric field at points inside the large sphere but outside the cavity,
$$\vec {E} = \dfrac {\rho \vec {r}}{3\epsilon_{0}}$$
and $$\vec {E}_{2} = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {q\vec {s}}{s^{3}} = \dfrac {\left (-\dfrac {4}{2} \pi \rho b^{3}\right )(\vec {R} - \vec {a})}{4\pi \epsilon_{0} |\vec {r} - \vec {a}|^{3}}$$
The resultant electric field is, therefore,
$$\vec {E} = \vec {E}_{1} + \vec {E}_{2} = \dfrac {\rho}{3\epsilon_{0}} \left [\vec {r} - \left (\dfrac {b}{|\vec {r} - \vec {a}|}\right )(\vec {r} - \vec {a})\right ]$$
(ii) Electric field at points outside the large sphere,
$$\vec {E}_{1} = \dfrac {Q_{total}}{4\pi \epsilon_{0}r^{3}} \vec {r} = \dfrac {\left (\dfrac {4}{3} \pi R^{3}\rho \right )}{4\pi \epsilon_{0} r^{3}} \vec {r} = \dfrac {R^{3}\rho}{3\epsilon_{0}r^{3}} \vec {r}$$
$$\vec {E}_{2} = \dfrac {q}{4\pi \epsilon_{0}r^{3}} \vec {r} = \dfrac {\left (-\dfrac {4}{3}\pi b^{3} \rho \right )}{4\pi \epsilon_{0}(|\vec {r} - \vec {a}|)^{3}} (\vec {r} - \vec {a})$$
$$= \dfrac {-\rho b^{3}}{3\epsilon_{0}(|\vec {r} - \vec {a}|)^{3}} (\vec {r} - \vec {a})$$
The resultant electric field,
$$\vec {E} = \vec {E}_{1} + \vec {E}_{2} = \dfrac {\rho}{3\epsilon_{0}} \left [\left (\dfrac {R}{r}\right )^{3} \vec {r} - \left (\dfrac {b}{|\vec {r} - \vec {a}|}\right )^{3} (\vec {r} - \vec {a})\right ]$$
$$E(\vec {r}) = \left\{\begin{matrix}\dfrac {\rho \vec {a}}{3\epsilon_{0}}\text {Electric field inside the cavity}\\ \dfrac {\rho}{3\epsilon_{0}} \left [\vec {r} = \left (\dfrac {b}{|\vec {r} - \vec {a}|}\right )^{3} (\vec {r} - \vec {a})\right ]\\ \text {Electric field outside the cavity but inside the large sphere}\\ \dfrac {\rho}{3\epsilon_{0}} \left [\left (\dfrac {R}{r}\right )^{3} \vec {r} - \left (\dfrac {b}{|\vec {r} - \vec {a}|}\right )^{3} (\vec {r} - \vec {a})\right ]\\ \text {Electric field outside the large sphere}
\end{matrix}\right.$$.