Electric field from a sphere with a cavity

A sphere of radius $a$ is filled with positive charge with uniform density $\rho$ . Then a smaller sphere of radius $a/2$ is carved out, as shown in the figure below, and left empty. What are the direction and magnitude of the electric field at $A$ ? At $B$ ?

Solution

This problem can be solved by using the principle of superposition. For instance, consider a point charge $+q$ at some point $P$ in space. It creates an electric field everywhere. However, if you place a negative charge $-q$ also at $P$ , it exactly cancels the electric field created by the original charge $+q$ . Therefore, it appears that there is no charge anywhere, even though there is in fact a $+q$ and $-q$ sitting atop the other one.

Using this idea, the given charge distribution can be realized as the sum of a cavity-free sphere of radius $a$ and charge density $\rho$ , with another sphere of radius $a/2$ with charge $-\rho$ (note the negative sign).

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The electric field can now be determined by a simple application of Gauss’s law, since sphere I and sphere II are symmetric objects. Then, the field due to each solid sphere is

(1) $\begin{equation*} \begin{aligned} {\bf E}^I_A &= 0 ,\\ {\bf E}^{II}_A &= \frac{\frac{4 \pi}{3} \left( \frac{a}{2} \right)^3 (-\rho)} {4 \pi \varepsilon_0 \left( \frac{a}{2} \right)^2} {\bf \hat{z}} = -\frac{\rho a}{6 \varepsilon_0} {\bf \hat{z}} , \end{aligned} \end{equation*}$

and the net electric field at $A$ is

(2) $\begin{align*} &{\bf E}_A = {\bf E}^I_A + {\bf E}^{II}_A \nonumber \\ \implies& \boxed{{\bf E}_A = -\frac{\rho a}{6 \varepsilon_0} {\bf \hat{z}}} . \end{align*}$

Similarly, for the point $B$

(3) $\begin{equation*} \begin{aligned} {\bf E}^{I}_B &= \frac{\frac{4 \pi}{3} a^3 \rho} {4 \pi \varepsilon_0 a^2} {\bf \hat{z}} = \frac{\rho a}{3 \varepsilon_0} {\bf \hat{z}} , \\ {\bf E}^{II}_A &= \frac{\frac{4 \pi}{3} \left( \frac{a}{2} \right)^3 (-\rho)} {4 \pi \varepsilon_0 a^2} {\bf \hat{z}} = -\frac{\rho a}{24 \varepsilon_0} {\bf \hat{z}}, \end{aligned} \end{equation*}$

which means the point $B$ experiences a field

(4) $\begin{align*} &{\bf E}_B = {\bf E}^I_B + {\bf E}^{II}_B \nonumber \\ \implies& \boxed{{\bf E}_B = \frac{7 \rho a}{24 \varepsilon_0} {\bf \hat{z}}} . \end{align*}$

Bonus problem: If a long cylindrical wire of radius $a$ had a portion of radius $a/2$ removed, such that its cross section looks like the figure above, what would be the magnetic field at points $A$ and $B$ ? Assume that the wire is carrying current of density $j$ out of the plane of the paper.

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