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?

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Related problems:
Electric field in a hollow region
Charge at one corner of a cube
Flux from a charged shell


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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