P20B2015 – Concentric Metal Sphere & Shell

Given the spherical symmetry of the charge distribution, we expect the electric field to be radial. We apply Gauss’s Law to find the electric field at point $C$ and choose a sphere of radius $r>3R$ centered on the solid sphere as our Gaussian surface.

The electric flux through the Gaussian surface is equal to

\oiint_{sphere}{\overrightarrow{E}\cdot d\overrightarrow{A}}=\oiint_{sphere}{E{\mathrm{cos} \left(0\right)\ }dA}=E\cdot 4\pi r^2

The charge enclosed by the Gaussian surface is equal to the total charge of the sphere and the shell i.e.

Q_{enc}=Q_{sphere}+Q_{shell}=-Q+6Q=5Q

By Gauss’s Law, we derive the electric field outside the metallic shell as follows

\begin{aligned}
\oiint_{sphere}{\overrightarrow{E}\cdot d\overrightarrow{A}}=\frac{Q_{enc}}{{\varepsilon }_0}\ \ \ \ &\Rightarrow \ \ \ \ \ E\cdot 4\pi r^2=\frac{5Q}{{\varepsilon }_0} \\
\\
&\Rightarrow \ \ \ \ \ E=\frac{5Q}{4\pi {\varepsilon }_0r^2}
\end{aligned}

At point C, where $r=4R$, the electric field is therefore equal to

\boxed{E_C=\frac{5Q}{64\pi {\varepsilon }_0r^2}}

Since the enclosed charge is positive, the electric field points radially outward at point C.

Answer: b

The electric field at point A is always zero, because point A is located inside conducting material. Whatever the charge of the metal sphere is, we will always have $E_A=0$.

The electric field at point B changes because the charge enclosed by a Gaussian sphere of radius $r=2R$ (so that point B is a point of the Gaussian surface) changes from $-Q$ to $+Q$. This causes the electric field to change direction and point radially outward instead of inward. The magnitude of the electric field, however, does not change.

The electric field at point C does not change because the charge enclosed by a Gaussian sphere of radius $r=4R$ remains the same, equal to $Q_{encl}=Q+4Q=5Q$.

The charge on the outer surface shell remains constant. Initially, the charge of $-Q$ of the sphere induces a charge $+Q$ on the inner surface of the shell, leaving $5Q$ of charge on the outer surface. After the charge is changed to $-Q$, the sphere induces a charge $-Q$ on the inner surface of the shell, causing the outer surface of the shell to have a charge $4Q-\left(-Q\right)=5Q$.