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Suppose I have a neutral spherical conductor with a cavity inside. Suppose there's a $+q$ point charge inside the cavity.

I know that the electric field $\vec{E}$ is zero within the conductor, also know that there will be a $-q$ net charge in the inner surface of conductor with the cavity.

But I still don't understand if there can be an non-zero electric field inside the cavity due to $+q$.

Gauss law to a gaussian surface inside the cavity says it does:

$$ \oint \vec{E} \cdot d\vec{A} = \frac{q}{\epsilon_0} \qquad \longrightarrow \qquad E \, A = \frac{1}{4\pi\epsilon_0} \, \frac{+q}{r^2}, $$

where $r$ is the distance from $+q$ to a point inside the cavity.

Is my reasoning right? I mean, the electric field is always zero inside the conductor, but there can be a non-zero electric field inside the cavity?

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1 Answer 1

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Let the inner and outer radii of a spherical conductor with a cavity be $R_1$ and $R_2$ respectively. The electric scalar potential distribution is given by the following equation. \begin{equation} \frac{4\pi\epsilon_0}{q}\phi(r) = \begin{cases} \frac{1}{r}+\left(-\frac{1}{R_1}+\frac{1}{R_2}\right)\;\;\;\text{(for }r<R_1 \text{)} \\ \frac{1}{R_2} \;\;\;\text{(for } R_1\leq r<R_2 \text{)} \\ \frac{1}{r} \;\;\;\text{(for } R_2\leq r \text{)}. \end{cases} \end{equation} Note that $\phi(r)$ is continuos at $r=R_1$ and $r=R_2$. Electric field for this case is given as \begin{equation} \vec{E}=-\nabla\phi= \begin{cases} \frac{q}{4\pi\epsilon_0}\frac{1}{r^2}\vec{e}_r\;\;\;\text{(for }r<R_1 \text{)} \\ 0\vec{e}_r \;\;\;\text{(for } R_1\leq r<R_2 \text{)} \\ \frac{q}{4\pi\epsilon_0}\frac{1}{r^2}\vec{e}_r \;\;\;\text{(for } R_2\leq r \text{)}, \end{cases} \end{equation} where $\vec{e}_r$ is the unit vector. As seen in this solution, what you wrote "the electric field is always zero inside the conductor, but there can be a non-zero electric field inside the cavity" is correct.

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