Question

What does Gauss's law give for a closed surface when the electric field inside the conductor is zero?

Options

• \[\oint\vec{E}\cdot d\vec{A}=q\varepsilon_0\], so zero flux gives \(q=\varepsilon_0\).

• \[\oint\vec{E}\cdot d\vec{A}=\sigma\], so zero flux gives \(\sigma=0\).

• \[\oint\vec{E}\cdot d\vec{A}=\frac{q}{\varepsilon_0}\], so zero flux gives \(q=0\).

• \[\oint\vec{E}\cdot d\vec{A}=\frac{\varepsilon_0}{q}\], so zero flux gives \(q=\varepsilon_0\).

Answer 1

Gauss's law relates electric flux to enclosed charge through \(\oint\vec{E}\cdot d\vec{A}=\frac{q}{\varepsilon_0}\). When the interior field and flux are zero, the enclosed charge is zero.