A capacitor of capacitance \( C \), charged to a potential \( V \), is enclosed in a closed surface. The flux of the electric field through the closed surface enclosing the capacitor is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → zero
By Gauss’s law, the flux through a closed surface depends only on the net charge enclosed: \( \phi = q_{\mathrm{enc}}/\varepsilon_{0} \).
A charged capacitor carries charge \( +Q = +CV \) on one plate and \( -Q = -CV \) on the other, so the whole capacitor has zero net charge.
\( q_{\mathrm{enc}} = +CV + (-CV) = 0 \)
\( \phi = 0/\varepsilon_{0} = 0 \)
Why the other options are incorrect
Option (2) \( 0.5\,C\,V\,\varepsilon_{0}^{-1} \) — Incorrect. This is half the flux due to the charge of one plate; the other plate carries an equal and opposite charge.
Option (3) \( C\,V\,\varepsilon_{0}^{-1} \) — Incorrect. This is \( Q/\varepsilon_{0} \), the flux due to the positive plate alone; the negative plate cancels it.
Option (4) \( 2\,C\,V\,\varepsilon_{0}^{-1} \) — Incorrect. This adds the magnitudes of the two plate charges, but charges add with their signs.