The force between the plates of a parallel plate capacitor of capacitance C, distance between plates d and potential difference V is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{CV^2}{2d}$
Electric field (E) between the plates -
$E=\frac{V}{d}$
∴ Force, $F =qE=q\frac{V}{d}$
$=(CV)(\frac{V}{d})$ $[Q=CV]$
$=\frac{CV^2}{d}$
∴ force between the plates = $\frac{CV^2}{2d}$