CUET Preparation Today
The force between the plates of a parallel plate capacitor of capacitance C, distance between plates d and potential difference V is |
$\frac{CV^2}{2d}$ $\frac{CV^2}{d}$ $\frac{C^2V^2}{4πε_0d}$ $\frac{1}{2}ε_0\frac{}{}$ |
$\frac{CV^2}{2d}$ |
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 = $\frac{CV^2}{2d}$ |
