Question

A parallel plate air capacitor has capacity 'C' distance of separation between plates is 'd' and potential difference 'V' is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is:(a) $\frac{CV^{2}}{2d}$(b) $\frac{CV^{2}}{d}$(c) $\frac{C^{2}V^{2}}{2d^{2}}$(d) $\frac{C^{2}V^{2}}{d^{2}}$

Answer 1

Answer:

B) CV²/2d

Explanation:

Capacity of the capacitor = C

Distance of separation = d

Force of attraction between the plates of the parallel plate air capacitor is -

F = Q²/2ε0A

where Q is the charge on the capacitor, ε0 is the permittivity of free space andl is the area of each plate.

But since Q = CV, then the potential energy stored in the plates of capacitor will be -

C = ε0A/d or ε0A = Cd

Therefore, F = C²V²/2cd

= CV²/2d

Thus, the force of attraction between the plates of the parallel plate air capacitor is CV²/2d

Answer 2

Hey Mate!

✓✓ Your Answer ✓✓

################

Good Question

**********************

Option :↓Choosen below↓

_____________________

A parallel plate air capacitor has capacity 'C' distance of separation between plates is 'd' and potential difference 'V' is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is:

(a) $\frac{CV^{2}}{2d}$

.........