A parallel plate capacitor is charged from a cell then it is disconnected from the cell. Now, the separation between its plates is increased then the force of attraction between the plates:

Question

A parallel plate capacitor is charged from a cell then it is disconnected from the cell. Now, the separation between its plates is increased then the force of attraction between the plates: (A) will decrease (B) will increase (C) will increase or decrease depending on charge (D) will remains same

Tardigrade Admin · Accepted Answer

Since the capacitor is isolated, charge on it will not change. Force on its plate, F=(q2/2 ϵ 0 A), will not change.

Q. A parallel plate capacitor is charged from a cell then it is disconnected from the cell. Now, the separation between its plates is increased then the force of attraction between the plates:

will decrease

will increase

will increase or decrease depending on charge

will remains same

Since the capacitor is isolated, charge on it will not change.
Force on its plate, $F = \frac{q ^{2}}{2 ϵ _{0} A},$ will not change.

Q. A parallel plate capacitor is charged from a cell then it is disconnected from the cell. Now, the separation between its plates is increased then the force of attraction between the plates:

will decrease

will increase

will increase or decrease depending on charge

will remains same

Since the capacitor is isolated, charge on it will not change. Force on its plate, F=2ϵ0​Aq2​, will not change.

Since the capacitor is isolated, charge on it will not change.
Force on its plate, $F = \frac{q ^{2}}{2 ϵ _{0} A},$ will not change.