A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery, the distance between the plates of the capacitors is decreased using insulating handle. The electric field between the plates

Question

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery, the distance between the plates of the capacitors is decreased using insulating handle. The electric field between the plates (A) Decreases (B) Increases (C) Remains same (D) First increases then decreases

Tardigrade Admin · Accepted Answer

Because battery is disconnected so charge will remain same. So according to E = (σ/ϵo) = (q/ϵoA) electric field is independent of distance.

Q. A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery, the distance between the plates of the capacitors is decreased using insulating handle. The electric field between the plates

Decreases

Increases

Remains same

First increases then decreases

Because battery is disconnected so charge will remain same. So according to E = $\frac{σ}{ϵ _{o}} = \frac{q}{ϵ _{o} A}$ electric field is independent of distance.

Q. A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery, the distance between the plates of the capacitors is decreased using insulating handle. The electric field between the plates

Decreases

Increases

Remains same

First increases then decreases

Because battery is disconnected so charge will remain same. So according to E = ϵo​σ​=ϵo​Aq​ electric field is independent of distance.

Because battery is disconnected so charge will remain same. So according to E = $\frac{σ}{ϵ _{o}} = \frac{q}{ϵ _{o} A}$ electric field is independent of distance.