A parallel plate capacitor is connected to a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The potential difference $V$ across the plates remains constant because the capacitor is connected to a battery.The electric field $E$ between the

Vedclass · Accepted Answer

decreases and the charge on the plates also decreases.

Vedclass · Answer

(A) The potential difference $V$ across the plates remains constant because the capacitor is connected to a battery.The electric field $E$ between the plates is given by $E = V/d$,where $d$ is the distance between the plates.As the plates are moved further apart,$d$ increases,so $E = V/d$ decreases.The capacitance $C$ of a parallel plate capacitor is given by $C = \epsilon_0 A / d$.As $d$ increases,the capacitance $C$ decreases.The charge $Q$ on the plates is given by $Q = CV$.Since $C$ decreases and $V$ is constant,the charge $Q$ on the plates also decreases.Therefore,both the electric field intensity and the charge on the plates decrease.

$A$ parallel plate capacitor is connected to a battery,which maintains a constant potential difference. If the plates of the capacitor are moved further apart,the electric field intensity...

Similar Questions

The distance between the plates of a parallel plate capacitor is $4 \ mm$ and the potential difference is $60 \ V$. If the distance between the plates is increased to $12 \ mm$,then:

If the distance between the plates of a parallel plate capacitor of capacity $10 \mu F$ is doubled,then the new capacity will be: (in $\mu F$)

Charges of $2Q$ and $-Q$ are placed on two plates of a parallel plate capacitor. If the capacitance of the capacitor is $C$,the potential difference between the plates is:

The plates of a parallel plate capacitor of capacity $C_1$ are moved closer together until they are half their original separation. The new capacitance $C_2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module