A parallel plate capacitor after charging is kept connected to a battery and the plates are pulled apart with the help of insulating handles. Now which of the following quantities will decrease ?

Question

A parallel plate capacitor after charging is kept connected to a battery and the plates are pulled apart with the help of insulating handles. Now which of the following quantities will decrease ? (A) Charge (B) Capacitance (C) Energy stored (D) All of these

Tardigrade Admin · Accepted Answer

V remains constant C=(A ε0/d) ⇒ d increases C decreases q=C V q decreases U=(1/2) C V2 U decreases

Q. A parallel plate capacitor after charging is kept connected to a battery and the plates are pulled apart with the help of insulating handles. Now which of the following quantities will decrease ?

Charge

Capacitance

Energy stored

All of these

$V$ remains constant
$C = \frac{A ε _{0}}{d}$
$\Rightarrow d$ increases
$C$ decreases
$q = C V$
$q$ decreases
$U = \frac{1}{2} C V^{2}$
$U$ decreases

Q. A parallel plate capacitor after charging is kept connected to a battery and the plates are pulled apart with the help of insulating handles. Now which of the following quantities will decrease ?

Charge

Capacitance

Energy stored

All of these

V remains constant C=dAε0​​ ⇒d increases C decreases q=CV q decreases U=21​CV2 U decreases

$V$ remains constant
$C = \frac{A ε _{0}}{d}$
$\Rightarrow d$ increases
$C$ decreases
$q = C V$
$q$ decreases
$U = \frac{1}{2} C V^{2}$
$U$ decreases