If the plates of a parallel plate capacitor c | vedclass

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Question

Given $V^{\prime \prime}=V=$ Constant

($i$)

$C^{\prime}=\frac{\varepsilon_0 A}{d^{\prime}}, C=\frac{\varepsilon_0 A}{d}$

$d^{\prime}>d$

$ C^{\p

Vedclass · Accepted Answer

$A, C$ and $E$ only

Vedclass · Answer

Given $V^{\prime \prime}=V=$ Constant

($i$)

$C^{\prime}=\frac{\varepsilon_0 A}{d^{\prime}}, C=\frac{\varepsilon_0 A}{d}$

$d^{\prime}>d$

$ C^{\prime}>C$

Hence, final capacitance greater than initial capacitance,

($ii$)

$U^{\prime}=\frac{1}{2} C^{\prime} V^2$

$U=\frac{1}{2} C V^2$

$U^{\prime}>U$

Hence final energy is greater than initial energy

($iii$)

$\frac{Q^{\prime}}{V^{\prime}}=C^{\prime} 	ext { and } \frac{Q}{V}=C$

$\frac{Q^{\prime}}{V^{\prime}} 
eq \frac{Q}{V}$

($iv$) Product of charge and voltage

$X^{\prime}=Q^{\prime} V=C^{\prime} V^2$

$X=Q V=C V^2$

$X^{\prime}>X$

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