A capacitor stores 60 mu C charge when connec | vedclass

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Question

The capacitance before filled with dielectric is $C=\frac{A \epsilon_{0}}{d}$

The capacitance after filled with dielectric is $C^{\prime}=\frac{A k

Vedclass · Accepted Answer

$3$

Vedclass · Answer

The capacitance before filled with dielectric is $C=\frac{A \epsilon_{0}}{d}$

The capacitance after filled with dielectric is $C^{\prime}=\frac{A k \epsilon_{0}}{d}=k C$

Initial charge, $Q=60 \mu C$ and total charge after filled with dielectric $Q^{\prime}=120+60=$ $180 \mu C$

Here, $Q=C V$ and $Q^{\prime}=C^{\prime} V=k C V$

$\frac{Q^{\prime}}{Q}=\frac{k C V}{C V}$

$k=\frac{Q^{\prime}}{Q}=\frac{180}{60}=3$

A capacitor stores $60\ \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric , a charge of $120\ \mu C$ flows through the battery. The dielectric constant of the material inserted is :

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