A capacitor stores 60,mu C charge w | vedclass

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Question

$\mathrm{q}_{0}=\mathrm{C}_{0} \mathrm{V}$

$q=C V$ or $q=K C_{0} V$

$	herefore \mathrm{K}=\frac{\mathrm{q}}{\mathrm{q}_{0}}=\frac{120+60}{60}=3

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$\mathrm{q}_{0}=\mathrm{C}_{0} \mathrm{V}$

$q=C V$ or $q=K C_{0} V$

$	herefore \mathrm{K}=\frac{\mathrm{q}}{\mathrm{q}_{0}}=\frac{120+60}{60}=3$

Because, the capacitor plates were initially charged to $60\, \mu C$ and on inserting the dielectric an additional charge of $120\, \mu \mathrm{C}$ also flows through the battery.

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

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