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2. Electric Potential and Capacitance
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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 :
A
$1$
B
$2$
C
$3$
D
none
Solution
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 $Q^{\prime}=C^{\prime} V=k C V$
$\frac{Q^{\prime}}{Q}=\frac{k C V}{C V}$
$k=\frac{Q^{\prime}}{Q}=\frac{180}{60}=3$
Standard 12
Physics
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