Question

A parallel plate capacitor with plates separated by air acquires 1 $\mu$C of charge when connected to a battery of 500V. The plates still connected to the battery are then immersed in benzene ($k=2.25$). Then a charge flows from the battery is :

• A. $1.25\mu$ C
• B. $2.28\mu$ C
• C. $1/4\mu$ C
• D. $4.56\mu$ C

Answer 1

The correct option is A $1.25\mu$ C

Initial charge $Q_1=1\mu c$
$V_1=500V.$
As the plates are still connected voltage is constant.
$\therefore$ when plates are dipped in benzene capacitance increases by $k=\left(\frac{k{\epsilon }_{0}A}{d}\right)$
$\therefore$ As voltage is constant Q also increases by k.
$\therefore Q_2=k{Q}_1$
charge flown from battery$=k{Q}_1-Q_1$
$1\mu c(2\cdot 25-1)$
$1\times {10}^{-6}(1\cdot 25)$
$=1\cdot 25\mu c.$