Question

An air capacitor of capacity $C=10\mu F$ is connected to a constant voltage battery of $12V$. Now, the space between the plates is filled with a Liquid of dielectric constant $5$. The additional Charge that flows now from battery to the Capacitor is:

• A. $24\mu C$
• B. $480\mu C$
• C. $600\mu C$
• D. $120\mu C$

Answer 1

Answer: (B)

The capacitance $(C)$ of a conductor is defined as the ratio of the charge $(Q)$ given to the rise in the potential $(V)$ of the conductor.
$\therefore Q=CV$
When $C_1=10\mu F,$
$V=12$ volt
$\therefore Q_1=(10\times 12)\mu C=120\mu C$
$C_2=K{C}_1=5\times 10=50\mu F$
Also, $Q_2=50\times 12=600\mu C$
Hence, additional charge
$\Delta Q=Q_2-Q_1$
$=(600-120)\mu C=480\mu C$