Question

An air capacitor of capacity $ C=10\, \mu F $ is connected to a constant voltage battery of $12\,V$. 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.
$\therefore Q=C V$
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} $
$=(600-120) \mu C=480\, \mu C$