Question

A capacitor of $10\mu F$ capacitance is charged by a $12V$ battery. Now the space between the plates of capacitors is filled with a dielectric of dielectric constant $K=3$ and again it is charged. The magnitude of the charge is :

• A. $120\mu C$
• B. $240\mu C$
• C. $360\mu C$
• D. $480\mu C$

Answer 1

The correct option is C $360\mu C$

$\text{Step 1: Initial charge [Refer Fig. 1]}$

$Q_1=C_1{V}_1=10\mu F\times 12V$

$=120\mu C$

$\text{Step 2: Final capacitance after insertion of dielectric [Refer Fig. 2]}$

$C_2=\frac{k{ϵ}_{0}A}{d}=k{C}_1=3\times 10\mu F$

$C_2=30\mu F$

$\text{Step 3: Final charge [Refer Fig. 3]}$

As the battery remains connected, voltage across capacitor remains same.

i.e., $V_2=V_1=12V$

$Q_2=C_2{V}_2$

$Q_2=30\mu F\times 12V$
$=360\mu C$

Hence, option C is correct.