Question

Two parallel plate capacitors of capacitance $4\mu \text{F}$ and $8\mu \text{F}$ are connected in parallel and charged to a voltage of $10\text{V}$ by a battery. The battery is then disconnected and the space between the plates of the $4\mu \text{F}$ capacitor is completely filled with a material of dielectric constant $3$. The potential difference across the capacitors now becomes

• A. $5\text{V}$
• B. $6\text{V}$
• C. $8\text{V}$
• D. $12\text{V}$

Answer 1

The correct option is B $6\text{V}$

Given:
$C_1=4\mu \text{F};C_2=8\mu \text{F};V=10\text{V}$

Equivalent capacitance of parallel combination will be

$C_{eq}=C_1+C_2=4+8=12\mu \text{F}$

So, net charges on the two capacitors will be

$Q=C_{eq}V=12\times 10=120\mu \text{C}$

After introduction of dielectric in $4\mu \text{F}$ capacitor, new capacitance will be

$C_1^{\prime }=K{C}_1=3\times 4\mu \text{F}=12\mu \text{F}$

So,

$C_{eq}^{\prime }=C_1^{\prime }+C_2=12+8=20\mu \text{F}$

Let $V^{\prime }$ be the voltage across capacitor after dielectric insertion.

$Q^{\prime }=C_{eq}^{\prime }{V}^{\prime }=20V^{\prime }\mu \text{C}$

In this case, the charge before and after the introduction of dielectric will remain the same.

$\therefore 20V^{\prime }=120$

$\Rightarrow V^{\prime }=6\text{V}$

Hence, option $\left(b\right)$ is correct.