Question

A parallel plate capacitor of capacitance $100\mu \text{F}$ is connected to a power supply of $200\text{V}$. A dielectric slab of dielectric constant $5$ is now inserted into the gap between the plates. The workdone by power supply will be:

• A. $-12\text{J}$
• B. $16\text{J}$
• C. $-24\text{J}$
• D. $8\text{J}$

Answer 1

The correct option is B $16\text{J}$

Before insertion of dielectric slab,

$C_i=100\mu \text{F};V=200\text{V}$

$Q_i=100\mu \text{F}\times 200\text{V}=20000\mu \text{C}$

$\therefore Q_i=20\text{mC}$

After the dielectric introduced:

$C_f=K{C}_i=5\times 100=500\mu \text{F}$

$V^{\prime }=V=200\text{Volt}$

$(\because$ battery remains connected $)$

So,

$Q_f=500\mu \text{F}\times 200\text{V}=100000\mu \text{C}$

$\therefore Q_f=100\text{mC}$

Therefore, the charge supplied by battery is,

$\Delta Q=Q_f-Q_i=100-20=80\text{mC}$

So, workdone by battery will be

$W_b=\Delta Q\times V$

$\Rightarrow W_b=(80\text{mC}\times 200\text{V})=16000\text{mJ}$

$\therefore W_b=16\text{J}$

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

Why this question? Tip: In this problem +ve charge is released from high potential (+ve) terminal of battery, hence $W_b>0$.