Question

Two identical parallel plate capacitors are connected in series and then joined with a battery of $100\text{V}$. A slab of dielectric constant $4$ is inserted between the plates of the second capacitor. The potential difference across the capacitors $\left(1\right)$ and $\left(2\right)$ will be respectively

• A. $50\text{V},50\text{V}$
• B. $80\text{V},20\text{V}$
• C. $20\text{V},80\text{V}$
• D. $75\text{V},25\text{V}$

Answer 1

The correct option is B $80\text{V},20\text{V}$

Let us assume the capacitance of each capacitor with air as dielectric is $C$.

On inserting the dielectric, capacitance of the second capacitor becomes $4C$.

The equivalent capacitance for series combination:

$C_{eq}=\frac{C\times 4C}{C+4C}=\frac{4C}{5}$

Charge on each capacitor:

$Q=\frac{4C}{5}\times 100=80C$

Thus, potential difference across first capacitor:

$V_1=\frac{80C}{C}=80\text{V}$

And, potential difference across second capacitor,

$V_2=\frac{80C}{4C}=20\text{V}$

Hence, option (b) is the correct answer.

Key Concept: Same charge is present on capacitors connected in series.