Question

Two identical parallel plate capacitors are connected in series to a battery of $100\text{V}$. A dielectric slab of dielectric constant $4$ is inserted between the plates of the second capacitor to fill the space between its plates, completely. The potential difference across the capacitors will now 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 $C$ be the initial capacitance of both capacitor.

And we know that when dielectric is inserted between the plates, the capacitance increases by $K$ times. So, the capacitance of second capacitor will be

$\therefore C^{\prime }=KC=4C$


So, their equivalent capacitance will be,

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

In series combination, charge on each capacitor remains the same, so the charge on each capacitor,

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

Now, Voltage across $C$ is,

$V_C=\frac{Q}{C}=\frac{80C}{C}=80\text{V}$

And similarly, voltage across $4C$ is

$V_{4C}=\frac{Q}{C^{\prime }}=\frac{80C}{4C}=20\text{V}$

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

Why this Question? Concept:When dielectric is inserted between the plates of the capacitor, its capacitance increases to initial capacitance times dielectric constant.