Question

Two identical parallel-plate capacitors are connected in series and then joined across a battery of 100 V. A slab of dielectric constant K = 3 is inserted between the plates of the first capacitor. Then, the potential difference across the capacitors will respectively be

• A. 25 V, 75 V
• B. 75 V, 25 V
• C. 20 V, 80 V
• D. 50 V, 50 V

Answer 1

The correct option is A 25 V, 75 V

From given,Initially,
$V_1+V_2=100\dots \dots \dots \left(1\right)$ and
$Q_1=Q_2$ as they are in series.

$\Rightarrow KC.V_1=C{V}_2$
$3V_1=V_2\dots \dots \dots \left(2\right)$

So from (1) and (2)
$V_1+3V_1=100$
$4V_1=100$
$V_1=\frac{100}{4}=25V$
$25+V_2=100$
$V_2=75V$