Question

A battery of 100 V is connected to series combination of two identical parallel-plate condensers. If dielectric of constant 4 is slipped between the plates of second condenser, then the potential difference on the condensers will respectively become :

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

Answer 1

The correct option is A 80 V, 20 V

Let the capacitance of each capacitors before applied dielectric is $C$.

When dielectric is inserted in the second capacitor , the capacitance of first capacitor will remain same i.e $C_1=C$ and the capacitance of second capacitor becomes $C_2=4C$

As they are in series, equivalent capacitance, $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{C.4C}{C+4C}=\frac{4}{5}C$

Equivalent charge on the circuit is $Q_{eq}=C_{eq}V=\frac{4}{5}C\times 100=80C$

Now the potential difference on first condenser is $V_1=\frac{Q_{eq}}{C_1}=\frac{80C}{C}=80V$

Now the potential difference on second condenser is $V_2=\frac{Q_{eq}}{C_2}=\frac{80C}{4C}=20V$