Q122 / 170 · Manipal (MET) 2017 · Physics
Two identical air core capacitors are connected in series to a voltage source of 15 V . If one of the capacitors is filled with a medium of dielectric constant 4, the new potential across this capacitor is
- A5 V
- B8 V
- Correct answerC10 V
- D12 V
Solution
Let capacitance of each capacitor is $C$. Then equivalent capacitance in series is

$\begin{aligned} \quad & \frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{C}=\frac{2}{C} \\ \Rightarrow & C^{\prime}=\frac{C}{2} \end{aligned}$ Charge $Q=C^{\prime} V=\frac{C}{2} \times 15...(i)$ When filled with dielectric $\begin{gathered} C_1=4 C, C_2=C \\ \frac{1}{C^{\prime}}=\frac{1}{4 C}+\frac{1}{C}=\frac{5}{4 C} \\ \end{gathered}$ $\Rightarrow C^{\prime}=\frac{4 C}{5}$ Since, charge is conserved $Q=C^{\prime} V^{\prime}=\frac{4 C}{5} V^{\prime}...(ii)$ From Eqs. (i) and (ii), we get $\begin{gathered} \frac{C}{2} \times 15=\frac{4 C}{5} V^{\prime} \\ \Rightarrow V^{\prime}=\frac{15 \times 5}{4 \times 2}=9.4 \mathrm{~V} \approx 10 \mathrm{~V} \end{gathered}$

$\begin{aligned} \quad & \frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{C}=\frac{2}{C} \\ \Rightarrow & C^{\prime}=\frac{C}{2} \end{aligned}$ Charge $Q=C^{\prime} V=\frac{C}{2} \times 15...(i)$ When filled with dielectric $\begin{gathered} C_1=4 C, C_2=C \\ \frac{1}{C^{\prime}}=\frac{1}{4 C}+\frac{1}{C}=\frac{5}{4 C} \\ \end{gathered}$ $\Rightarrow C^{\prime}=\frac{4 C}{5}$ Since, charge is conserved $Q=C^{\prime} V^{\prime}=\frac{4 C}{5} V^{\prime}...(ii)$ From Eqs. (i) and (ii), we get $\begin{gathered} \frac{C}{2} \times 15=\frac{4 C}{5} V^{\prime} \\ \Rightarrow V^{\prime}=\frac{15 \times 5}{4 \times 2}=9.4 \mathrm{~V} \approx 10 \mathrm{~V} \end{gathered}$
Hence, the correct answer is option: 'C'.