Q145 / 160 · TS EAMCET 2024 (09 May Shift 2) · Physics
If half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant 4 , the capacitance is $C_1$. If one third of the space between the plates of the capacitor is filled with the medium of dielectric constant 4 , the capacitance is $C_2$. If in both cases, the dielectric is placed parallel to the plate of the capacitor, then $C_1: C_2=$
- A$2: 3$
- B$4: 3$
- Correct answerC$6: 5$
- D$7: 5$
Solution
Capacitance of capacitor when dielectric of thickness t and dielectric constant $K$ is placed between plates
$C=\frac{\epsilon_0 A}{d-t+\frac{t}{K}}$
In first case.
$C_1=\frac{\epsilon_0 A}{d-d / 2+\frac{d}{8}}=\frac{8 \epsilon_0 A}{5 d}$
In second case $\begin{aligned} & C_2=\frac{\epsilon_0 A}{d-d / 3+\frac{d}{12}}=\frac{12 \epsilon_0 A}{9 d} \\ & \therefore \frac{C_1}{C_2}=\frac{8 \epsilon_0 A}{5 d} \times \frac{9 d}{12 \epsilon_0 A}=\frac{6}{5} \end{aligned}$
In second case $\begin{aligned} & C_2=\frac{\epsilon_0 A}{d-d / 3+\frac{d}{12}}=\frac{12 \epsilon_0 A}{9 d} \\ & \therefore \frac{C_1}{C_2}=\frac{8 \epsilon_0 A}{5 d} \times \frac{9 d}{12 \epsilon_0 A}=\frac{6}{5} \end{aligned}$
Hence, the correct answer is option: 'C'.