Q146 / 160 · AP EAMCET 2023 (17 May Shift 2) · Physics
The space between the plates of a parallel plate capacitor is filled with a mica sheet of thickness $1 \times 10^{-3} \mathrm{~m}$ and a fiber sheet of thickness $0.5 \times 10^{-3} \mathrm{~m}$. The dielectric constants of mica and fiber are 8 and 2.5 respectively. If the fiber breaks down at an electric field of $6.4 \times 10^{-6} \mathrm{Vm}^{-1}$ then the maximum voltage that can be applied to the capacitor is
- A$3400 \mathrm{~V}$
- Correct answerB$5200 \mathrm{~V}$
- C$2700 \mathrm{~V}$
- D$4800 \mathrm{~V}$
Solution
Thickness, $\mathrm{d}_1=1 \times 10^{-3} \mathrm{~m}$ and $\mathrm{d}_2=0.5 \times 10^{-3} \mathrm{~m}$ Dielectric constant, $\mathrm{K}_1=8$ and $\mathrm{K}_2=2.5$ Electric field; $\mathrm{E}_2=6.4 \times 10^{-6} \mathrm{~V} / \mathrm{m}$
$\begin{aligned} \mathrm{E}_1 & =\frac{\sigma}{\mathrm{K}_1 \epsilon_0} \\ \mathrm{E}_2 & =\frac{\sigma}{\mathrm{K}_2 \epsilon_0} \\ \frac{\mathrm{E}_1}{\mathrm{E}_2} & =\frac{\mathrm{K}_2}{\mathrm{~K}_1} \\ \Rightarrow \mathrm{E}_1 & =\frac{\mathrm{E}_2 \mathrm{~K}_1}{\mathrm{~K}_2}=\frac{6.4 \times 10^{-6} \times 8}{2.5}=20.48 \times 10^6 \mathrm{v} / \mathrm{m}\end{aligned}$
Hence, the correct answer is option: 'B'.