JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2019
- Hard
- MCQ
Voltage rating of a parallel plate capacitor is \(500\,V\). Its dielectric can withstand a maximum electric field of \({10^6}\,\frac{V}{m}\). The plate area is \(10^{-4}\, m^2\) . What is the dielectric constant if the capacitance is \(15\, pF\) ? (given \({ \in _0} = 8.86 \times {10^{ - 12}}\,{C^2}\,/N{m^2}\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(8.5\)
Step-by-step Solution
Detailed explanation
\(A=10^{-4}\, \mathrm{m}^{2}\)
\(\mathrm{E}_{\max }=10^{6} \,\mathrm{V} / \mathrm{m}\)
\(\mathrm{C}=15\, \mu \mathrm{F}\)
\(\mathrm{C}=\frac{\mathrm{k} \varepsilon_{0} \mathrm{A}}{\mathrm{d}} ; \quad \frac{\mathrm{Cd}}{\varepsilon_{0} \mathrm{A}}=\mathrm{k}\)
\(\mathrm{k}=\frac{15 \times 10^{-12} \times 500 \times 10^{-6}}{8.86 \times 10^{-12} \times 10^{4}}=\frac{15 \times 5}{8.86}=8.465\)
\(k \approx 8.5\)
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