Voltage rating of a parallel plate capacitor is 500,V . Its dielectric can withstand a maximum elect

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2. Electric Potential and Capacitance

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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 ${ \in _0} = 8.86 \times {10^{ - 12}}\,{C^2}\,/N{m^2}$)

A

$3.8$

B

$6.2$

C

$4.5$

D

$8.5$

(JEE MAIN-2019)

Solution

$A=10^{-4}\, \mathrm{m}^{2}$

$\mathrm{E}_{\max }=10^{6} \,\mathrm{V} / \mathrm{m}$

$\mathrm{C}=15\, \mu \mathrm{F}$

$\mathrm{C}=\frac{\mathrm{k} \varepsilon_{0} \mathrm{A}}{\mathrm{d}} ; \quad \frac{\mathrm{Cd}}{\varepsilon_{0} \mathrm{A}}=\mathrm{k}$

$\mathrm{k}=\frac{15 \times 10^{-12} \times 500 \times 10^{-6}}{8.86 \times 10^{-12} \times 10^{4}}=\frac{15 \times 5}{8.86}=8.465$

$k \approx 8.5$

Standard 12

Physics

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