Capacitors

JEE Main level, single correct, medium

94 of 180

Question 94 of 180

A parallel plate capacitor is formed by two plates each of area 30πcm230\pi {cm}^{2} separated by 1mm1mm. A material of dielectric strength 3.6×107Vm−13.6\times {10}^{7}V{m}^{-1} is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7×10−6C7\times {10}^{-6}C, the value of dielectric constant of the material is :
[Use 14πϵo=9×109Nm2C−2\dfrac{1}{4\pi {\epsilon }_{o}}=9\times {10}^{9}N{m}^{2}{C}^{-2}]

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Answer: (D) 2.332.33

The dielectric strength given is 3.6×107Vm−13.6\times {10}^{7}V{m}^{-1}.and the separation is 1mm1mm. Therefore, maximum potential difference that can be applied will be,
Vmax=3.6×107×d{V}_{max}=3.6\times {10}^{7}\times d.
Now,
C=qmaxVmax⇒kϵ0Ad=7×10−73.6×107×dk=dϵ0×30π×10−4×7×10−73.6×107×d=2.33C=\dfrac{{q}_{max}}{{V}_{max}} \Rightarrow \dfrac{k{\epsilon }_{0}A}{d}=\dfrac{7\times {10}^{-7}}{3.6\times {10}^{7}\times d} k=\dfrac{d}{{\epsilon }_{0}\times 30\pi \times {10}^{-4}}\times \dfrac{7\times {10}^{-7}}{3.6\times {10}^{7}\times d}=2.33

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