The dielectric strength given is 3.6×107Vm−1.and the separation is 1mm. Therefore, maximum potential difference that can be applied will be,
Vmax=3.6×107×d.
Now,
C=Vmaxqmax⇒dkϵ0A=3.6×107×d7×10−7k=ϵ0×30π×10−4d×3.6×107×d7×10−7=2.33
JEE Main 2022 — Physics Electromagnetism
A parallel plate capacitor is formed by two plates each of area 30πcm2 separated by 1mm. A material of dielectric strength 3.6×107Vm−1 is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7×10−6C, the value of dielectric constant of the material is :
[Use 4πϵo1=9×109Nm2C−2]
Held on 24 Jun 2022 · Verified 6 Jul 2026.
1.66
1.75
2.25
2.33
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