A parallel plate capacitor is formed by two plates each of area 30π cm230\pi\ \mathrm{cm^2}30π cm2 separated by 1 mm. A material of dielectric strength 3.6×107 Vm−13.6\times10^7\ \mathrm{Vm^{-1}}3.6×107 Vm−1 is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7×10−67\times10^{-6}7×10−6 C, the value of dielectric constant of the material is :
[Use 14πε0=9×109 Nm2C−2\dfrac1{4\pi\varepsilon_0}=9\times10^9\ \mathrm{Nm^2C^{-2}}4πε01=9×109 Nm2C−2]
Official answer: D
Maximum voltage. V=EmaxdV=E_{\max}dV=Emaxd, so Qmax=CV=Kε0Ad⋅Emaxd=Kε0AEmaxQ_{\max}=CV=\dfrac{K\varepsilon_0A}d\cdot E_{\max}d=K\varepsilon_0AE_{\max}Qmax=CV=dKε0A⋅Emaxd=Kε0AEmax.
Numbers. ε0A=30π×10−44π×9×109=3036×10−13\varepsilon_0A=\dfrac{30\pi\times10^{-4}}{4\pi\times9\times10^9}=\dfrac{30}{36}\times10^{-13}ε0A=4π×9×10930π×10−4=3630×10−13, and ε0AEmax=3036×10−13×3.6×107=3×10−6\varepsilon_0AE_{\max}=\dfrac{30}{36}\times10^{-13}\times3.6\times10^7=3\times10^{-6}ε0AEmax=3630×10−13×3.6×107=3×10−6 C.
Dielectric constant. K=7×10−63×10−6≈2.33K=\dfrac{7\times10^{-6}}{3\times10^{-6}}\approx2.33K=3×10−67×10−6≈2.33.
(D)