Q

# I have a doubt, kindly clarify. - Electrostatics - JEE Main-8

Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of  $10^{6}\; V/m$. The plate area is $10^{-4}\;m^{2}.$ What is the dielectric constant if the capacitance is $15\; pF$ ?

(given $\epsilon _{0}=8.86\times 10^{-12}C^{2}/Nm^{2}$)

• Option 1)

$8.5$

• Option 2)

$6.2$

• Option 3)

$4.5$

• Option 4)

$3.8$

Views

$Area =10^{-4}m^{2}$

$C = 15\; PF$

$Voltage=500\; v$

$E=\frac{v}{d}\; \; \; \; \; (1)$

$C=\frac{k\varepsilon _{o}A}{d}\; \; \; \; \; \; \; (2)$

from (1) & (2)

$k=\frac{VC}{A\varepsilon _{o}E}=\frac{500\times 15\times 10^{12}}{10^{-4}\times 10^{6}\times 8.85\times 10^{-12}}$

$k\approx 8.5$

Option 1)

$8.5$

Option 2)

$6.2$

Option 3)

$4.5$

Option 4)

$3.8$

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