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Two square plates of capacitor are of length 32 cm each, Between the plates, material having thickness $640 μ m$ is sandwiched whose dielectric strength is $20 {MVm}^{- 1}$ . The maximum charge that can be put on the capacitor is (Permittivity of free space $= 9 \times 10^{- 12}$ SI unit & dielectric constant of material is 4)

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$4.6 μ C$

$5 μ C$

$73.7 μ C$

$18.4 μ C$

answer is C.

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Detailed Solution

Net electric field inside the dielectric

$\begin{array}{l} E = \frac{Q}{{kAε}_{0}} \\ \frac{Q}{4 \times {(\frac{32}{100})}^{2} \times 9 \times 10^{- 12}} = 20 \times 10^{6} \\ \to Q - 73 .7 μC \end{array}$

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