A capacitor of capacitance 15 ,nF having diel | vedclass

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Question

The expression for capacitance is given by,

$C=\frac{A \varepsilon_{0} \varepsilon_{r}}{d}$ $......(I)$

Substitute $15 	imes 10^{-9}$ for $C, 8

Vedclass · Accepted Answer

$6.7$

Vedclass · Answer

The expression for capacitance is given by,

$C=\frac{A \varepsilon_{0} \varepsilon_{r}}{d}$ $......(I)$

Substitute $15 	imes 10^{-9}$ for $C, 8.85 	imes 10^{-12}$ for $\varepsilon_{0}$ and $2.5$ for $\varepsilon,$ in equation $(I).$

$15 	imes 10^{-9}=\frac{A\left(8.85 	imes 10^{-12}ight)(2.5)}{d} \quad \ldots \ldots$ $(II)$

The expression for electric field in given by,

$E=\frac{V}{d}$

$30 	imes 10^{6}=\frac{30}{d}$

$d=10^{-6} m$

Substitute $10^{-6} m$ for $d$ in equation $( II )$.

$15 	imes 10^{-9}=\frac{A\left(8.85 	imes 10^{-12}ight)(2.5)}{10^{-6}}$

$A=\frac{15 	imes 10^{-9} 	imes 10^{-6}}{8.85 	imes 10^{-12} 	imes 2.5}$

$=6.7 	imes 10^{-4} m ^{2}$

A capacitor of capacitance $15 \,nF$ having dielectric slab of $\varepsilon_{r}=2.5$ dielectric strength $30 \,MV / m$ and potential difference $=30\; volt$ then the area of plate is ....... $ \times 10^{-4}\; m ^{2}$

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