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Standard XII

Physics

Capacitance of a Random Conductor

A capacitor o...

Question

A capacitor of capacitance $C = 15 p F$ is charged with voltage $V = 500 V$ . The electric field inside the capacitor with dielectric is $10^{6} V / m$ and the area of the plate is $10^{- 4} m^{2}$ , then the dielectric constant of the medium is :( $ε_{0} = 8.85 \times 10^{- 12}$ in S.I.units)

12.47

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8.47

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10.85

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14.85

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Solution

The correct option is B $8.47$
$E = \frac{V}{d} = \frac{V C}{A K ε_{0}}$
$K = \frac{V C}{A ε_{0} E} = \frac{500 \times 15 \times 10^{- 12}}{10^{- 4} \times 10^{6} \times 8.85 \times 10^{- 12}} = 8.47$

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A capacitor of capacitance C=15pF is charged with voltage V=500V. The electric field inside the capacitor with dielectric is 106V/m and the area of the plate is 10−4m2, then the dielectric constant of the medium is :(ε0=8.85×10−12 in S.I.units)

The correct option is B 8.47E=Vd=VCAKε0 K=VCAε0E=500×15×10−1210−4×106×8.85×10−12=8.47

A capacitor of capacitance $C = 15 p F$ is charged with voltage $V = 500 V$ . The electric field inside the capacitor with dielectric is $10^{6} V / m$ and the area of the plate is $10^{- 4} m^{2}$ , then the dielectric constant of the medium is :( $ε_{0} = 8.85 \times 10^{- 12}$ in S.I.units)

The correct option is B $8.47$
$E = \frac{V}{d} = \frac{V C}{A K ε_{0}}$
$K = \frac{V C}{A ε_{0} E} = \frac{500 \times 15 \times 10^{- 12}}{10^{- 4} \times 10^{6} \times 8.85 \times 10^{- 12}} = 8.47$