A parallel-plate capacitor of capacitance 40 F is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K = 2. Due to the introduction of dielectric material, the extra

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Solution

Q.
A parallel-plate capacitor of capacitance 40 $μ$ F is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K = 2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are [2025]

1 2 mC and 0.2 J

2 8 mC and 2.0 J

3 4 mC and 0.2 J

4 2 mC and 0.4 J

Ans.
(3)

$C = 40 μ F, C' = k C = 80 μ F$

V = 100 V

Q = CV = 4 mC

$Q' = C' V = 8 mC$

$∴$ Extra charge, $△ Q$ = 4 mC

$U_{1} = \frac{1}{2} {CV}^{2}, U_{2} = \frac{1}{2} C' V^{2}$

$△ U = \frac{1}{2} (C' - C) V^{2} = \frac{1}{2} \times 40 \times 10^{4} \times 10^{- 6} = 0.2 J$

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