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A parallel-plate capacitor having plate area 400 cm²and separation between the plates 1.0 mm is connected to a power supply of 100 V. A dielectric slab of thickness 0.5 mm and dielectric constant 5.0 is inserted into the gap. Find the increase in electrostatic energy (of the order of $μJ$ )

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answer is 1.18.

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Detailed Solution

The capacitance of the capacitor is

initially

$C_{1} = \frac{A ε_{0}}{d}$

After introducing a dielectric slab

$C_{2} = \frac{A ε_{0}}{d - t + \frac{t}{k}}$

$\begin{array}{l} E_{final} - E_{initial} = \frac{1}{2} C_{2} V_{2}^{2} - \frac{1}{2} C_{1} V_{1}^{2} \end{array}$

$C_{2} - C_{1} = A ε_{0} (\frac{1}{d - t + \frac{t}{k}} - \frac{1}{d}) = A ε_{0} (\frac{d - (d - t + \frac{t}{k})}{d \times (d - t + \frac{t}{k})})$

$C_{2} - C_{1} = A ε_{0} (\frac{1 - (1 - 0.5 + 0.5 / 5)}{d (1 - 0.5 + 0.5 / 5)}) = \frac{2 A ε_{0}}{3 d}$

$= \frac{1}{2} \times 10000 [\frac{8.85 \times 10^{- 12} \times 400 \times 10^{- 4} \times 2}{3 \times 10^{- 3}}] = 1.18 \times 10^{- 6} J$

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