A parallel -plate capacitor having plate area `400 cm ^2` and separation between the plate 1.0mm is connected to a power supply of `100V.` A dielectric slab of thickness1.0mm and dielectric constant 5.0 is inserted into the gap .(a)Find the increase in electrostatic energy .(b) If the power supply is now disconnected and the dielectric slab is taken out , find the further increase in energy. (c) Why does the energy increase in inserting the slab as well as in taking it out?

Explanation

To solve the problem step-by-step, we will break it down into parts (a), (b), and (c) as requested.

### Given Data:
- Plate area \( A = 400 \, \text{cm}^2 = 400 \times 10^{-4} \, \text{m}^2 = 0.04 \, \text{m}^2 \)
- Separation between plates \( d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \)
- Voltage \( V = 100 \, \text{V} \)
- Dielectric thickness \( t = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \)
- Dielectric constant \( K = 5.0 \)
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