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`        A parallel-plate capacitor having plate area 400 cm^2 and 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. (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?`
3 years ago

Navjyot Kalra
654 Points
```										Sol. A = 400 cm2 = 4 × 10^–2 m^2
d = 1 cm = 1× 10^–3 m
V = 160 V
t = 0.5 = 5 × 10^–4 m
k = 5
C = ε base 0A/d-t+t/k = 8.85 * 10^-12 * 4 * 10^-2/10^-3 -5 * 10^-4 +5*10^-4/5 = 35.4 * 10^-4/10^-3 -0.5

```
3 years ago
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