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`        A parallel-plate capacitor has plate area 20 cm^2 , plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy of the capacitor 8.9 μs after the connections are made.`
3 years ago

Kevin Nash
332 Points
```										Sol. A 20 cm^2, d = 1 mm, K = 5, e = 6 V
R = 100 * 10^3 Ω, t = 8.9 * 10^-5 s
C = KE base 0A/d = 5 * 8.85 * 10^-12 *20 * 10^-4/1 * 10^-3
= 10 * 8.85 * 10^-3 * 10^-12/10^-3 = 88.5 * 10^-12
q = EC(1 – e^-tRC)
= 6 * 88.5 * 10^-12 (1 – e^-89*10^-6/88.5*10^-12*10^4) = 530.97
Energy = ½ * 500.97 * 530/88.5 * 10^-12
= 530.97 * 530.97/88.5 * 2 *10^12

```
3 years ago
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