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`        A parallel-plate capacitor with the plate area 100 cm^2 and the separation between the plates 1.0 cm is connected across a battery of emf 24 volts. Find the force of attraction between the plates.`
3 years ago

396 Points
```										Sol. A = 100 cm2 = 10^–2 m^2
d = 1 cm = 10^–2 m
V = 24 V base 0
∴ The capacitance C = ε base 0A/d = 8.85 * 10^-12 * 10^-2/10^-2 = 8.85 * 10^-12
∴ The energy stored C base 1 = (1/2) CV^2 = (1/2) × 10^–12 × (24)^2 = 2548.8 × 10^–12
∴ The forced attraction between the plates = C base 1/d = 2548.8 * 10^-12/10^-2 = 2.54 * 10^-7 N.

```
3 years ago
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