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Shane Macguire Grade: upto college level
`        A charge of + 2.0 × 10^-8 C is placed on the positive plate and a change of – 1.0 × 10^-8 C on the negative plate of a parallel-plate capacitor of capacitance 1.2 × 10^-3 μF. Calculate the potential difference developed between the plates. `
3 years ago

Navjyot Kalra
654 Points
```										Sol. q base 1 = +2.0 × 10^–8 c q base 2 = –1.0 × 10^–8 c
C = 1.2 × 10^–3 μF = 1.2 × 10^–9 F
net q = q base 1 – q base 2/2 = 3.0 * 10^-8/2
V = q/c = 3 * 10^-8/2 * 1/1.2 * 10^-9 = 12.5 V

```
3 years ago
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