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`        A charge of 1 μC is given to one plate of  a parallel-plate of an isolated parallel-plate capacitance 0.1 μF and a charge of 2 μC is given to the other plate. Find the potential difference developed between the plates.`
3 years ago

Deepak Patra
474 Points
```										Sol. q base 1 = 1 μC = 1 × 10^–6 C C = 0.1 μF = 1 × 10^–7F
Q base 2 = 2 μC = 2 × 10^–6 C
net q = q base 1 – q base 2/2 = (1 - 2) * 10^-6/2 = - 0.5 * 10^-6 C
potential ‘V’ = q/c = 1 * 10^-7/5 * 10^-7 = - 5 V
but potential can never be (-)ve. So, V = 5 V

```
3 years ago
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