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`         A 100 pF capacitor is charged to a potential difference of 24 V. It is connected to an uncharged capacitor of capacitance 20 pF. What will be the new potential difference across the 100 pF capacitor?`
5 years ago 396 Points
```							Sol. Given that
C = 100 PF = 100 * 10^-12 F C base cq = 20 PF = 20 * 10^-12 F
V = 24 V q = 24 * 100 * 10^-12 = 24 * 10^-10
Q base 2 = ?
Let q base 1 = The new charge 100 PF V base 1 = The Voltage.
After the flow of charge, potential is same in the two capacitor
V base 1 = q base 2/C base 2 = q base 1/C base 1
= q – q base 1/C base 2 = q base 1/C base 1
= 24 * 10^-10 –q base 1/24 * 10^-12 = q base 1/100 * 10^-12
= 24 * 10^-10 – q base 1 = q base 1/5
= 6q base 1 = 120 * 10^-10
= q base 1 = 120/6 * 10^-10 = 20 * 10^10
∴ V base 1 = q base 1/ C base 1 = 20 *10^-10/100 * 10^-12 = 20 V

```
5 years ago
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