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Two plates ( area =s ) charged to + q(1)...

Two plates ( area =s ) charged to `+ q_(1) ` and `+q_(2)(q_(2) lt q_(1))` are brought closer to form a capacitor of capacitance C. The potential difference across the plates is

A

`(q_(1)-q_(2))/(2C)`

B

`(q_(1)-q_(2))/( C)`

C

`(q_(1)-q_(2))/(4 C )`

D

`(2(q_(1)-q_(2)))/ ( C)`

Text Solution

AI Generated Solution

To find the potential difference across the plates of a capacitor formed by two charged plates, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Charges**: We have two plates with charges \( +q_1 \) and \( +q_2 \), where \( q_2 < q_1 \). 2. **Effective Charge Calculation**: When two plates are brought together to form a capacitor, the effective charge on the plates can be determined. The charge on the inside surface of the first plate will be \( Q_1 \) and on the second plate will be \( Q_2 \). The effective charge that contributes to the capacitance is given by: \[ ...
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