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Two identical metal plates are given poi...

Two identical metal plates are given poistive charges `Q_1` and `Q_2` `(ltQ_1)` respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potencial difference between them is

A

`(Q_1+Q_2)//2C`

B

`(Q_1+Q_2)//C`

C

`(Q_1-Q_2)//C`

D

`(Q_1-Q_2)//2C`

Text Solution

AI Generated Solution

To solve the problem of finding the potential difference between two identical metal plates with charges \( Q_1 \) and \( Q_2 \) (where \( Q_1 > Q_2 \)), we can follow these steps: ### Step 1: Understand the Configuration We have two parallel plates, one with charge \( Q_1 \) and the other with charge \( Q_2 \). The area of each plate is \( A \), and the separation between the plates is \( D \). **Hint:** Visualize the setup of the parallel plates and the charges on them. ### Step 2: Calculate the Electric Field ...
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