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

Options

A. (a) `(Q_1+Q_2)//(2C)`
B. (b) `(Q_1+Q_2)//C`
C. (c) `(Q_1-Q_2)//C`
D. (d) `(Q_1-Q_2)//(2C)`

Explanation

To find the potential difference between two identical metal plates with positive charges \( Q_1 \) and \( Q_2 \) (where \( Q_1 > Q_2 \)), we can follow these steps:

### Step 1: Understand the Electric Field
When two charged plates are brought close together, they create an electric field between them. The electric field \( E \) due to a charged plate is given by the formula:
\[ E = \frac{Q}{\epsilon_0 A} \]
where \( Q \) is the charge on the plate, \( \epsilon_0 \) is the permittivity of free space, and \( A \) is the area of the plate.

### Step 2: Calculate Electric Fields from Each Plate
For the plate with charge \( Q_1 \):
\[ E_1 = \frac{Q_1}{\epsilon_0 A} \]
For the plate with charge \( Q_2 \):
\[ E_2 = \frac{Q_2}{\epsilon_0 A} \]

### Step 3: Determine the Net Electric Field
Since both plates are positively charged, the electric fields will point away from each plate. The net electric field \( E \) between the plates can be calculated as:
\[ E = E_1 - E_2 = \frac{Q_1}{\epsilon_0 A} - \frac{Q_2}{\epsilon_0 A} = \frac{Q_1 - Q_2}{\epsilon_0 A} \]

### Step 4: Relate Electric Field to Potential Difference
The potential difference \( V \) between the plates is related to the electric field by the formula:
\[ V = E \cdot d \]
where \( d \) is the distance between the plates.

### Step 5: Substitute the Expression for Electric Field
Substituting the expression for \( E \) into the potential difference formula gives:
\[ V = \left(\frac{Q_1 - Q_2}{\epsilon_0 A}\right) \cdot d \]

### Step 6: Use the Capacitance Formula
The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\epsilon_0 A}{d} \]
From this, we can express \( \frac{1}{C} \):
\[ \frac{1}{C} = \frac{d}{\epsilon_0 A} \]

### Step 7: Substitute Capacitance into the Potential Difference Equation
Now we can rewrite the potential difference in terms of capacitance:
\[ V = (Q_1 - Q_2) \cdot \frac{1}{C} \]

### Final Result
Thus, the potential difference \( V \) between the plates is:
\[ V = \frac{Q_1 - Q_2}{C} \]