A pararllel plate capacitor has plates of area `A` and separation `d` and is charged to potential diference `V`. The charging battery is then disconnected and the plates are pulle apart until their separation is `2d`. What is the work required to separate the plates?

Options

A. `2epsilon_(0)AV^(2)//d`
B. `epsilonAV^(2)//d`
C. `3epsilon_(0)AV^(2)//2d`
D. `epsilon_(0)AV^(2)//2d`

Explanation

To solve the problem of determining the work required to separate the plates of a parallel plate capacitor after it has been charged and disconnected from the battery, we can follow these steps:

### Step 1: Determine Initial Capacitance
The capacitance \( C_1 \) of a parallel plate capacitor is given by the formula:
\[ C_1 = \frac{\varepsilon_0 A}{d} \]
where:
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( d \) is the separation between the plates.

### Step 2: Calculate Initial Energy Stored
The energy \( U_1 \) stored in the capacitor when it is charged to a potential difference \( V \) is given by:
\[ U_1 = \frac{Q^2}{2C_1} \]
where \( Q \) is the charge on the capacitor. Since \( Q = C_1 V \), we can substitute \( Q \):
\[ U_1 = \frac{(C_1 V)^2}{2C_1} = \frac{C_1 V^2}{2} \]

### Step 3: Determine New Capacitance After Separation
When the plates are pulled apart to a new separation of \( 2d \), the new capacitance \( C_2 \) becomes:
\[ C_2 = \frac{\varepsilon_0 A}{2d} \]

### Step 4: Calculate New Energy Stored
The energy \( U_2 \) stored in the capacitor after the plates are separated is:
\[ U_2 = \frac{Q^2}{2C_2} \]
Substituting \( Q = C_1 V \) again:
\[ U_2 = \frac{(C_1 V)^2}{2C_2} = \frac{C_1 V^2}{2C_2} \]
Now substituting \( C_2 = \frac{C_1}{2} \):
\[ U_2 = \frac{C_1 V^2}{2 \cdot \frac{C_1}{2}} = \frac{C_1 V^2}{C_1} = \frac{V^2}{2} \]

### Step 5: Calculate Work Done
The work done \( W \) in separating the plates is the difference in energy stored:
\[ W = U_2 - U_1 \]
Substituting the expressions for \( U_1 \) and \( U_2 \):
\[ W = \frac{V^2}{2} - \frac{C_1 V^2}{2} \]
Now substituting \( C_1 = \frac{\varepsilon_0 A}{d} \):
\[ W = \frac{V^2}{2} - \frac{\frac{\varepsilon_0 A}{d} V^2}{2} \]
Factoring out \( \frac{V^2}{2} \):
\[ W = \frac{V^2}{2} \left( 1 - \frac{\varepsilon_0 A}{d} \right) \]

### Final Expression
After simplifying, we find:
\[ W = \frac{\varepsilon_0 A V^2}{2d} \]

### Conclusion
Thus, the work required to separate the plates of the capacitor when the separation is increased from \( d \) to \( 2d \) is:
\[ W = \frac{1}{2} \frac{\varepsilon_0 A V^2}{d} \]