The plate separation in a parallel plate condenser and plate area is A. If it is charged to V volt battery is diconnected then the work done increasing the plate separation to 2d will be

Options

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

Answer

To solve the problem of finding the work done in increasing the plate separation of a parallel plate capacitor from \(d\) to \(2d\) after it has been charged to \(V\) volts and disconnected from the battery, we can follow these steps:

### Step 1: Understand the Initial Conditions
A parallel plate capacitor is charged to a voltage \(V\) and the battery is then disconnected. The charge \(Q\) on the capacitor can be expressed as:
\[ Q = CV \]
where \(C\) is the capacitance of the capacitor.

### Step 2: Determine the Initial Capacitance
The capacitance \(C\) of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon_0 A}{d} \]
where:
- \(\varepsilon_0\) is the permittivity of free space,
- \(A\) is the area of the plates,
- \(d\) is the separation between the plates.

### Step 3: Calculate the Initial Charge
Substituting the expression for \(C\) into the equation for \(Q\):
\[ Q = \left(\frac{\varepsilon_0 A}{d}\right)V \]

### Step 4: Determine the New Capacitance After Separation
When the plate separation is increased to \(2d\), the new capacitance \(C_2\) becomes:
\[ C_2 = \frac{\varepsilon_0 A}{2d} \]

### Step 5: Calculate the Initial and Final Potential Energy
The potential energy \(U\) stored in a capacitor is given by:
\[ U = \frac{Q^2}{2C} \]
- Initial potential energy \(U_1\):
\[ U_1 = \frac{Q^2}{2C_1} = \frac{Q^2}{2\left(\frac{\varepsilon_0 A}{d}\right)} = \frac{Q^2 d}{2\varepsilon_0 A} \]
- Final potential energy \(U_2\):
\[ U_2 = \frac{Q^2}{2C_2} = \frac{Q^2}{2\left(\frac{\varepsilon_0 A}{2d}\right)} = \frac{Q^2 \cdot 2d}{2\varepsilon_0 A} = \frac{Q^2 d}{\varepsilon_0 A} \]

### Step 6: Calculate the Work Done
The work done \(W\) in increasing the plate separation is the difference between the final and initial potential energies:
\[ W = U_2 - U_1 = \frac{Q^2 d}{\varepsilon_0 A} - \frac{Q^2 d}{2\varepsilon_0 A} = \frac{Q^2 d}{2\varepsilon_0 A} \]

### Step 7: Substitute for \(Q\)
Now, substitute \(Q = \frac{\varepsilon_0 A}{d} V\) into the expression for work done:
\[ W = \frac{\left(\frac{\varepsilon_0 A}{d} V\right)^2 d}{2\varepsilon_0 A} = \frac{\varepsilon_0^2 A^2 V^2}{d^2} \cdot \frac{d}{2\varepsilon_0 A} \]
This simplifies to:
\[ W = \frac{\varepsilon_0 A V^2}{2d} \]

### Final Answer
Thus, the work done in increasing the plate separation to \(2d\) is:
\[ W = \frac{\varepsilon_0 A V^2}{2d} \]