A parallel plate capacitor has plate area `A` and separation `d`. It is charged to a potential difference `V_(0)`. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

Options

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

Explanation

To solve the problem of finding the work required to separate the plates of a parallel plate capacitor when the separation is increased from `d` to `3d`, we can follow these steps:

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

When the capacitor is charged to a potential difference \( V_0 \), the charge \( Q \) on the capacitor can be calculated as:
\[ Q = C V_0 = \frac{\varepsilon_0 A}{d} V_0 \]

### Step 2: Calculate the Initial Energy Stored in the Capacitor
The energy \( U_i \) stored in the capacitor initially is given by:
\[ U_i = \frac{1}{2} C V_0^2 \]
Substituting the expression for \( C \):
\[ U_i = \frac{1}{2} \left(\frac{\varepsilon_0 A}{d}\right) V_0^2 \]

### Step 3: Determine the New Capacitance After Separation
When the plates are pulled apart to three times the initial separation, the new separation becomes \( 3d \). The new capacitance \( C' \) is:
\[ C' = \frac{\varepsilon_0 A}{3d} \]

### Step 4: Calculate the New Voltage Across the Capacitor
Since the battery is disconnected, the charge \( Q \) remains constant. The new voltage \( V' \) across the capacitor can be found using the relationship:
\[ Q = C' V' \]
Substituting for \( Q \):
\[ \frac{\varepsilon_0 A}{d} V_0 = \frac{\varepsilon_0 A}{3d} V' \]
From this, we can solve for \( V' \):
\[ V' = 3 V_0 \]

### Step 5: Calculate the Final Energy Stored in the Capacitor
The final energy \( U_f \) stored in the capacitor after the plates are separated is:
\[ U_f = \frac{1}{2} C' (V')^2 \]
Substituting for \( C' \) and \( V' \):
\[ U_f = \frac{1}{2} \left(\frac{\varepsilon_0 A}{3d}\right) (3 V_0)^2 \]
\[ U_f = \frac{1}{2} \left(\frac{\varepsilon_0 A}{3d}\right) (9 V_0^2) = \frac{9}{6} \left(\frac{\varepsilon_0 A}{d}\right) V_0^2 = \frac{3}{2} \left(\frac{\varepsilon_0 A}{d}\right) V_0^2 \]

### Step 6: Calculate the Work Done to Separate the Plates
The work done \( W \) to separate the plates is the change in energy:
\[ W = U_f - U_i \]
Substituting the expressions for \( U_f \) and \( U_i \):
\[ W = \left(\frac{3}{2} \frac{\varepsilon_0 A}{d} V_0^2\right) - \left(\frac{1}{2} \frac{\varepsilon_0 A}{d} V_0^2\right) \]
\[ W = \left(\frac{3}{2} - \frac{1}{2}\right) \frac{\varepsilon_0 A}{d} V_0^2 = \frac{2}{2} \frac{\varepsilon_0 A}{d} V_0^2 = \frac{\varepsilon_0 A}{d} V_0^2 \]

### Final Answer
The work required to separate the plates is:
\[ W = \frac{\varepsilon_0 A}{d} V_0^2 \]