A parallel plate capacitor has area of each plate `A`, the separation between the plates is `d`. It is charged to a potential `V` and then disconnected from the battery. The amount of work done in the filling the capacitor Completely with a dielectric constant `k` is.

Options

A. `(1)/(2)(epsilon_(0)AV^(2))/(d)[1-(1)/(k^(2))]`
B. `(1)/(2)(V^(2) epsilon_(0)A)/(kd)`
C. `(1)/(2)(V^(2)epsilon_(0)A)/(k^(2)d)`
D. `(1)/(2)(epsilon_(0)AV^(2))/(d)[1-(1)/(K)]`

Explanation

To find the amount of work done in filling a parallel plate capacitor with a dielectric, we can follow these steps:

### Step 1: Understand the Initial Conditions
The capacitor has:
- Area of each plate: \( A \)
- Separation between the plates: \( d \)
- Initial potential difference: \( V \)

### Step 2: Calculate the Initial Capacitance
The capacitance \( C_0 \) of the capacitor without the dielectric is given by the formula:
\[ C_0 = \frac{\varepsilon_0 A}{d} \]
where \( \varepsilon_0 \) is the permittivity of free space.

### Step 3: Calculate the Initial Charge
The charge \( Q \) stored in the capacitor when charged to potential \( V \) is:
\[ Q = C_0 V = \frac{\varepsilon_0 A V}{d} \]

### Step 4: Determine the New Capacitance with Dielectric
When a dielectric with dielectric constant \( k \) is fully inserted, the new capacitance \( C \) becomes:
\[ C = k C_0 = k \frac{\varepsilon_0 A}{d} \]

### Step 5: Calculate the New Potential Difference
Since the capacitor is disconnected from the battery, the charge \( Q \) remains constant. The new potential difference \( V' \) across the capacitor with the dielectric is given by:
\[ V' = \frac{Q}{C} = \frac{Q}{k C_0} = \frac{\frac{\varepsilon_0 A V}{d}}{k \frac{\varepsilon_0 A}{d}} = \frac{V}{k} \]

### Step 6: Calculate the Work Done in Inserting the Dielectric
The work done \( W \) in filling the capacitor with the dielectric can be calculated using the change in energy stored in the capacitor. The initial energy \( U_0 \) stored in the capacitor is:
\[ U_0 = \frac{1}{2} C_0 V^2 = \frac{1}{2} \left(\frac{\varepsilon_0 A}{d}\right) V^2 \]
The new energy \( U \) stored after inserting the dielectric is:
\[ U = \frac{1}{2} C V'^2 = \frac{1}{2} \left(k \frac{\varepsilon_0 A}{d}\right) \left(\frac{V}{k}\right)^2 = \frac{1}{2} \left(k \frac{\varepsilon_0 A}{d}\right) \left(\frac{V^2}{k^2}\right) = \frac{1}{2} \frac{\varepsilon_0 A V^2}{dk} \]

### Step 7: Calculate the Work Done
The work done \( W \) in inserting the dielectric is the difference in energy:
\[ W = U - U_0 = \frac{1}{2} \frac{\varepsilon_0 A V^2}{dk} - \frac{1}{2} \frac{\varepsilon_0 A V^2}{d} \]
Factoring out the common terms:
\[ W = \frac{1}{2} \frac{\varepsilon_0 A V^2}{d} \left(\frac{1}{k} - 1\right) \]

### Final Result
Thus, the amount of work done in filling the capacitor completely with a dielectric constant \( k \) is:
\[ W = \frac{1}{2} \frac{\varepsilon_0 A V^2}{d} \left(\frac{1}{k} - 1\right) \]