A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

Options

A. zero
B. `1/2(K-1)CV^(2)`
C. `(CV^(2)(K-1))/(K)`
D. `(K-1)CV^(2)`

Answer

To solve the problem step by step, let's analyze the situation of a parallel plate capacitor with a dielectric slab, and the work done during the process of removing and reinserting the dielectric.

### Step 1: Understand the Initial Condition
The initial capacitance \( C \) of the parallel plate capacitor with a dielectric of dielectric constant \( K \) is given. The energy stored in the capacitor when charged to a potential \( V \) is given by the formula:

\[ U_i = \frac{1}{2} C V^2 \]

### Step 2: Remove the Dielectric
When the dielectric slab is removed, the capacitance changes. The new capacitance \( C' \) when the dielectric is removed is:

\[ C' = \frac{C}{K} \]

The energy stored in the capacitor after the dielectric is removed, while keeping the potential \( V \) constant, is:

\[ U' = \frac{1}{2} C' V^2 = \frac{1}{2} \left(\frac{C}{K}\right) V^2 = \frac{1}{2} \frac{C V^2}{K} \]

### Step 3: Reinsert the Dielectric
Now, when the dielectric slab is reinserted, the capacitance returns to its original value \( C \). The energy stored in the capacitor with the dielectric reinserted is:

\[ U_f = \frac{1}{2} C V^2 \]

### Step 4: Calculate the Net Work Done
The net work done \( W \) by the system during the entire process (removing and reinserting the dielectric) can be calculated as the difference between the final energy and the initial energy:

\[ W = U_f - U_i \]

Substituting the values we found:

\[ W = \left(\frac{1}{2} C V^2\right) - \left(\frac{1}{2} C V^2\right) = 0 \]

### Conclusion
The net work done by the system in this process is:

\[ W = 0 \]