The potential energy of a charged parallel plate capacitor is `U_(0)`. If a slab of dielectric constant `K` is inserted between the plates, then the new potential energy will be

Options

A. `(U_(0))/(K)`
B. `U_(0)K^(2)`
C. `(U_(0))/(K^(2))`
D. `U_(0)^(2)`

Answer

To find the new potential energy of a charged parallel plate capacitor after inserting a dielectric slab with dielectric constant \( K \), we can follow these steps:

### Step 1: Understand the initial conditions
The initial potential energy \( U_0 \) of a charged parallel plate capacitor can be expressed using the formula:
\[ U_0 = \frac{Q^2}{2C} \]
where \( Q \) is the charge on the capacitor and \( C \) is the capacitance of the capacitor before the dielectric is inserted.

### Step 2: Insert the dielectric
When a dielectric slab with dielectric constant \( K \) is inserted into the capacitor, the capacitance of the capacitor increases. The new capacitance \( C' \) can be expressed as:
\[ C' = K \cdot C \]

### Step 3: Determine the new potential energy
Since the capacitor is isolated (not connected to a battery), the charge \( Q \) remains constant. The new potential energy \( U' \) after inserting the dielectric can be calculated using the formula:
\[ U' = \frac{Q^2}{2C'} \]
Substituting \( C' = K \cdot C \) into the equation gives:
\[ U' = \frac{Q^2}{2(K \cdot C)} = \frac{1}{K} \cdot \frac{Q^2}{2C} \]

### Step 4: Relate the new potential energy to the initial potential energy
We know from Step 1 that \( U_0 = \frac{Q^2}{2C} \). Therefore, we can express \( U' \) in terms of \( U_0 \):
\[ U' = \frac{1}{K} U_0 \]

### Conclusion
The new potential energy of the capacitor after inserting the dielectric slab is:
\[ U' = \frac{U_0}{K} \]