The potential enery 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, we can follow these steps:

### Step 1: Understand the Initial Potential Energy
The initial potential energy \( U_0 \) of a charged parallel plate capacitor is given by the formula:
\[ U_0 = \frac{Q^2}{2C} \]
where \( Q \) is the charge on the capacitor and \( C \) is the capacitance.

### Step 2: Recognize the Effect of Dielectric
When a dielectric slab with dielectric constant \( K \) is inserted between the plates of the capacitor, the capacitance changes. The new capacitance \( C' \) can be expressed as:
\[ C' = K \cdot C \]
This means the capacitance increases by a factor of \( K \).

### Step 3: Determine the New Potential Energy
The new potential energy \( U' \) of the capacitor with the dielectric inserted can be calculated using the same formula for potential energy, but with the new capacitance:
\[ U' = \frac{Q^2}{2C'} \]
Substituting \( C' \) into the equation gives:
\[ U' = \frac{Q^2}{2(K \cdot C)} = \frac{1}{K} \cdot \frac{Q^2}{2C} = \frac{U_0}{K} \]

### Step 4: Conclusion
Thus, the new potential energy of the capacitor after inserting the dielectric slab is:
\[ U' = \frac{U_0}{K} \]

### Final Answer
The new potential energy will be \( \frac{U_0}{K} \).

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