The energy and capacity of a charged parallel plate capacitor are U and C respectively. Now a dielectric slab of `in_(r)= 6` is inserted in it then energy and capacity becomes (Assuming charge on plates remains constant)

Options

A. 6U, 6C
B. U, C
C. `U/6, 6C`
D. U, 6C

Answer

To solve the problem, we need to determine the new capacitance and energy of a charged parallel plate capacitor after inserting a dielectric slab with a dielectric constant \( K = 6 \). The charge on the plates remains constant during this process.

### Step 1: Determine the initial capacitance
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{A \epsilon_0}{D} \]
where:
- \( A \) is the area of the plates,
- \( \epsilon_0 \) is the permittivity of free space,
- \( D \) is the separation between the plates.

### Step 2: Calculate the new capacitance with the dielectric
When a dielectric slab is inserted, the capacitance increases by a factor equal to the dielectric constant \( K \):
\[ C' = K \cdot C \]
Substituting \( K = 6 \):
\[ C' = 6C \]

### Step 3: Determine the initial energy stored in the capacitor
The energy \( U \) stored in a capacitor is given by the formula:
\[ U = \frac{Q^2}{2C} \]
where \( Q \) is the charge on the plates.

### Step 4: Calculate the new energy with the dielectric
Since the charge \( Q \) remains constant, the new energy \( U' \) can be expressed as:
\[ U' = \frac{Q^2}{2C'} \]
Substituting \( C' = 6C \):
\[ U' = \frac{Q^2}{2 \cdot 6C} = \frac{Q^2}{12C} \]

Now, we can relate \( U \) to \( C \) using the original energy formula:
\[ U = \frac{Q^2}{2C} \]
From this, we can express \( U' \) in terms of \( U \):
\[ U' = \frac{U}{6} \]

### Final Results
Thus, the new capacitance and energy after inserting the dielectric are:
- New capacitance: \( C' = 6C \)
- New energy: \( U' = \frac{U}{6} \)

### Summary of Results
- \( C' = 6C \)
- \( U' = \frac{U}{6} \)