A parallel capacitor of capacitance `C` is charged and disconnected from the battery. The energy stored in it is `E`. If a dielectric slab of dielectric constant `6` is inserted between the plates of the capacitor then energy and capacitance will become.

Options

A. 6E, 6C
B. E, C
C. `(E)/(6)`,6C
D. E, 6C

Explanation

To solve the problem step by step, we will analyze the situation involving the capacitor before and after inserting the dielectric slab.

### Step 1: Understand the initial conditions
- A parallel plate capacitor with capacitance \( C \) is charged and disconnected from the battery.
- The energy stored in the capacitor is \( E \).

**Hint:** Recall the formula for energy stored in a capacitor: \( E = \frac{1}{2} \frac{Q^2}{C} \).

### Step 2: Determine the charge on the capacitor
When the capacitor is charged, the charge \( Q \) stored in it can be expressed as:
\[ Q = C \cdot V \]
where \( V \) is the voltage across the capacitor.

**Hint:** Remember that the charge remains constant when the capacitor is disconnected from the battery.

### Step 3: Express the initial energy in terms of charge and capacitance
The energy stored in the capacitor can be rewritten using the charge:
\[ E = \frac{1}{2} \frac{Q^2}{C} \]

**Hint:** This formula shows how energy is related to charge and capacitance.

### Step 4: Insert the dielectric slab
When a dielectric slab with dielectric constant \( K = 6 \) is inserted between the plates of the capacitor, the new capacitance \( C' \) becomes:
\[ C' = K \cdot C = 6C \]

**Hint:** The presence of a dielectric increases the capacitance by a factor equal to the dielectric constant.

### Step 5: Analyze the energy after inserting the dielectric
Since the charge \( Q \) remains constant, the new energy \( E' \) stored in the capacitor with the dielectric can be expressed as:
\[ E' = \frac{1}{2} \frac{Q^2}{C'} \]
Substituting \( C' = 6C \):
\[ E' = \frac{1}{2} \frac{Q^2}{6C} = \frac{1}{6} \left( \frac{1}{2} \frac{Q^2}{C} \right) = \frac{E}{6} \]

**Hint:** Notice how the energy decreases when the dielectric is inserted, due to the increase in capacitance.

### Step 6: Summarize the results
- The new capacitance \( C' \) is \( 6C \).
- The new energy \( E' \) is \( \frac{E}{6} \).

**Final Answer:**
- Capacitance becomes \( 6C \).
- Energy becomes \( \frac{E}{6} \).