A parallel plate capacitor is charged completely and then disconnected from the battery. IF the separation between the plates is reduced by 50% and the space between the plates if filled with a dielectric slab of dielectric constant 10, then the potential difference between the plates

Options

A. Decrease by 95%
B. Increase by 95%
C. Decrease by 50%
D. Increases by 50%

Answer

To solve the problem step by step, we will analyze the situation of the parallel plate capacitor before and after the changes made.

### Step 1: Understand the Initial Conditions
Initially, we have a parallel plate capacitor with:
- Capacitance \( C_0 \)
- Charge \( Q = C_0 V_0 \) (where \( V_0 \) is the initial potential difference)
- Plate separation \( D \)

### Step 2: Analyze the Change in Plate Separation
When the separation between the plates is reduced by 50%, the new separation \( D' \) becomes:
\[ D' = \frac{D}{2} \]
The capacitance of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon_0 A}{D} \]
Thus, when the distance \( D \) is halved, the new capacitance \( C' \) becomes:
\[ C' = \frac{\varepsilon_0 A}{D'} = \frac{\varepsilon_0 A}{D/2} = 2 \cdot \frac{\varepsilon_0 A}{D} = 2C_0 \]

### Step 3: Insert the Dielectric Slab
Now, we insert a dielectric slab with a dielectric constant \( K = 10 \). The capacitance with the dielectric becomes:
\[ C'' = K \cdot C' = 10 \cdot (2C_0) = 20C_0 \]

### Step 4: Calculate the New Potential Difference
Since the capacitor was disconnected from the battery, the charge \( Q \) remains constant. The new charge is still:
\[ Q = C_0 V_0 \]
The new potential difference \( V' \) across the capacitor can be calculated using the new capacitance:
\[ V' = \frac{Q}{C''} = \frac{C_0 V_0}{20C_0} = \frac{V_0}{20} \]

### Step 5: Determine the Change in Potential Difference
The initial potential difference was \( V_0 \) and the new potential difference is \( V' = \frac{V_0}{20} \). The change in potential difference can be calculated as:
\[ \text{Decrease in potential} = V_0 - V' = V_0 - \frac{V_0}{20} = V_0 \left(1 - \frac{1}{20}\right) = V_0 \left(\frac{19}{20}\right) \]
This indicates that the potential difference has decreased by 95%.

### Final Answer
The potential difference between the plates after reducing the separation and inserting the dielectric slab is:
\[ V' = \frac{V_0}{20} \]
This means the potential difference has decreased by 95%.

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