A parallel plate capacitor of capacitance C is charged to a potential V and then disconnected with the battery. If separation between the plates is decreased by `50%` and the space between the plates is filled with a dielectric slab of dielectric constant `K=10`. then

Options

A. P.d. between the plates increases by `95%`
B. P.d. between the plates decreases by `50%`
C. P.d. between the plates increases by `50%`
D. P.d. between the plates decreases by `95%`

Explanation

To solve the problem step by step, we will analyze the situation of a parallel plate capacitor when the separation between the plates is decreased and a dielectric slab is introduced.

### Step 1: Understand the initial conditions
The initial capacitance \( C \) of the capacitor is given by the formula:
\[ C = \frac{K \epsilon_0 A}{D} \]
where:
- \( K \) is the dielectric constant (initially \( K = 1 \) since there is no dielectric),
- \( \epsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( D \) is the separation between the plates.

### Step 2: Determine the charge on the capacitor
When the capacitor is charged to a potential \( V \), the charge \( Q \) on the capacitor can be expressed as:
\[ Q = C \cdot V \]
Since the capacitor is disconnected from the battery, the charge \( Q \) remains constant.

### Step 3: Analyze the new conditions after decreasing the plate separation
When the separation between the plates is decreased by 50%, the new separation \( D' \) is:
\[ D' = \frac{D}{2} \]

### Step 4: Introduce the dielectric slab
Now, we fill the space between the plates with a dielectric slab of dielectric constant \( K = 10 \). The new capacitance \( C' \) can be calculated as:
\[ C' = \frac{K \epsilon_0 A}{D'} \]
Substituting \( D' \):
\[ C' = \frac{10 \epsilon_0 A}{\frac{D}{2}} = \frac{20 \epsilon_0 A}{D} \]

### Step 5: Relate the new capacitance to the initial capacitance
Since the initial capacitance \( C \) was:
\[ C = \frac{\epsilon_0 A}{D} \]
We can express the new capacitance \( C' \) in terms of \( C \):
\[ C' = 20 C \]

### Step 6: Calculate the new potential difference
Since the charge \( Q \) remains constant, we can relate the new potential difference \( V' \) to the new capacitance:
\[ Q = C' \cdot V' \implies V' = \frac{Q}{C'} \]
Substituting \( Q = C \cdot V \) and \( C' = 20C \):
\[ V' = \frac{C \cdot V}{20C} = \frac{V}{20} \]

### Step 7: Calculate the percentage decrease in potential
The percentage decrease in potential can be calculated as:
\[ \text{Percentage Decrease} = \frac{V - V'}{V} \times 100 \]
Substituting \( V' = \frac{V}{20} \):
\[ \text{Percentage Decrease} = \frac{V - \frac{V}{20}}{V} \times 100 = \frac{20V - V}{20V} \times 100 = \frac{19V}{20V} \times 100 = 95\% \]

### Conclusion
Thus, the potential difference between the plates decreases by 95%.