A capacitor with plate separation d is charged to V volts. The battery is disconnected and a dielectric slab of thickness `(d)/(2)` and dielectric constant '2' is inserted between the plates. The potential difference across its terminals becomes

Options

A. V
B. 2 V
C. `(4V)/(3)`
D. `(3V)/(4)`

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understanding the Initial Conditions
A capacitor is charged to a voltage \( V \) volts. When charged, the charge \( Q \) on the capacitor can be expressed as:
\[ Q = C \cdot V \]
where \( C \) is the capacitance of the capacitor.

### Step 2: Disconnecting the Battery
Once the battery is disconnected, the charge \( Q \) remains constant. This means that the charge on the capacitor does not change when we insert the dielectric slab.

### Step 3: Inserting the Dielectric Slab
A dielectric slab of thickness \( \frac{d}{2} \) and dielectric constant \( K = 2 \) is inserted between the plates of the capacitor. The new configuration of the capacitor includes two regions:
1. The region with the dielectric slab (thickness \( \frac{d}{2} \)).
2. The region without the dielectric (thickness \( \frac{d}{2} \)).

### Step 4: Calculating the New Capacitance
The capacitance of a capacitor can be calculated using the formula:
\[ C = \frac{\varepsilon_0 A}{d} \]
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

When the dielectric slab is inserted, the total capacitance \( C' \) can be calculated as the sum of the capacitances of the two regions:

1. **Capacitance of the region with the dielectric slab:**
\[ C_1 = \frac{K \varepsilon_0 A}{\frac{d}{2}} = \frac{2 \varepsilon_0 A}{\frac{d}{2}} = \frac{4 \varepsilon_0 A}{d} \]

2. **Capacitance of the region without the dielectric:**
\[ C_2 = \frac{\varepsilon_0 A}{\frac{d}{2}} = \frac{2 \varepsilon_0 A}{d} \]

### Step 5: Finding the Total Capacitance
Since the two capacitances are in series, the total capacitance \( C' \) can be calculated using the formula for capacitors in series:
\[ \frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values:
\[ \frac{1}{C'} = \frac{1}{\frac{4 \varepsilon_0 A}{d}} + \frac{1}{\frac{2 \varepsilon_0 A}{d}} \]
\[ \frac{1}{C'} = \frac{d}{4 \varepsilon_0 A} + \frac{d}{2 \varepsilon_0 A} = \frac{d}{4 \varepsilon_0 A} + \frac{2d}{4 \varepsilon_0 A} = \frac{3d}{4 \varepsilon_0 A} \]
Thus,
\[ C' = \frac{4 \varepsilon_0 A}{3d} \]

### Step 6: Calculating the New Potential Difference
The new potential difference \( V' \) across the capacitor can be expressed as:
\[ V' = \frac{Q}{C'} \]
Substituting \( C' \):
\[ V' = \frac{Q}{\frac{4 \varepsilon_0 A}{3d}} = \frac{3dQ}{4 \varepsilon_0 A} \]
Since \( Q = C \cdot V \) and \( C = \frac{\varepsilon_0 A}{d} \), we can substitute \( Q \):
\[ V' = \frac{3d \left(\frac{\varepsilon_0 A}{d} V\right)}{4 \varepsilon_0 A} = \frac{3V}{4} \]

### Conclusion
The new potential difference across the terminals of the capacitor after inserting the dielectric slab is:
\[ V' = \frac{3V}{4} \]