Two identical parallel plate capacitors are connected in series to a battery of `100V`. A dielectric slab of dielectric constant `4.0` is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively

Options

A. `50V, 50V`
B. `80V, 20V`
C. `20V, 80V`
D. `75V, 25V`

Answer

To solve the problem, we need to determine the potential difference across two identical parallel plate capacitors connected in series, with a dielectric slab inserted in the second capacitor. Here’s a step-by-step solution:

### Step 1: Understand the Configuration
We have two identical capacitors, \( C_1 \) and \( C_2 \), connected in series to a battery of \( 100V \). A dielectric slab with a dielectric constant \( K = 4.0 \) is inserted in the second capacitor \( C_2 \).

### Step 2: Determine Capacitance with Dielectric
The capacitance of a capacitor with a dielectric is given by:
\[ C = K \cdot C_0 \]
where \( C_0 \) is the capacitance without the dielectric. Since both capacitors are identical, we can denote:
- \( C_1 = C_0 \)
- \( C_2 = K \cdot C_0 = 4C_0 \)

### Step 3: Charge on Capacitors in Series
In a series connection, the charge \( Q \) on both capacitors is the same:
\[ Q = C_1 V_1 = C_2 V_2 \]
Substituting the values of capacitance:
\[ Q = C_0 V_1 = 4C_0 V_2 \]

### Step 4: Relate Voltages
From the equation above, we can express \( V_1 \) in terms of \( V_2 \):
\[ V_1 = 4 V_2 \]

### Step 5: Total Voltage Equation
The total voltage across the series combination is the sum of the individual voltages:
\[ V = V_1 + V_2 \]
Given that \( V = 100V \), we can substitute \( V_1 \):
\[ 100 = 4V_2 + V_2 \]
This simplifies to:
\[ 100 = 5V_2 \]

### Step 6: Solve for \( V_2 \)
Now, we can solve for \( V_2 \):
\[ V_2 = \frac{100}{5} = 20V \]

### Step 7: Solve for \( V_1 \)
Using the relationship \( V_1 = 4V_2 \):
\[ V_1 = 4 \times 20 = 80V \]

### Final Result
Thus, the potential differences across the capacitors are:
- \( V_1 = 80V \)
- \( V_2 = 20V \)

### Summary
The potential difference across the first capacitor is \( 80V \) and across the second capacitor is \( 20V \).

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