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

Explanation

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

### Step 1: Understand the Initial Setup
We have two identical capacitors, both with capacitance \( C \), connected in series to a \( 100V \) battery.

### Step 2: Calculate Initial Voltage Distribution
In a series connection, the total voltage is divided across the capacitors. Since both capacitors are identical, the voltage across each capacitor is the same. Therefore, the potential difference across each capacitor before inserting the dielectric is:
\[ V_1 = V_2 = \frac{100V}{2} = 50V \]

### Step 3: Insert the Dielectric
Now, we insert a dielectric slab with a dielectric constant \( K = 4.0 \) into the second capacitor. The capacitance of the second capacitor becomes:
\[ C_2' = K \cdot C = 4C \]

### Step 4: Set Up the Voltage Equation
In a series circuit, the charge \( Q \) on both capacitors is the same. The relationship between charge, capacitance, and voltage is given by:
\[ Q = C_1 V_1 = C_2' V_2 \]
Where:
- \( C_1 = C \) (first capacitor)
- \( C_2' = 4C \) (second capacitor with dielectric)

### Step 5: Relate the Voltages
From the charge equations, we can express \( V_1 \) and \( V_2 \):
\[ V_1 = \frac{Q}{C} \quad \text{and} \quad V_2 = \frac{Q}{4C} \]
From these, we can relate \( V_1 \) and \( V_2 \):
\[ V_1 = 4V_2 \]

### Step 6: Write the Total Voltage Equation
The total voltage across both capacitors is still \( 100V \):
\[ V_1 + V_2 = 100V \]

### Step 7: Substitute and Solve
Substituting \( V_1 = 4V_2 \) into the total voltage equation:
\[ 4V_2 + V_2 = 100V \]
\[ 5V_2 = 100V \]
\[ V_2 = \frac{100V}{5} = 20V \]

### Step 8: Calculate \( V_1 \)
Now, substituting \( V_2 \) back to find \( V_1 \):
\[ V_1 = 4V_2 = 4 \times 20V = 80V \]

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

### Conclusion
The potential difference across the capacitors will now be respectively \( 80V \) and \( 20V \).

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