Two identical capacitors are connected in series. Charge on each capacitor is `q_0`. A dielectric slab is now introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge in each capacitor will now be

Options

A. `(2q_0)/( 1+(1)/(K))`
B. `(q_0)/(1 +(1)/(K))`
C. `(2q_0)/(1+K) `
D. `(q_0)/(1+K)`

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the Initial Setup
We have two identical capacitors, C1 and C2, connected in series. The charge on each capacitor is given as \( q_0 \). The voltage across the series combination of capacitors can be expressed as:
\[ V = V_1 + V_2 \]
where \( V_1 \) and \( V_2 \) are the voltages across capacitors C1 and C2, respectively.

### Step 2: Relate Charge and Voltage
For capacitors in series, the charge on each capacitor is the same, so:
\[ V_1 = \frac{q_0}{C} \quad \text{and} \quad V_2 = \frac{q_0}{C} \]
Thus, the total voltage across the series combination is:
\[ V = \frac{q_0}{C} + \frac{q_0}{C} = \frac{2q_0}{C} \]

### Step 3: Introduce the Dielectric
Now, a dielectric slab with dielectric constant \( K \) is introduced between the plates of one of the capacitors (let's say C2). The capacitance of C2 will now change to:
\[ C_2' = K \cdot C \]

### Step 4: Analyze the New Configuration
Now, the new charge on C1 remains \( q' \) (which we will find), and the charge on C2 (with the dielectric) will be \( q' \) as well. The voltage across C2 with the dielectric becomes:
\[ V_2' = \frac{q'}{K \cdot C} \]
Thus, the total voltage across the series combination is now:
\[ V = \frac{q'}{C} + \frac{q'}{K \cdot C} \]

### Step 5: Set Up the Equation
We can factor out \( q' \):
\[ V = q' \left( \frac{1}{C} + \frac{1}{K \cdot C} \right) = q' \left( \frac{1 + \frac{1}{K}}{C} \right) \]
This simplifies to:
\[ V = \frac{q'}{C} \left( 1 + \frac{1}{K} \right) \]

### Step 6: Relate New Charge to Old Charge
From the initial setup, we know:
\[ V = \frac{2q_0}{C} \]
Equating the two expressions for \( V \):
\[ \frac{2q_0}{C} = \frac{q'}{C} \left( 1 + \frac{1}{K} \right) \]
Cancelling \( C \) from both sides gives:
\[ 2q_0 = q' \left( 1 + \frac{1}{K} \right) \]
Solving for \( q' \):
\[ q' = \frac{2q_0}{1 + \frac{1}{K}} = \frac{2q_0 K}{K + 1} \]

### Conclusion
The new charge on each capacitor after the introduction of the dielectric slab is:
\[ q' = \frac{2q_0 K}{K + 1} \]