There are two indetical capacitor , the first one is uncharged and filled with dielectric of constant K while the other one is charged to potential v having air between its plates, if two capacitors are joined end to end, the common potential will be

Options

A. `(V)/(K - 1)`
B. `(KV)/(K + 1)`
C. `(KV)/(K - 1)`
D. `(V)/(K + 1)`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the Capacitors
We have two identical capacitors:
- Capacitor 1 (C1) is uncharged and filled with a dielectric of constant K.
- Capacitor 2 (C2) is charged to a potential V and has air between its plates.

### Step 2: Determine the Capacitance Values
- The capacitance of Capacitor 1 (C1) with dielectric is given by:
\[ C_1 = K \cdot C \]
where C is the capacitance of the capacitor without the dielectric.

- The capacitance of Capacitor 2 (C2) with air is:
\[ C_2 = C \]

### Step 3: Identify Initial Conditions
- The initial voltage across Capacitor 1 (V1) is:
\[ V_1 = 0 \quad (\text{since it is uncharged}) \]

- The initial voltage across Capacitor 2 (V2) is:
\[ V_2 = V \quad (\text{since it is charged to potential V}) \]

### Step 4: Apply the Formula for Common Potential
When the two capacitors are connected end to end, the common potential (V_common) can be calculated using the formula:
\[ V_{\text{common}} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \]

### Step 5: Substitute the Values
Substituting the values we have:
\[ V_{\text{common}} = \frac{(K \cdot C)(0) + (C)(V)}{(K \cdot C) + C} \]

This simplifies to:
\[ V_{\text{common}} = \frac{0 + CV}{KC + C} \]
\[ V_{\text{common}} = \frac{CV}{C(K + 1)} \]

### Step 6: Simplify the Expression
We can cancel C from the numerator and denominator:
\[ V_{\text{common}} = \frac{V}{K + 1} \]

### Step 7: Conclusion
Thus, the common potential when the two capacitors are connected end to end is:
\[ \boxed{\frac{V}{K + 1}} \]

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