A capacitor of capacitance `C_(0)` is charged to a patential `V_(0)` and then isolated. A small capacitor C is then charged from `C_(0)`, discharged and charge again, the process being is decreased to V. Find the value of C.

Explanation

To solve the problem step by step, we need to analyze the charging and discharging process of the capacitors involved.

### Step 1: Initial Charge on Capacitor C₀
The initial charge \( Q_0 \) on the capacitor \( C_0 \) when charged to a potential \( V_0 \) is given by the formula:
\[ Q_0 = C_0 V_0 \]

### Step 2: Charge Transfer to Capacitor C
When the small capacitor \( C \) is charged from \( C_0 \), the charge on \( C \) after the first charging process can be expressed as:
\[ Q_1 = \frac{C_0}{C + C_0} Q_0 \]
Substituting \( Q_0 \):
\[ Q_1 = \frac{C_0}{C + C_0} (C_0 V_0) = \frac{C_0^2 V_0}{C + C_0} \]

### Step 3: Potential After First Charging
The potential \( V_1 \) across the capacitor \( C \) after the first charging process can be calculated using:
\[ V_1 = \frac{Q_1}{C} = \frac{C_0^2 V_0}{C(C + C_0)} \]

### Step 4: Repeating the Process
If we repeat the charging and discharging process \( n \) times, we can express the potential after \( n \) processes as:
\[ V_n = \frac{C_0^2}{C + C_0} V_{n-1} \]
Continuing this process, we find that:
\[ V_n = \left(\frac{C_0}{C + C_0}\right)^n V_0 \]

### Step 5: Setting the Final Potential
According to the problem, after \( n \) processes, the potential decreases to \( V \):
\[ V = \left(\frac{C_0}{C + C_0}\right)^n V_0 \]

### Step 6: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{C_0}{C + C_0} = \left(\frac{V}{V_0}\right)^{\frac{1}{n}} \]

### Step 7: Solving for C
Cross-multiplying and rearranging to isolate \( C \):
\[ C_0 = \left(\frac{V}{V_0}\right)^{\frac{1}{n}} (C + C_0) \]
\[ C_0 = \left(\frac{V}{V_0}\right)^{\frac{1}{n}} C + \left(\frac{V}{V_0}\right)^{\frac{1}{n}} C_0 \]
\[ C_0 - \left(\frac{V}{V_0}\right)^{\frac{1}{n}} C_0 = \left(\frac{V}{V_0}\right)^{\frac{1}{n}} C \]
\[ C_0 \left(1 - \left(\frac{V}{V_0}\right)^{\frac{1}{n}}\right) = \left(\frac{V}{V_0}\right)^{\frac{1}{n}} C \]
Finally, solving for \( C \):
\[ C = C_0 \frac{1 - \left(\frac{V}{V_0}\right)^{\frac{1}{n}}}{\left(\frac{V}{V_0}\right)^{\frac{1}{n}}} \]

### Final Result
Thus, the value of \( C \) is:
\[ C = C_0 \left(\frac{V_0}{V} - 1\right) \]