A capacitor is charge until it stores an energy of 1J. A second uncharged capacitor is connected to it, so that charge districutes equally. The final energy stored in the second capacitor is

Options

A. 1J
B. `0.5 J`
C. `0.25 J`
D. `0.125J`

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Initial Condition
We have a capacitor (let's call it C1) that is charged to store an energy of 1 Joule. The energy stored in a capacitor can be expressed using the formula:
\[ U = \frac{Q^2}{2C} \]
where \( U \) is the energy stored, \( Q \) is the charge, and \( C \) is the capacitance.

### Step 2: Relate Energy to Charge and Capacitance
From the given information, we can set up the equation:
\[ \frac{Q^2}{2C_1} = 1 \quad \text{(1)} \]
This equation relates the charge \( Q \) on the capacitor C1 to its capacitance \( C_1 \).

### Step 3: Connect the Second Capacitor
Now, a second uncharged capacitor (C2) is connected to C1. When they are connected, the charge will redistribute equally between the two capacitors. Assuming that both capacitors have the same capacitance \( C_1 = C_2 \), the total charge \( Q \) will be shared equally:
\[ Q_1 = \frac{Q}{2} \quad \text{and} \quad Q_2 = \frac{Q}{2} \]
where \( Q_1 \) is the charge on C1 and \( Q_2 \) is the charge on C2 after redistribution.

### Step 4: Calculate the Energy Stored in the Second Capacitor
The energy stored in the second capacitor (C2) can be calculated using the same energy formula:
\[ U_2 = \frac{Q_2^2}{2C_2} \]
Substituting \( Q_2 = \frac{Q}{2} \) and \( C_2 = C_1 \):
\[ U_2 = \frac{\left(\frac{Q}{2}\right)^2}{2C_1} \]
\[ U_2 = \frac{Q^2/4}{2C_1} = \frac{Q^2}{8C_1} \quad \text{(2)} \]

### Step 5: Substitute from Equation (1)
From equation (1), we know:
\[ \frac{Q^2}{2C_1} = 1 \implies Q^2 = 2C_1 \]
Now substitute \( Q^2 \) into equation (2):
\[ U_2 = \frac{2C_1}{8C_1} = \frac{2}{8} = \frac{1}{4} \quad \text{(3)} \]

### Step 6: Final Result
Thus, the final energy stored in the second capacitor is:
\[ U_2 = 0.25 \text{ Joules} \]

### Conclusion
The final energy stored in the second capacitor after charge redistribution is **0.25 Joules**.

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