A condenser of capacity `10 mu F` is charged to a potential of `500 V`. Its terminals are then connected to those of an uncharged condenser of capacity `40 mu F`. The loss of energy in connecting them together is.

Options

A. 1 J
B. `2.5 J`
C. 10 J
D. 12 J

Explanation

To solve the problem step by step, we will calculate the loss of energy when two capacitors are connected together.

### Step 1: Calculate the initial charge on the charged capacitor
The first capacitor has a capacitance \( C_1 = 10 \, \mu F \) and is charged to a potential \( V_1 = 500 \, V \).

The charge \( Q_1 \) on the capacitor can be calculated using the formula:
\[ Q_1 = C_1 \times V_1 \]
Substituting the values:
\[ Q_1 = 10 \times 10^{-6} \, F \times 500 \, V = 5000 \, \mu C \]

### Step 2: Determine the final voltage after connecting the uncharged capacitor
The second capacitor has a capacitance \( C_2 = 40 \, \mu F \) and is initially uncharged. When the two capacitors are connected together, charge is conserved.

Let \( V_f \) be the final voltage across both capacitors after they are connected. The total charge before connecting them is equal to the total charge after they are connected:
\[ Q_1 = Q_{total} = (C_1 + C_2) \times V_f \]
Thus,
\[ Q_1 = Q_1 + Q_2 \]
Where \( Q_2 = C_2 \times V_f \).

Substituting the values:
\[ 5000 \, \mu C = (10 \, \mu F + 40 \, \mu F) \times V_f \]
\[ 5000 \, \mu C = 50 \, \mu F \times V_f \]
Now, solving for \( V_f \):
\[ V_f = \frac{5000 \, \mu C}{50 \, \mu F} = 100 \, V \]

### Step 3: Calculate the initial energy stored in the charged capacitor
The initial energy \( U_i \) stored in the first capacitor can be calculated using the formula:
\[ U_i = \frac{1}{2} C_1 V_1^2 \]
Substituting the values:
\[ U_i = \frac{1}{2} \times 10 \times 10^{-6} \times (500)^2 \]
\[ U_i = \frac{1}{2} \times 10 \times 10^{-6} \times 250000 = 1.25 \, J \]

### Step 4: Calculate the final energy stored in both capacitors
The final energy \( U_f \) stored in both capacitors can be calculated as:
\[ U_f = \frac{1}{2} C_1 V_f^2 + \frac{1}{2} C_2 V_f^2 \]
Substituting the values:
\[ U_f = \frac{1}{2} \times 10 \times 10^{-6} \times (100)^2 + \frac{1}{2} \times 40 \times 10^{-6} \times (100)^2 \]
\[ U_f = \frac{1}{2} \times 10 \times 10^{-6} \times 10000 + \frac{1}{2} \times 40 \times 10^{-6} \times 10000 \]
\[ U_f = 0.05 \, J + 0.2 \, J = 0.25 \, J \]

### Step 5: Calculate the loss of energy
The loss of energy \( \Delta U \) when the capacitors are connected is given by:
\[ \Delta U = U_i - U_f \]
Substituting the values:
\[ \Delta U = 1.25 \, J - 0.25 \, J = 1.00 \, J \]

### Final Answer
The loss of energy in connecting the two capacitors together is \( 1.00 \, J \).

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