To solve the problem step by step, we will calculate the initial energy stored in the first capacitor, find the new voltage after connecting the second capacitor, calculate the energy stored in the system after the connection, and finally determine the loss of energy.
### Step 1: Calculate the initial energy stored in the first capacitor (U1)
The formula for the energy stored in a capacitor is given by:
\[
U = \frac{1}{2} C V^2
\]
Where:
- \( C \) is the capacitance in farads (F)
- \( V \) is the voltage in volts (V)
For the first capacitor:
- \( C_1 = 4 \, \mu F = 4 \times 10^{-6} \, F \)
- \( V_1 = 200 \, V \)
Substituting these values into the formula:
\[
U_1 = \frac{1}{2} \times 4 \times 10^{-6} \times (200)^2
\]
Calculating \( (200)^2 \):
\[
(200)^2 = 40000
\]
Now substituting:
\[
U_1 = \frac{1}{2} \times 4 \times 10^{-6} \times 40000
\]
\[
U_1 = 2 \times 10^{-6} \times 40000 = 8 \times 10^{-2} \, J
\]
### Step 2: Calculate the new voltage (V2) after connecting the second capacitor
When the charged capacitor is connected to an uncharged capacitor, charge conservation applies. The total charge before connection is equal to the total charge after connection.
The charge \( Q \) on the first capacitor is:
\[
Q = C_1 \times V_1 = 4 \times 10^{-6} \times 200 = 8 \times 10^{-4} \, C
\]
After connecting the second capacitor \( C_2 = 2 \, \mu F = 2 \times 10^{-6} \, F \), the total capacitance \( C_{total} \) becomes:
\[
C_{total} = C_1 + C_2 = 4 \times 10^{-6} + 2 \times 10^{-6} = 6 \times 10^{-6} \, F
\]
The new voltage \( V_2 \) can be calculated using the formula:
\[
V_2 = \frac{Q}{C_{total}} = \frac{8 \times 10^{-4}}{6 \times 10^{-6}} = \frac{8}{6} \times 10^{2} = \frac{4}{3} \times 100 = \frac{400}{3} \, V
\]
### Step 3: Calculate the energy stored in the system after connection (U2)
Using the total capacitance and the new voltage:
\[
U_2 = \frac{1}{2} C_{total} V_2^2
\]
Substituting the values:
\[
U_2 = \frac{1}{2} \times 6 \times 10^{-6} \times \left(\frac{400}{3}\right)^2
\]
Calculating \( \left(\frac{400}{3}\right)^2 \):
\[
\left(\frac{400}{3}\right)^2 = \frac{160000}{9}
\]
Now substituting:
\[
U_2 = \frac{1}{2} \times 6 \times 10^{-6} \times \frac{160000}{9}
\]
\[
U_2 = \frac{3 \times 10^{-6} \times 160000}{9} = \frac{480000}{9} \times 10^{-6} = \frac{16}{3} \times 10^{-2} \, J
\]
### Step 4: Calculate the loss of energy (ΔU)
The loss of energy is given by:
\[
\Delta U = U_1 - U_2
\]
Substituting the values we calculated:
\[
\Delta U = 8 \times 10^{-2} - \frac{16}{3} \times 10^{-2}
\]
Finding a common denominator (3):
\[
\Delta U = \frac{24}{3} \times 10^{-2} - \frac{16}{3} \times 10^{-2} = \frac{8}{3} \times 10^{-2} \, J
\]
Calculating the numerical value:
\[
\Delta U \approx 2.67 \times 10^{-2} \, J
\]
### Final Answer
The loss of energy during the process is approximately:
\[
\Delta U \approx 0.0267 \, J
\]
