Verified answer

Explanation:

The initial energy stored in the 4mf capacitor is:

E = 1/2 * C * V^2

E = 1/2 * 4 * 10^-6 * (200)^2

E = 8 * 10^-2 J (to 2 decimal places)

When the 4mf capacitor is connected to the 2mf capacitor, they form a circuit with a total capacitance of 6mf. The total charge on the capacitors will stay the same, but the voltage across them will be split in proportion to their capacitances, so:

Q = C1 * V1 = C2 * V2

V2 = C1/C2 * V1

V1 + V2 = 200

Substituting C1 = 4mf and C2 = 2mf gives:

V1 = 2/3 * 200 = 133.33 V

V2 = 1/3 * 200 = 66.67 V

The final energy stored in the capacitors can be calculated using:

E = 1/2 * C * V^2

E = 1/2 * 4 * 10^-6 * (133.33)^2 + 1/2 * 2 * 10^-6 * (66.67)^2

E = 2.963 J (to 3 decimal places)

The loss of energy during the process is the difference between the initial and final energies:

Loss of energy = 8 * 10^-2 - 2.963

Loss of energy = 0.07037 J (to 3 decimal places)