A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V. A 5.0-µF capacitor is similarly charged so that the potential difference between its plates is 5.0 V. The two charged capacitors are then connected to each other in parallel with positive plate connected to positive plate and negative plate connected to negative plate. Find the charch that flows from one capacitor to the other when the capacitors are connected (17 microcoulombs) and the energy that is dissipated when the two capacitors are connected together (42 microjoules)

Explanation:

Given that,

First Capacitor is 10 µF

C_1 = 10 µF

Potential difference is

V_1 = 10 V.

The charge on the plate is

q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC

q_1 = 100 µC

A second capacitor is 5 µF

C_2 = 5 µF

Potential difference is

V_2 = 5V.

Then, the charge on the capacitor 2 is.

q_2 = C_2 × V_2

q_2 = 5µF × 5 = 25 µC

Then, the average capacitance is

q = (q_1 + q_2) / 2

q = (25 + 100) / 2

q = 62.5µC

B. The two capacitor are connected together, then the equivalent capacitance is

Ceq = C_1 + C_2.

Ceq = 10 µF + 5 µF.

Ceq = 15 µF.

The average voltage is

V = (V_1 + V_2) / 2

V = (10 + 5)/2

V = 15 / 2 = 7.5V

Energy dissipated is

U = ½Ceq•V²

U = ½ × 15 × 10^-6 × 7.5²

U = 4.22 × 10^-4 J

U = 422 × 10^-6

U = 422 µJ

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