Value of the 2nd capacitor

Question

A 7.7µF capacitor is charged by a 125 V battery and then disconnected from the battery. When the charged capacitor (C₁) is then placed in a circuit with a second, initially uncharged capacitor C₂, the final voltage on each capacitor is 15 V. What is the value of C₂?

Answer 1

At first, Q = (125 Volts)(7.7E-6 Farads) or 9.625E-4 Coulombs. After placement in a circuit with C_2, V₁ will equal Q₁/C₁ and V₂ will equal Q₂/C₂ . Moreover, Q = Q₁ + Q₂ and V₁ = V₂ = 15.

Then 15 = Q₁/7.7E-6 or Q₁ = 1.155E-4 Coulombs and Q₂ = Q − Q₁ or (9.625E-4 − 1.155E-4) or 8.47E-4 Coulombs.

Finally, V₂ or 15 = Q₂/C₂ or 8.47E-4/C₂, giving C₂ equal to 5.64666666667E-5 Farads equivalent to 56.467

Micro-Farads (μF).