Problem 16: Two parallel plate capacitors A and B having capacitance of 2 µF and 6 µF are charged separately to the same potential of 120 V. Now positive plate of A is connected to the negative plate of B and the negative plate of A is connected to the positive of B. Find the final charges on each capacitor.

Solution

First we calculate the charge on each capacitor when they are charged separately.

Let Q₁ is the charge on the first capacitor and Q₂ on the second, then

Q₁ = C₁V which implies, Q₁ = 2 × 120 = 240 μF     …     (1)

Q₂ = C₂V which implies, Q₂ = 6 × 120 = 720 μF     …     (2)

Now the capacitors are connected as follow;

Now difference in charges on the capacitors = (720 × 10^-6 – 240 ×10^-6) = 480× 10^-6 C

This charge will adjust on the capacitors according to their capacitances. To find the charge, we should know about the voltage across the capacitors.

Now, equivalent capacitance, C_e = C_A + C_B = 2μ + 6μ = 8 × 10^-6 F

Voltage, V = Q/C_e = (480 × 10^-6)/(8 × 10^-6) = 60 V

Therefore, charge on capacitor A = 2 × 10^-6 × 60 = 120 × 10^-6 = 120 μF

Charge on capacitor B = 6 × 10^-6 × 60 = 360 × 10^-6 = 360 μF