**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 Q_{1} is the charge on the first capacitor and Q_{2} on the second, then

Q_{1} = C_{1}V which implies, Q_{1} = 2 × 120 = 240 μF … (1)

Q_{2} = C_{2}V which implies, Q_{2} = 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

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