A 3 μF capacitor is charged to a potential of 300 V and a 2 μF capacitor is charged to 200 V. The capacitors are then connected in parallel with plates of opposite polarity joined together. What amount of charge will flow through the connecting wire when the plates are so connected?

A
$1300 μC$
B
$700 μC$
C
$250 μC$
D
$600 μC$

Solution:

Charge on the first capacitor $Q_{1} = C_{1} V_{1} = 3 \times 10^{- 6} \times 300$
$= 900 \times 10^{- 6} C$
Charge on the second one $Q_{2} = C_{2} V_{2} = 2 \times 10^{- 6} \times 200$
$= 400 \times 10^{- 6} C$
When plates with opposite polarity are joint, charge will be $900 - 400 = 500 μ C$
The capacity will be $C = C_{1} + C_{2} = 5 μF$
Common potential $= \frac{500 μ C}{5}$
$= 100 V$
Final charge on first capacitor $= 300 μ C$
Charge flow $= 900 μ C - 300 μ C$
$= 600 μ C$

A $3 μF$ capacitor is charged to a potential of $300 V$ and a $2 μF$ capacitor is charged to $200 V$ . The capacitors are then connected in parallel with plates of opposite polarity joined together. What amount of charge will flow through the connecting wire when the plates are so connected?

Solution:

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