A  $3\text{\hspace{0.17em}μF}$ capacitor is charged to a potential of $300\text{\hspace{0.17em}V}$ and a $2\text{\hspace{0.17em}μF}$ capacitor is charged to $200\text{\hspace{0.17em}V}$. 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?

Charge on the first capacitor  $Q_1=C_1{V}_1=3\times {10}^{-6}\times 300$
$=900\times {10}^{-6}\text{\hspace{0.17em}C}$ 
Charge on the second one  $Q_2=C_2{V}_2=2\times {10}^{-6}\times 200$
$=400\times {10}^{-6}\text{\hspace{0.17em}C}$ 
When plates with opposite polarity are joint, charge will be  $900-400=500\mu \text{C}$
The capacity will be  $C=C_1+C_2=5\text{\hspace{0.17em}μF}$
Common potential  $=\frac{500\mu \text{C}}{5}$
$=100\text{\hspace{0.17em}V}$ 
Final charge on first capacitor  $=300\mu \text{C}$
Charge flow  $=900\mu \text{C}-300\mu \text{C}$
$=600\mu \text{C}$