Answer:
0.099C
Explanation:
First, we need to get the common potential voltage using the formula
[tex]V=\frac {C_2V_2-C_1V_1}{C_1+C_2}[/tex]
Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then
[tex]C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}[/tex]
Therefore
[tex]V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437[/tex]
Charge, Q is given by CV hence for the first capacitor charge will be [tex]Q_1=C_1V[/tex]
Here, [tex]Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C[/tex]
