Answer:
Explanation:
Given
First capacitor magnitude [tex]C=1.75\ \mu F[/tex]
Second capacitor magnitude [tex]C=6.00\ \mu F[/tex]
Voltage of battery [tex]V=3.00\ V[/tex]
Both capacitor are connected in series so net capacitor is given by
[tex]\frac{1}{C_{net}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}[/tex]
[tex]\frac{1}{C_{net}}=\frac{1}{1.75}+\frac{1}{6}[/tex]
[tex]C_{net}=\frac{6\times 1.75}{1.75+6}[/tex]
[tex]C_{net}=1.35\ \mu F[/tex]
So charge Across each capacitor is given by
[tex]Q=C_{net}\times V[/tex]
[tex]Q=1.35\times 10^{-6}\times 3[/tex]
[tex]Q=4.064\ \mu C[/tex]
