Answer:[tex]Q=31.368\ \mu C[/tex]
Explanation:
Given
capacitance [tex]C=30\ muF[/tex]
Resistance [tex]R=110\ Omega [/tex]
Applied Voltage [tex]V=4\ V[/tex]
[tex]t=0.001\ s[/tex]
Charge on Capacitor in a R-C circuit is given by
[tex]Q=Q_0(1-e^{\frac{-t}{RC}})[/tex]
[tex]Q=CV(1-e^{\frac{-t}{RC}})[/tex]
[tex]Q=30\times 10^{-6}\times 4(1-e^{\frac{-0.001}{110\times 30\times 10^{-6}}})[/tex]
[tex]Q=120\times 10^{-6}\times 0.2614[/tex]
[tex]Q=31.368\ \mu C[/tex]
