Answer:
[tex]8.0\mu C[/tex]
Explanation:
We are given that
[tex]f=1.6 Hz[/tex]
[tex]q=3.0\mu C=3.0\times 10^{-6} C[/tex]
[tex]1\mu C=10^{-6} C[/tex]
Current,I=[tex]75\mu A=75\times 10^{-6} A[/tex]
[tex]1\mu A=10^{-6} A[/tex]
We have to find the maximum charge of the capacitor.
Charge on the capacitor,[tex]q=q_0cos\omega t[/tex]
[tex]\omega=2\pi f=2\pi\times 1.6=3.2\pi rad/s[/tex]
[tex]3\times 10^{-6}=q_0cos3.2\pi t[/tex]....(1)
[tex]I=\frac{dq}{dt}=-q_0\omega sin\omega t[/tex]
[tex]75\times 10^{-6}=-q_0(3.2\pi)sin3.2\pi t[/tex]....(2)
Equation (2) divided by equation (1)
[tex]-3.2\pi tan3.2\pi t=\frac{75\times 10^{-6}}{3\times 10^{-6}}=25[/tex]
[tex]tan3.2\pi t=-\frac{25}{3.2\pi}=-2.488[/tex]
[tex]3.2\pi t=tan^{-1}(-2.488)=-1.188rad[/tex]
[tex]q_0=\frac{q}{cos\omega t}=\frac{3\times 10^{-6}}{cos(-1.188)}=8.0\times 10^{-6}=8\mu C[/tex]
Hence, the maximum charge of the capacitor=[tex]8.0\mu C[/tex]
