Answer:
Q= 22.3 μC
Explanation:
Given that
C= 20 μF
Qo= 100 μC
R= 100 kΩ
t= 3 s
T= R C
T= 100 x 1000 x 20 x 10⁻⁶ s
T=2 s
We know that charge on the capacitor is given as
[tex]Q=Q_0e^{\dfrac{-t}{T}}[/tex]
[tex]Q=100\times 10^{-6}\times e^{\dfrac{-3}{2}}[/tex]
Q= 0.0000223 C
Q= 22.3 μC
