Question is incomplete. Missing part:
Find the charge on the capacitor at the following times:
1) t = 0 mu S
2) t = 1 mu S
3) t = 50 mu S
1) [tex]22.4 \mu C[/tex]
We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:
[tex]C=\frac{Q_0}{V_0}[/tex]
where
C is the capacitance
Q0 is the initial charge stored
V0 is the initial potential difference across the capacitor
When the capacitor is connected to the battery, we have:
[tex]C=4.4\mu F = 4.4\cdot 10^{-6}F[/tex]
[tex]V_0 = 5.1 V[/tex]
Solving for [tex]Q_0[/tex],
[tex]Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C[/tex]
So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.
2) [tex]20.0\mu C[/tex]
To find the charge on the capacitor at any other time t, we use the equation:
[tex]Q(t) = Q_0 e^{-\frac{t}{RC}}[/tex]
where
[tex]Q_0 = 22.4 \mu C[/tex]
t is the time
[tex]R=2.0 \Omega[/tex] is the resistance
[tex]C=4.4\mu F[/tex] is the capacitance
Therefore, at time [tex]t=1 \mu s[/tex], we have:
[tex]Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C[/tex]
3) [tex]0.08 \mu C[/tex]
As before, we use again the equation:
[tex]Q(t) = Q_0 e^{-\frac{t}{RC}}[/tex]
However, here the time to consider is
[tex]t=50 \mu C[/tex]
Substituting into the formula,
[tex]Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C[/tex]
