Charge on a Charging Capacitor as a Percentage of the Maximum Charge

A 4.0 μF capacitor is connected in series with a 2.0 kΩ resistor across a 20-V DC source and an open switch. 
If the switch is closed at t = 0s, what is the charge on the capacitor at 9ms?

This is an RC circuit. When you see the letters R and C next to each other, you should be thingking Time COnstant τ. This value is equal to the resistance, R, multiplied by the capacitance, C. τ=RC

τ = 4.0 μF x 2.0 kΩ = (0.000004)(2,000) = 0.008 = 8 ms

This means that at 8 ms, the capacior will be 63.2% charged, it will have 63.2% of its maximum voltage and its current will be 26.7% of the maximum.

So what about at 9ms? Well theres a simple equation that will help us find that out. 

Do you know which one it is...? It's this one:

The charge, Q, on the capacitor at 9 ms is 0.000054 C

So then what percentage of the maximum charge is on the capacitor at this point? Well let's first find out the maximum charge that can be placed on this capacitor at this voltage using the equation C = Q/V : Q = CV = 4.0 μF x 20-V = (0.000004)(20) = 0.00008 C. 

Now we divide the charge on the capacitor at 9 ms by the maximum charge that can be on the capacitor at 20-V and multiply by 100%: (5.4/8) x 100% = 67.5%