Capacitor charge in an RC circuit
You should view E as the difference between the current charge of the capacitor and its final value at t=∞. Since the capacitor isn't fully discharged when the pin turns on in your graph, you have some initial charge (or voltage) on the capacitor. 

For most intents and purposes, you can take t=5T as the final value since you'll be within 99%; 1-e^-5=0.993. You should wait 5 time constants (Ts) between toggling the digital pin if you want to see a charge where Vc(0) ≈ 0.

You can plot this in your calculator for a sanity check. Suppose Vcc is 5V and you have 2.3V of charge on the capacitor initially (at t=0), your equation should be:

Vc(t)=(5-2.3)*(1-e^t/T)+2.3. 


You will see the same charging curve.

I'm swapping the terms "charge" and "voltage" a little recklessly here. Voltage = Charge/ Capacitance. They are somewhat interchangeable in terms of this explanation since there is a direct correlation.