Question

once we connect two charged plates the charge on them1 will
decrease exponentially because the charges flow from one plate to
the other due to the potential difference between the two plates.
The final answer we arrived today for how much charge Q(t) is left
on the plates after closing the switch at t = 0 was Q(t) =
Q_{0}e ^{− t/RC} , where Q_{0} is the
initial charge on the capacitor plates, R is the resistance of the
light bulb, and C is the capacitance of the capacitor.

(a) Show that the the time-constant RC has units of time. Comment on why it must have units of time for the above answer to make sense

(b) Show that if Q(t) decays exponentially, the current flowing through the circuit I(t) will decay exponentially too with the same time-constant RC.

(c) Assuming that the brightness of the lightbulb is proportional to the current through it, show that we can estimate the time-constant from measuring the (exponential) decay of the bulb brightness as a function of time without knowing the constant of proportionality.

Answer #1

A battery with potential different E charges an ideal circular
parallel-plate capacitor of capacitance C, plate radius r0 and
separation between the plates d, through a wire with resistance R.
The total charge on each plate as a function of time is : Q(t) =
CE(1-eˆ(-t/RC)). Consider the surface charge density uniform on the
plates.
1. Find the electric flux between the plates as a function of
time.
2. The rate of change of the electric flux between the plates...

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