The electrodes of a capacitor of capacitance `C = 2.00 muF` carry opposite charges `q_(0) = 1.00 mC`. Then the electores are interconnected through a resistanace `R = 5.0 M Omega`. Find:
(a) the charge flowing thorugh that resistance during a time interval `tau = 2.00 s`,
(b) the amount of heat generated in the resistance during the same interval.

Explanation

(a) Let at any moment of time, charge on the plates be `(q_(0) - q)` then current through teh resistor, `i = -(d (q_(0) - q))/(dt)`, because the capacitor is discharging.
or, `i = (dq)/(dt)`
Now, applying loop rule in the circuit,
`iR = (q_(0) - q)/(C) = 0`
or, `(dq)/(dt) R = (q_(0) - q)/(C)`
or, `(dq)/(q_(0) - q) = (1)/(RC) dt`
At `t = 0, q = 0` and at `t = tau, q = q`
So, In `(q_(0) - q)/(q_(0)) = (-tau)/(RC)`
Thus `q = q_(0) (1 - e^(-tau//R C)) = 0.18 mC` (b) Amount of heat generated = decrement in capacitance energy
`= (1)/(2) (q_(0)^(2))/(C) - (1)/(2) ([q_(0) - q_(0) (1 - e^(-tau//R C))]^(2))/(C)`
`= (1)/(2) (q_(0)^(2))/(C)[1 - e^(- (2tau)/(R C))] = 82 mJ`