Question

The electric field between the plates of a parallel-plate capacitor of capacitance`2.0(mu)F` drops to one third of its initial value in `(4.4 mu)s` when the plates are connected by a thin wire. Find the resistance of the wire.

Answer 1

`q = q_(0) //3`, form `q = q_(0) e^(-t//tau)` we have
`q_(0) //3 = q_(0) r^(-t//tau)`
or `t = r tau in 3 = RC` in 3 or `R = (t)/(C in 3) = (4.4xx10^(-3))/(2xx10^(-6)in3) = (2200)/(in 3) Omega`