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

2

Explanation

`q=q_(0)e^(-t//tay) R=epsilon_(0)e^(-t//tau) (t)/(ta) =ln 2 rc =(t)/(ln 3)`
`R =(4.4)/((1.1)2)=2 Omega`.