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.

Explanation

The electric field between the plates is
`E=(Q)/(A epsilon_(0))=Q_(0)/(A epsilon_(0))e^(-t//RC)`
`E=E_(0)e^(-t//RC)`
In the given problem, `E=(1)/(3)E_(0) at t =4.4(mu)s.
Thus,`1/3=e^-((4.4(mu)s)/(RC))`
`or, (4.4(mu)s)/(RC)=In 3=1.1`
`R=(4.4(mu)s)/(1.1xx2.0(mu)F)=2.0(Omega).`