A capacitor charged to `50 V` is discharged by connecting the two plates at `t=0`.If the potential difference across the plates drops to 1.0 V at t=10 ms,what will be the potential difference at t=20 ms?

Explanation

The potential difference at time t is given by
`V=Q//C=(Q_(0)//C)e^(-t//RC)`.
`or, V=V_(0)e^(-t//RC)`
Acorrding to the given data,
`1 V=(50V)e^(-10 ms//RC)`
`or, e^(-10 ms//RC)=(1)/(50).`
The potential dofference at t-20 ms is
`V=V_(0)e^(-t//RC)`
(50V)e^(-20ms//RC)=(50V)(e^(-10ms//RC))^(2).`
`0.02V.`