Question

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?

Answer 1

We know that in a discharging circuit charge on the
capacitor at any tiem is given by `q = Qe^(-t//tau).` Because `q prop V`, so
same relation can be written in the form of potential difference.
`V = V_0e^(-t//tau)` where `V_0 = 50V`
Given at` t = 10ms, v =1 V`, putting in the above equation
`1 = 50e^(-10ms//tau) (i)`
Let at `t=20ms`, `V = V_2`, then
`V_2 = 50(e^(-10ms//tau))^2 (ii)`
From Eqs. (i) and (ii),
`V_2 = 50((1)/(50))^2 = 0.02V`
Alternatively 2 : if we take the instant t =10ms as `t = 0`, then
potential difference at this instant, which is 1 V, becomes `V_0.`
And we have to find potential difference at the end of next 10ms.
`V_2 = (1.0)e^(-10ms//tau) (iii)`
From Eqs. (i) and (iii), `V_2 = (1)/(50) V = 0.02V`