A charge of `20 muC` produces an electric field. Two points are 10 cm and 5 cm from this charge. Find the values of potentials at these points and calculate work done to take an electron from one point to the other.

Answer

1.8xx10^(6)V, 3.6xx10^(6) V, 2.88xx10^(-13)J`

Explanation

Here, `q = 20 muC = 20xx10^(-6)C`.
If A and B are two points at distances `r_(1) = 10 cm`
`= 0.1 m and r_(2) = 5cm = 0.05m`
then `V_(A) = (q)/(4pi in_(0) r_(1)) = (20xx10^(-6)xx9xx10^(9))/(0.1)`
`= 1.8xx10^(6) V`
`V_(B) = (q)/(4pi in_(0) r_(2)) = (20xx10^(-6)xx9xx10^(9))/(0.05)`
`= 3.6xx10^(6) V`
`W_(AB) = (V_(B) - V_(A))e`
`= (3.6xx10^(6) - 1.8xx10^(6))xx1.6xx10^(-19)`
`= 2.88xx10^(-13) J`