two point charges `2 muC` and `-4 muC` are situated at points `(-2m, 0m)` and `(2m, 0m)` respectively. Find out potential at point `C( 4m, 0m)` and `D (0 m, sqrt(5) m)`.

Explanation

Potential at point C
`V_(C)=V_(q_(1))+V_(q_(2))=(K(2muC))/6+(K(-4muC))/2=(9xx10^(9)xx2xx10^(-6))/6-(9xx10^(9)xx4xx10^(-6))/2=-15000 V`.
Similarly, `V_(D)=V_(q_(1))+V_(q_(2))=(K(2muC))/sqrt((sqrt(5))^(2)+2^(2))+(K(-4muC))/sqrt((sqrt(5))^(2)+2^(2))=(K(2 muC))/3+(K(-4muC))/3=-6000 V`.