Question

1.electric potensionl at any point due to a point charge is 60volt and intensity of electric field is 20N/C.calculate the strenght of charge and distance of that point.​

Answer 1

Answer:

distance= 3m

strength of charge = 20 nC

Explanation:

potential = kq/r = 60 V

electric field = kq/r^2 = 20 N/C

solve both the equations to find r and q