Question

A point source q1=20nC is located at S(1,4).Determine some point near S, where the magnitude of the electric field E is equal to 30V/m

Answer 1

Electric field can be given as

$E=\frac{kq}{r^2}\\30=\frac{kq}{r^2}\\r^2=\frac{9\times20}{30}\\r=2.45m$

From 20nC charge at distance 2.45m in every direction electric field Will be 30V/m.

Some point where electric field will be 30V/m can be (1, 6.45) (3.45, 4) (-1.45, 4) (1, 1.55)