Question

Two charges are placed on the x axis. One charge (q1= +8.5 µC) is at x1= +3.0 cm and the other (q2= - 21 µC) is at x1= +9.0 cm. Find the net electric field (magnitudeand direction) at
(a) x = 0 cm
(b) x = +6.0 cm

Answer 1

a).For x=0cm
E=(1/4πε₀)[(8.5/9)-(21/81)]*10^-6=(1/4πε₀)[(76.5-21)/81]*10^-6=(1/4πε₀)*0.685*10^-6 {field due to q1 is outward while that due to q2 is inward.}

b).Here the field due to both the charges is towards q2.

So E=(1/4πε₀)[(8.5/9)+(21/9)]*10^-6=(1/4πε₀)*3.28*10^-6