Question

A +5.7C X 10 to the -6 power point charge is on the x-axis at x=-3.0m,and a +2.0C X 10 to the -6 power point charge is on the x-axis at x=+1.0m.Determine the net electric field (magnitude and direction) on the y-axis at y=+2.0m.

Answer 1

If point P (0, +2) and point A(-3, 0), then PA = sqrt(2^2+3^2) =sqrt(13) = 3.6 m,
If point P (0, +2) and point B(+2, 0), then PB = sqrt(2^2+1^2) =sqrt(5) =2.236 m.
Electric field due to the point charge q1 is
E1=(1/4πε°)• q1/(PA)^2 =9•10^9•5.7•10^-6/13=3946 V/m
An angle PAB=α, sin α =2/3.6 =0.555, cos α = 3/3.6 =0.833
Projections of E1 are
E1(x) =E1•cos α = 3946•0.833= 3287 V/m (directed +x)
E1(y) =E1•sin α = 3946•0.555=2192 V/m (directed +y)

Electric field due to the point charge q2 is
E2=(1/4πε°)• q2/(PB)^2 =9•10^9•2•10^-6/5=3600 V/m
An angle PBA=β, sin β =2/2.236 =0.894, cos β= 1/2.236 =0.447
Projections of E2 are
E2(x) =E2•cos β = 3600•0.894= 3218 V/m (directed -x)
E2(y) =E2•sin β = 3600•0.447=1609 V/m (directed +y)
E(x) = E1(x)-E2(x) =3287 – 3218=69V/m
E(y) = E1(y)+E2(y) =2192 + 1609 = 3801V/m
E =sqrt(E(x)^2+ E(y)^2)) = 3802 V/m