26: Electric Field due to Multiple Point Charges

INTRO:
Two point charges are placed on the x axis. The first charge, q₁ = 8.00 nC , is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q₂ = 6.00 nC , is placed a distance 9.00 m from the origin along the negative x axis.

Part A: 
Calculate the electric field at point A, located at coordinates (0 m, 12.0m).
Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures.

SOLUTION:
Find the contributions to the electric field at point A separately for q₁ and q₂, then add them together using vector addition to find the total electric field at that point. You will need to use the Pythagorean theorem to find the distance of each charge from point A.

Triangle 1 ⇒ ΔOAq₁
a = 12 m; b = 16 m
∴c = 20 m
Triangle 2 ⇒ ΔOAq₂
a = 12 m; b = 9 m
∴c = 15 m

Now find the contributions to the electric field at point A
E_A1 = K⋅q₁/r² = K⋅(8.00 nC)/(20 m)² = 0.1798 N/C & points in direction <-16, 12>
E_A2 = K⋅q₂/r² = K⋅(6.00 nC)/(15 m)² = 0.2397 N/C & points in direction <9, 12>

Now calculate the x and y components of both electric fields 
to get the angle for E_A1, θ = tan^-1(12/16) = 36.87°
E_A1x = E_A1 cos(θ) = 0.1798 N/C⋅0.8 = 0.14384 N/C
E_A1y = E_A1 sin(θ) = 0.1798 N/C⋅0.6 = 0.10788 N/C
E_A1 = <-0.14384, 0.10788> N/C
to get the angle for E_A2, θ = tan^-1(12/9) = 53.13°
E_A2x = E_A2 cos(θ) = 0.2397 N/C⋅0.6 = 0.14382 N/C
E_A2y = E_A2 sin(θ) = 0.2397 N/C⋅0.8 = 0.19176 N/C 
E_A2 = <0.14382, 0.19176> N/C

SO, now that you have the vector components of the two electric fields, you are able to use vector addition to determine the net electric field at point A
E_NET = E_A1 + E_A2 = <-0.144, 0.108> N/C + <0.144 0.192> N/C = <0, 0.300> N/C

Part B:
An unknown additional charge q₃ is now placed at point B, located at coordinates (0 m, 15.0m).

Find the magnitude and sign of q₃ needed to make the total electric field at point A equal to zero.
Express your answer in nanocoulombs to three significant figures.

SOLUTION:
luckily q₃ is in line with point A so we don't need to worry about x and y components as with the previous part. 
givens:
r_AB = 3 m
E_AB = <0, 0.300> N/C

E_AB = K q₃/r_AB²
q₃ = E_AB⋅r_AB² / K
→0.300 N/C ⋅ (3 m)² / [4πε₀]
q₃ = 3×10^-10 C = 0.300 nC