Three point charges q1 = +2 μC, q2 = −3 μC, and q3 = +4 μC are located at the corners of a right angle triangle as shown below. Find the magnitude and direction of the resultant force on q3.

The magnitude and direction of the resultant force on q3 is;
F_net = 32.2 N in the direction of q3
We are given;
q1 = +2 μC = 2 × 10^(-6) C
q2 = -3 μC = -3 × 10^(-6) C
q3 = +4 μC = 4 × 10^(-6) C
Now,from the given image, charges q1 and q2 attract charge q3.
Formula for force on a charge is;
F = (kq1 * q2)/r²
Where;
k = 9 × 10^(9) N.m²/C²
Thus;
F_y = kq1*q3/(r1)²
We have;
q1 = 2 × 10^(-6) C
r1 = 5 cm = 0.05 m
q3 = 4 × 10^(-6) C
Thus;
F_y = (9 × 10^(9) × 2 × 10^(-6) × 4 × 10^(-6))/(0.05^(2))
F_y = 28.8 N
Similarly;
F_x = kq2*q3/(r2)²
r2 = 5/(tan 30)
r2 = 8.66 cm = 0.0866 m
F_x = (9 × 10^(9) × -3 × 10^(-6) × 4 × 10^(-6))/(0.0866^(2))
F_x = -14.4 N
Thus, magnitude of resultant force is;
F_net = √((28.8)² + (-14.4)²)
F_net = √1,036.8
F_net = 32.2 N
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