Three point charges, two positive and one negative, each having a magnitude of 20 μC are placed at the vertices of an equilateral triangle (30 cm on a side). What is the magnitude of the electrostatic force on the negative charge?
A) 80 N
B) 40 N
C) 69 N
D) 57 N
E) 75 N

Question

Quizplus · Accepted Answer

We can solve this problem using Coulomb's law, which states that the electrostatic force between two point charges is proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Using this law, we can find the force on the negative charge due to each of the other charges separately, and then add them up vectorially. 
Let's choose the negative charge at the origin, and the positive charges at (15 cm, 15√3 cm) and (-15 cm, 15√3 cm) respectively. The distance between the charges is 30 cm, so the force between each pair of charges is given by:
F = kq1q2/r^2
where k is the Coulomb constant (9×10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between them. For the positive charges, q1=q2=20 μC, and r=30 cm.
Fp = kq1q2/r^2 = (9×10^9)(20×10^-6)^2/(0.3)^2 = 600 N
Since the positive charges are identical, the vector sum of the forces on the negative charge due to them is just the sum of their individual forces, directed along the line connecting them:
Fp,total = 2Fp sin(60°) = 2×600×(√3/2) = 1044 N
For the negative charge, q=-20 μC, and the distance to each of the positive charges is also 30 cm. However, the direction of the force is along the line connecting the negative charge to each positive charge, but in the opposite direction. Therefore, the vector sum of the forces on the negative charge due to the positive charges is:
Fp,total = -2Fp sin(60°) = -2×600×(√3/2) = -1044 N
The total force on the negative charge is then the vector sum of the forces due to the positive charges and the negative charge:
Ftotal = Fp,total + Fn,total
where Fn,total is the force on the negative charge due to the other negative charge. Since the charges are identical in magnitude, but opposite in sign, the force between them is attractive and is given by:
Fn = kq1q2/r^2 = (9×10^9)(20×10^-6)^2/(0.3)^2 = 600 N
The direction of this force is along the line connecting the two negative charges. Since this line is perpendicular to the line connecting the negative charge to either positive charge, the vector sum of the forces is just the magnitude of the force due to the other negative charge:
Fn,total = Fn = 600 N
Finally, the total force on the negative charge is:
Ftotal = -1044 + 600 = -444 N
The magnitude of this force is just the absolute value, which is 444 N. Therefore, the answer is (C) 69 N.

Three Point Charges, Two Positive and One Negative, Each Having