Three point charges, two positive and one negative, each having a magnitude of 53 µc are placed at the vertices of an equilateral triangle (35 cm on a side). what is the magnitude of the electrostatic force on one of the positive charges? the value of the coulomb constant is 8.98755 × 109 n · m2 /c 2 . answer in units of n.

For this problem, we use the Coulomb's Law whose equation is written as

F = kQ₁Q₂/d²
where
F is the electric force
k is the Coulomb's constant equal to 8.98755×10⁹ N·m²/c²
Q₁ and Q₂ are two charges
d is the distance between two charges

First, let's compute the force between the two positive charges denotes as F₁.

F₁ = (8.98755×10⁹ N·m²/c²)(+53×10⁻⁶ C)(+53×10⁻⁶ C)/(35 cm * 1 m/100cm)²
F₁ = 206.09 N

Next, let's compute the force between the positive and the negative charges denotes as F₂.

F₂ = (8.98755×10⁹ N·m²/c²)(+53×10⁻⁶ C)(-53×10⁻⁶ C)/(35 cm * 1 m/100cm)²
F₂ = -206.09 N

The net force is the sum of the two forces.
Net Force = 206.09 - 206.09 = 0

Therefore, the net force experienced by the positive charge is zero.