Question

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.

Answer 1

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.

Answer 2

Answer:

The net force on the positive charge will be zero.

Explanation:

Positive charges = q1 = q2 = 53uC

Negative charge = q3 = -53uC

Length of side of equilateral triangle = r = 35cm = 0.35m  

Force on q1 due to q2 will be  

F12 = kq1q2/r^2

F12 = (9 × 10^9)(53u)(53u)/(0.35)^2

F12 = 206.38N

Force on q1 due to q3 will be  

F13 = kq1q3/r^2

F13 = (9 × 10^9)(53u)(-53u)/(0.35)^2

F13 = -206.38N

Net force = F = F12 + F13 = 206.38 – 206.38

F = 0N

Hence, the net force on the positive charge will be zero.