Hree point charges are located on the positive x-axis of a coordinate system. charge q1 = 1.0 nc is 2.0 cm from the origin, charge q2 = -2.5 nc is 4.0 cm from the origin and charge q3 = 6.0 nc located at the origin. what is the net force ((a)magnitude and (b) direction) on charge q1 = 1.0 nc exerted by the other two charges?

The concept applied here is the Coulomb's Law. This law is used conveniently in equation:

F = kQₐQₓ/d²
where
F is the electric force between two charges
k is the Coulomb's Law constant equal to 9×10⁹ N
Qₐ and Qₓ are the charges of two bodies 
d is the distance between the two bodies

The net force is the total amount of force acting on the object. For this problem, we focus on Q₁. To be consistent with units, let's make sure to use the SI units. For length, that would be in meters. For charges, that would be in Coulombs. The conversion for each is that 100 cm = 1 m and 10⁻9 C = 1 nC. 

Let's start with the electric force between Q₁ and Q₂.Let's denote this as F₁. Refer to the diagram in the attached picture to determine the distances between the charges.

F₁ = (9×10⁹)(1×10⁻⁹)(-2.5×10⁻⁹)/(0.02)²
F₁ = -5.625×10⁻⁵ N

Next, let's find the electric force between Q₁ and Q₃ denoted as F₂.

F₂ = (9×10⁹)(1×10⁻⁹)(6×10⁻⁹)/(0.02)²
F₂ = 1.35×10⁻⁴ N

Therefore, the net force is equal to
F = F₁ + F₂
F = -5.625×10⁻⁵ N + 1.35×10⁻⁴ N
F = 7.875×10⁻⁵ N

Therefore, the magnitude of the net force is 7.875×10⁻⁵ N. Since force is a vector quantity, we can determine the direction through the sign of its quantity. By convention, if F is +, the direction is towards to the right. If F is -, the direction is towards the left. Thus, the direction of the 7.875×10⁻⁵ N force is towards the right.

Less force and energy was applied