The word “electricity” comes from the Greek word “elekron” which means “amber”. Thales, the Greek philosopher observed that if a person rubbed amber (a petrified tree resin) with wool or fur, then the amber would attract small pieces of leaf or cloth. Dr. Gilberts (physician to the Queen of England) performed some experimental works on this effect in 1600. Later Col Charles coulomb (1736-1806) (a Colonel in French Army Engineers) performed an elaborate series of experiments to determine quantitatively the force exerted between two objects each having a static charge of electricity. The result of his experiment in popularly referred to as coulomb’s law.
Coulomb’s law state the force of attraction or repulsion between two point charges Q₁ and Q₂ (both measured in Coulomb, C) separated in a vacuum or free space by a distance which is large compared to their sizes is proportional to the product of the two charges and inversely proportional to the square of the distance between them.
Mathematically:
From the first statement:
Fα Q₁Q₂…….......equation (i)
And from the second statement:
F α 1/r²…………...equation (ii)
If we are to combine equations (i) and (ii) together, we have:
F α Q₁Q₂/ r²
F =kQ₁Q₂/ r² (measured in Newton)
Where k = constant of proportionality = 1/4πε
ε = 8.85 x 10-¹² Farad/meter = permittivity of free space
∴ k= 1/ (4π x 8.85 x 10-¹²) = 8.988 x E9 Nm²/C²
NOTE: The force is repulsive if the two charges in question are of the same signs (positive /positive – negative/negative). It is attractive if the two charges are of different signs (positive/negative). The formula above is the scalar representation of electric force between charges.
Example 1: What is the force between the two point charges Q₁ = +12 x E-9C and Q₂ = -18 x E-9C separated by a distance of 0.3m.
credit: 
solution:
F = K Q₁Q₂/ r²
F = (8.988 x 12 x E-9 x -18 x E-9) / (0.3)²
= 2.2 x E-5 N(attractive- since charges with different signs were given).
Example 2: What is the total force on Q₃ from Q₁ and Q₂ as shown below.
credit :
Solution:
F₁₃ = kQ₁Q₃/ (5)² (repulsive): Q₃ away from Q₁
= 8.988 x E9 x 6.00 x E-9 x 5 x E9/ (4)²
F₁₃= 1.079 x E-8N
F₂₃= kQ₂Q₃/ (4)² (attractive): Q₃ moves towards Q₂
= 8.988 x E9 x 2.0 x E-9 x 5 x E9/ (4)²
F₂₃= 5.617 x E-9N
Angle = Θ= tan-¹(3/4) = 36.9°
credit:
∑Fy = F₁₃sin (36.9°) = 1.079 x E-8sin (36.9°) = 6.476 x E-9N
∑Fx = -F₂₃ + F₁₃cos (36.9°) = 5.617 x E-9 cos (36.9°) = 3.011 x E-9N
F = √(F²x + F²y)
F= (6.476 x E-9)² + (3.011 x E-9)²
F =7.12 x E-9 N
∅ = tan-¹(Fy/Fx) = tan-¹(6.476 x E-9/ 3.011 x E-9)
∅= 65.1°
**Note: **
When two positive charges are given, the force moves towards the charge with the least magnitude.
When both positive and negative charges are given, the force tends towards the negative charge.
Remember that electric force is a vector, therefore separate the components of the force into x and y components and add up as vectors.*
Quantities are classified into scalar (the one with magnitude but without direction) and vector (the one with both magnitude and direction). From the two examples solved so far, I considered only the magnitude of the force. So, to get Force in vector form, we simply multiply the scalar form of the force by the unit vector. In such a condition, the distance between the two charges would not be given and will be computed from the unit vector calculated.
F (vector) = F (scalar) . Unit vector
= (K Q₁Q₂/ r²) . R₁₂/|R₁₂|
R₁₂ = position vector taking the reference charge constant
|R₁₂| = magnitude of the position vector. This is what will be used as the distance between the two charges given.
Example 3: If Q₁ is at point (1, 2, 3) and 3 x 10-⁴C in magnitude and Q₂ is at point (2, 0, 5) and of -10- ⁴C. What is the force on Q₂?
Solution:
Looking at the question, we weren’t given the distance between the charges but we were given the location of the charges in space (i.e.: x, y and z coordinates).
We were asked to find the force on Q₂. This makes Q₂ the reference charge.
To get the position vector, we simply subtract the spatial coordinates of the other charge from that of the Q2: the reference charge.
R₁₂= position vector = (2-1)ax + (0-2)ay + (5-3)az
R₁₂ = ax – 2ay + 2az
|R₁₂| = magnitude = √[1² + (-2)² + 2²] =
√9 = 3.
Therefore, distance r = 3
Recall F = (K Q₁Q₂/ r²) . R₁₂/|R₁₂|
F = [8.988 x E9 x 3x10-⁴ x (-10-⁴)/3²] . [(ax – 2ay + 2az)/3] N
References:
KhanAcademy
