Coulomb's Law - Vector form 2

Description:

Let us consider,

The two charges q₁ and q₂ are placed at certain position with respect to the origin with the position vector r₁ and r₂ respectively.

Now, We need to calculate the force between the two charges q₁, q₂.

As we know, according to coulumb’s law force is always along the line going the two charges.

Join the two charges q₁ and q₂

(Considering both the charges are positive so the force experienced by either of the charge will be the force of repulsion.)

Consider q₂ is experiencing the force due to q₁, So the experienced force will be F₂₁

Let,

The distance between q₁ and q₂ be Δr→

Then, by vector addition

r₁→ + Δr→ = r₂→

Δr→ = r₂→ - r₁→ (From figure) ......(1)

So, Δr→ can also be written as,

Δr→ = r₂₁→ ......(2)

Now, putting (1) in vector equation of columb’s force

We get,

Similarly

F₂₁→ = 1/4πε_o q₁q₂/|r₂→ - r₁→| (r₂→ - r₁→)

F₁₂→ = 1/4πε_o q₁q₂/|r₁→ - r₂→| (r₂→ - r₁→)