Question

Two charges of +1 micro coulomb and +4 micro

coulomb are situated at a distance in air. The ratio of

the forces acting on them is:​

Answer 1

Answer:

1:1

Explanation:

F=kqQ/r^2

According to columbs law force on two charges are equal and opposite. Thus the ratio will be 1:1

Answer 2

Given,

Two charges of +1 micro coulomb and +4 micro coulomb are situated at a distance in the air.

To Find,

The ratio of the force acting on them.

Solution,

We can solve this mathematical problem using mathematical formulas.

Let q₁ be in charge of +1 micro coulomb.

Let q₂ be in charge of +4 micro coulomb.

So,

F₁₂ =  Force acting on q₁ due to q₂.

⇒ $\frac{k q_{1} q_{2} }{r}$

F₂₁ =  Force acting on q₂ due to q₁.

⇒$\frac{k q_{1} q_{2} }{r}$

The ratio of forces,

⇒ F₁₂: F₂₁

⇒ $\frac{k q_{1} q_{2} }{r}$ : $\frac{k q_{1} q_{2} }{r}$ ⇒ 1: 1

⇒ F₁₂ = F₂₁

Therefore, The ratio of the force acting on them is 1: 1.