Question

there are two charges +1mC and +5mC. Find the ratio of force acting on them​

Answer 1

Answer :-

We have got the charges :-

+1 mC and +5 mC

Let they have a distance of x m

Then as we know :-

$Force \: = \dfrac{kQq}{r^2}$

Now by substiting value of the charges :-

By 1mC on 5mC

$= \dfrac{k\times(1 \times 10^{-6}\:C)\times (5 \times 10^{-6} \:C)}{x^2}$

$= \dfrac{k \times 5 \times 10^{-12} \: C}{x^2}$

By 5mC on 1mC

$= \dfrac{k\times(5 \times 10^{-6}\:C)\times (1 \times 10^{-6} \:C)}{x^2}$

$= \dfrac{k \times 5 \times 10^{-12} \: C}{x^2}$

Now ratio of charge

$= \dfrac{k \times 5 \times 10^{-12} \: C}{x^2} : \dfrac{k \times 5 \times 10^{-12} \: C}{x^2}$

$= 1 : 1$

**Ratio of force is always Same if it's between two charges.