Question

Three charges Q1,Q2,and Q3 are arrange as in figure. Find the force on the charge Q3.In which direction will this force act?​

Answer 1

Answer:

50.4 newton

Explanation:

hope gonna help you

Answer 2

Answer:

The force on the charge $Q_{3}$  is 722.16N and the resultant force acts in the right direction.

Explanation:

Given:

$Q_{1}=6\times 10^{-6} C\\ Q_{2}=2\times 10^{-6} C\\Q_{3}=4\times 10^{-6}C$

$r_{13}=0.1m\\ r_{23}=0.01m$

Here,

As we all know, the coulomb's constant is denoted as K.

The distance between the charge $Q_{1}$ and $Q_{3}$ is denoted by $r_{13}$.

The distance between the charge $Q_{2}$ and $Q_{3}$ is denoted by $r_{23}$.

The force on the charge $Q_{3}$ due to charge $Q_{1}$ is denoted by $F_{1}$.

The force on the charge $Q_{3}$ due to charge $Q_{2}$ is denoted by $F_{2}$.

The resultant force on the charge $Q_{3}$ is denoted by $F_{r}$.

By the equation,

$F_{r}= F_{1}+ F_{2} \\F_{r}=\frac{KQ_{1} Q_{2} }{r_{13} ^{2} } +\frac{KQ_{3} Q_{2} }{r_{23} ^{2} } \\F_{r}=\frac{9\times10^{9} \times6\times10^{-6}\times4\times10^{-6} }{0.1 ^{2} } +\frac{9\times10^{9} \times2\times10^{-6}\times4\times10^{-6} }{0.01 ^{2} } \\F_{r}=722.16N$

The resultant force acts in the right direction.

So, the force on the charge $Q_{3}$  is 722.16N and the resultant force acts in the right direction.