Question

Three equal charges each 1microC are placed at the three corners of an equilateral triangle of side 3cm. Find the force experienced by each charge due to other three charges. ​

Answer 1

$\huge \sf \fbox{Solution :) }$

We know that , the coulomb's law is given by

$\large \sf \fbox{F = \frac{k (q_{1} \times q_{2}) }{ {(r)}^{2} } }$

Where , k = coulomb's constant , which is equal to 9 × (10)^9

Thus ,

Force on A due to B = K(10)^(-12)/(3)² Newton

Force on A due to B = K(10)^(-12)/(3)² Newton

It is observed that ,

$\sf \large \fbox{ F_{AB } = F_{ AC} = F \: (let)}$

We know that , the resultant between two vectors (A and B) is given by

$\sf \fbox{Resultant = \sqrt{ {(A)}^{2} + {(B)}^{2} + 2AB \cos( \theta) }}$

Thus , the resultant of force on A due B and C is

$\sf \mapsto Resultant = \sqrt{ {2(F)}^{2} + 2 {(F)}^{2} \times \cos(60) } \\ \\ \sf \mapsto Resultant = \sqrt{ {2(F)}^{2} + \cancel{ 2}{(F)}^{2} \times \frac{1}{ \cancel{2}} } \\ \\ \sf \mapsto Resultant = \sqrt{3{(F)}^{2} } \\ \\ \sf \mapsto Resultant = F \sqrt{3}$

Put the value of F , we get

$\sf \mapsto Resultant = \sqrt{3} \times \frac{k {(10)}^{ - 12} }{ {(3)}^{2} } \\ \\ \sf \mapsto Resultant = \sqrt{3} \times \frac{ \cancel9 \times {(10)}^{9 } \times {(10)}^{ - 12} }{ \cancel9} \\ \\ \sf \mapsto Resultant = \sqrt{3} \times {(10)}^{-3} \: \: Newton$

Hence , the force experienced by each charge due to other two charges is √3 × (10)^(-3) Newton