Question

Three point Charges of 5 micro coulomb each are placed at the vertices of an equilateral triangle wch has sides 10cm long Find the force On each charge.

Answer 1

the force acting on the each chatge is 22.5N

Answer 2

Answer:

The force on each charge is 38.97 N.

Explanation:

Given that,

Charge $q= 5\times10^{-6}\ C$

Distance $d = 10\ cm$

The force F₁₃ between the charges is

$F_{13}=\dfrac{9\times10^{9}\times5\times10^{-6}\times5\times10^{-6}}{(10\times10^{-2})^{2}}$

$F_{13}=22.5$

The force F₂₃ between the charges is

$F_{23}=\dfrac{9\times10^{9}\times5\times10^{-6}\times5\times10^{-6}}{(10\times10^{-2})^{2}}$

$F_{23}=22.5$

The net force is

$F=\sqrt{F_{1}^2+F_{2}^2+2F_{1}\times F_{2}cos60^{\circ}}$

$F= \sqrt{(22.5)^2+(22.5)^2+2\times22.5\times22.5\times\dfrac{1}{2}}$

$F= 38.97\ N$

Hence, The force on each charge is 38.97 N.