Answer:
At centroid
Explanation:
In the given equilateral triangle ABC with side of 1 m. The three equal charges e,e,e are placed at the A,B and C.
And the fourth charge 2e is put at point O which is called centroid.
Now we can calculate the distance AD by applying pythagorean theorem as,
[tex]AD^{2}=AB^{2}+BD^{2}[/tex]
Put the values and get.
[tex]AD^{2}=1^{2}+(\frac{1}{2} )^{2}\\AD=\sqrt{\frac{3}{4} } \\AD=\frac{\sqrt{3}}{2}[/tex]
Now calculate AO as,
[tex]AO=\frac{2}{3}\times \frac{\sqrt{3} }{2}\\AO=\frac{1}{\sqrt{3} }[/tex]
And the sides BO=CO=AO.
Now Force can be calculated as
[tex]F_{1}=\frac{2ke^{2} }{\frac{1}{\sqrt{3} } ^{2} }\\F_{1}=6ke^{2}[/tex]
And similarly,
[tex]F_{2}=F_{3}=6ke^{2}[/tex]
Now we can calculate resultant of [tex]F_{2}andF_{3}[/tex] in upward direction. as,
[tex]F_{net}=\sqrt{F_{2}^{2}+F_{3}^{2}+2F_{2}F_{3}cos120 } \\F_{net}=\sqrt{F_{2}^{2}+F_{2}^{2}+2F_{2}F_{2}(-\frac{1}{2})}\\F_{net}=6ke^{2}[/tex]
Therefore the resultant force on centroid O.
[tex]F=F_{1}-F_{net}\\F=6ke^{2}-6ke^{2}\\F=0[/tex]
Therefore the fourth charge 2e should be placed on centroid so that it experience zero force.
