Question

Three equal charges of magnitude 'e' are located at the vertices of an equilateral triangle of side 1m. Where should you place a charge of 2e' so that the '2e' does not experience any force?

Answer 1

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,

$AD^{2}=AB^{2}+BD^{2}$

Put the values and get.

$AD^{2}=1^{2}+(\frac{1}{2} )^{2}\\AD=\sqrt{\frac{3}{4} } \\AD=\frac{\sqrt{3}}{2}$

Now calculate AO as,

$AO=\frac{2}{3}\times \frac{\sqrt{3} }{2}\\AO=\frac{1}{\sqrt{3} }$

And the sides BO=CO=AO.

Now Force can be calculated as

$F_{1}=\frac{2ke^{2} }{\frac{1}{\sqrt{3} } ^{2} }\\F_{1}=6ke^{2}$

And similarly,

$F_{2}=F_{3}=6ke^{2}$

Now we can calculate resultant of $F_{2}andF_{3}$ in upward direction. as,

$F_{net}=\sqrt{F_{2}^{2}+F_{3}^{2}+2F_{2}F_{3}cos120 } \\F_{net}=\sqrt{F_{2}^{2}+F_{2}^{2}+2F_{2}F_{2}(-\frac{1}{2})}\\F_{net}=6ke^{2}$

Therefore the resultant force on centroid O.

$F=F_{1}-F_{net}\\F=6ke^{2}-6ke^{2}\\F=0$

Therefore the fourth charge 2e should be placed on centroid so that it experience zero force.