Question

Consider 3 charges q q and -q placed at the vertices of an equilateral triangle.what is the force on each charge? (NCERTexample 1.7)

Answer 1

Given :

Three charges of magnitude q , q and -q are placed at the vertices of an equilateral triangle .

To Find :

What is the value  of the force on each of the three charges = ?

Solution :

Let each side of the triangle = a

Force on charge 1 = Force due to charge 2 + Force due to charge 3

Now , Force due to 2 is :

$F_2 = \frac{Kq.q}{a^2}= \frac{Kq^2}{a^2}$

And force due to 3 is  :

$F_3 = \frac{Kq.(-q)}{a^2}= \frac{-kq^2}{a^2}$

Net force on charge 1 will be :

$F_{net1} = \sqrt {(\frac{Kq^2}{a^2})^2+ (\frac{-Kq^2}{a^2} )^2+ 2 \times \frac{Kq^2}{a^2}\times (\frac{-Kq^2}{a^2} ) \times cos120}$

       = $\sqrt {(\frac{Kq^2}{a^2})^2+ (\frac{-Kq^2}{a^2} )^2+ (\frac{Kq^2}{a^2} )^2 }$      ( since cos(120) = $\frac{-1}{2}$ )

       = $\frac{\sqrt3 Kq^2}{a^2}$

Similarly net force on charge 2 will also be equal to:

$F_{net2}=\frac{\sqrt3 Kq^2}{a^2}$

And on charge 3 :

$F_{net3} = \frac{Kq^2}{a^2}$

Hence , net force on charge 1 and 2 is $\frac{\sqrt3 Kq^2}{a^2}$ and on charge 3 is $\frac{Kq^2}{a^2}$ .