Question

What is the magnitude of the electric force on the charge at the bottom left-hand vertex of the triangle due to the other two charges

Answer 1

Answer:

$\frac{kq^{2} }{2a^2}$

Explanation:

The magnitude on the charge at the bottom-left corner due to the charge on the top vertex of the triangle will act along the +ve x-axis and the +ve y-axis.

From Coulomb's law the magnitude of the forces on the charge at the bottom-left corner, due to the charge on the top vertex of the triangle are $\frac{kq^{2} }{a^2}$cos 60, and

The magnitude on the charge due to the charge at the bottom-right corner will only act in the -ve x-axis, since they repel each other (like charges repel). The magnitude is  $\frac{kq^{2} }{a^2}$

The angle made by the upper charge to the charge we're considering is 60° with the horizontal.

The total force on the charge along the x-axis is

$F_{x}$ = $\frac{kq^{2} }{a^2}$cos 60 -

$F_{x}$  = $\frac{kq^{2} }{2a^2}$ - $\frac{kq^{2} }{a^2}$

==> -$\frac{kq^{2} }{2a^2}$

For the y-axis, we have

$F_{y}$ = $\frac{kq^{2} }{a}$sin 60

$F_{y}$ = $\frac{\sqrt{3}* kq^{2} }{2a^2}$

The resultant force is

$|F| = \sqrt{F_{x}^{2}+ F_{y}^2 }$

The common factors between the two x-axis force, and the y-axis force is

$\frac{kq^{2} }{2a^2}$, we put this outside the square root (squaring this and square rooting will give us the initial value)

$|F|$ = $\frac{kq^{2} }{2a^2}$$\sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3} }{2})^2 }$

$|F|$ =  $\frac{kq^{2} }{2a^2}$$\sqrt{\frac{1}{4} +\frac{3}{4} }$

==>  $\frac{kq^{2} }{2a^2}$$\sqrt{1}$

the magnitude of the electric force on the charge at the bottom left-hand vertex of the triangle due to the other two charges is

$|F|$  =  $\frac{kq^{2} }{2a^2}$