Question

Three point charges, each $+q$ are placed at the vertices of an equilateral triangle. What charge should be placed at its centroid so that all four charges are in equilibrium?

• A. $\frac{q}{\sqrt{2}}$
• B. $\frac{q}{\sqrt{3}}$
• C. $\frac{\sqrt{3}q}{2}$
• D. $\frac{2q}{\sqrt{3}}$

Answer 1

The correct option is B $\frac{q}{\sqrt{3}}$

Let the change at center be (-Q), to balance the + ve for change at vertices

$F_{net}=(F_1+F_3+2F_1{F}_{3}cos{60}^{\circ }{)}^{\frac{1}{2}}$

$F_{net}=\sqrt{3}F$ . . . . . .(1)

By pythagoras median destance = $\frac{a}{\sqrt{3}}$

$F=\frac{k{q}^2}{a^2}$ . . . . . . . .(2)

$F_2=\frac{k\left(q\right)(-Q)}{(\frac{a}{\sqrt{3}}{)}^2}$ . . . . . . . (3}

Eqvation (1), (2) and(3) . .. . . . .[$F_{net}=F_2$]

$\sqrt{3}\left(\frac{k{q}^2}{a^2}\right)$ = $\frac{kq(-Q)}{(\frac{a}{\sqrt{3}}{)}^2}$

$Q=\frac{-q}{\sqrt{3}}$