Question

There point charges of $+2\mu C,-3\mu C,-3\mu C$ are kept at the vertices A,B and C respectively of an equilateral triangle of side 20 cm as shown in the figure.What should be the sign and magnitude of the charge to be placed at the mid - point (M) of side BC so that the charge at A remains in equilibrium ?

Answer 1

Let $K=\frac{1}{\left(4\pi \epsilon \right)}=9\times {10}^9$N-m²/C²

The forces F1 and F2 on the charge at Q are along the sides AB and AC.

magnitude of$F1=F2=\frac{K\times 2\times 10⁻⁶\times 3\times {10}^{-6}}{{0.20}^2}=1.5\times {10}^8$K Newtons

Their components along the horizontal directions are equal to $F_{1}Cos60°$ and $F2Cos60°$ which are equal in magnitude and opposite in direction. They cancel.

Their components along AM will add to :

$2F_{1}Cos30°=2\times 1.5\times {10}^{-8}\times \frac{\sqrt{3}}{2}\times Knewtons=1.5\sqrt{3}\times {10}^{-8}$K Newtons

Let the charge placed at M be q Coulombs. $A{M}^2={0.20}^2-{0.10}^2=0.03$m²

Force due to q on charge at A will be along MA: F3

$F_3=Kq\times 2\times {10}^{-6}$

$0.03=6.667qK\times {10}^{-9}$ Newtons

For the charge at A to be stationary,

$6.667qK\times {10}^{-9}=1.5\surd 3\times {10}^{-8}K$

q = 3.897 Coulombs positive charge.