Question

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

Answer 1

Force due to A on C is $F_{AC}=K\frac{2\times (-3)\times {10}^{-12}}{{0.2}^2}=9\times {10}^9\frac{2\times (-3)\times {10}^{-12}}{{0.2}^2}=-1.35N$

Force due to A on B will be same. That is, $F_{AB}=F_{AC}=-1.35N$

Resultant force, $F^{/}$=$\sqrt{F_{AB}^2+F_{AC}^2+2F_{AB}{F}_{AC}cos{60}^0}=\sqrt{2{F_{AB}^2+2F}_{AB}^2\times \frac{1}{2}}=F_{AB}\times \sqrt{3}$

=2.33N

Let, at point P, charge is placed.

Force between the charges placed at P and A is $F^{//}$

For equilibrium condition, $F^{/}=F^{//}$--------- (1)

Where $F^1$ and $F^{//}$ both are opposite in direction.

Distance between A and P is = $\sqrt{{0.2}^2-{0.1}^2}m$ (applying Pythagoras)

AP= $0.173m$

From equation (1), 2.33= $k\frac{2\times {10}^{-6}\times Q}{{\left(0.173\right)}^2}$

=> Q=$\frac{2.33\times {\left(0.173\right)}^2}{2\times {10}^{-6}\times 9\times {10}^9}$

Q =$3.8\times {10}^{-6}$ Coulomb

Charge should be positive laced at point P