Question

Point charges +q , +q and -q are placed at the corners A , B and C respectively of an equilateral triangle ABC . The resultant force on '-q' placed at C acts as :

• A. in a direction bisecting angle ACB
• B. along BC
• C. in a directing making an angle of ${60}^{\circ }$ with BC
• D. along AC

Answer 1

The correct option is A in a direction bisecting angle ACB

Force on charge $C$ due to charge $A=\frac{-k{q}^2}{a^2}$

Force on charge $C$ due to charge $B=\frac{-k{q}^2}{a^2}$

$F_{horizontal}=F_1\sin {30}^o-F_2\sin {30}^o$

$=\frac{k{q}^2}{a^2}\sin {30}^o-\frac{k{q}^2}{a^2}\sin {30}^o$

$=0$

$F=F_{vertical}=F_1\cos {30}^o+F_2\cos {30}^o$

$=\frac{2k{q}^2}{a^2}\frac{\sqrt{3}}{2}=\frac{\sqrt{3}k{q}^2}{a^2}$

$\therefore$ Force is in a direction bisecting angle $ACB$