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 with BC
D
along AC

Solution

## The correct option is A in a direction bisecting angle ACBForce on charge $$C$$ due to charge $$A= \dfrac{-kq^2}{a^2}$$Force on charge $$C$$ due to charge $$B = \dfrac{-kq^2}{a^2}$$$$F_{horizontal} = F_1 \sin 30^o - F_2 \sin 30^o$$$$= \dfrac{kq^2}{a^2} \sin 30^o - \dfrac{kq^2}{a^2} \sin 30^o$$$$= 0$$$$F= F_{vertical} = F_1 \cos 30^o + F_2 \cos 30^o$$$$= \dfrac{2kq^2}{a^2} \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}kq^2}{a^2}$$$$\therefore$$ Force is in a direction bisecting angle $$ACB$$Physics

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