CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
along BC
loader
C
in a directing making an angle of 60 with BC
loader
D
along AC
loader

Solution

The correct option is A in a direction bisecting angle ACB
Force 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 $$ACB$$

1710074_1811688_ans_bf54f9f42f1048c7b2af9fa17460c85c.png

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image