# This is electric charges and fields chapter.Answer is 3rd optionExplain with diagram sir.

Venkat
273 Points
4 years ago
Charge at B due charge at A
$F_{1}=\frac{1}{4\pi \varepsilon _{0}}\frac{(2\mu*C)(2\mu*C) }{AB^{2}} \\F_{1}=\frac{1}{4\pi \varepsilon _{0}}\frac{(4*10^{-12}) }{0.03^{2}}$
Note that the force F1 is a vector

Charge at B due to C.
$F_{2}=\frac{1}{4\pi \varepsilon _{0}}\frac{(2\mu*C)(2\mu*C) }{BC^{2}} \\F_{2}=\frac{1}{4\pi \varepsilon _{0}}\frac{(4*10^{-12}) }{0.04^{2}}$
Note that the force F2 is a vector.

Let us take the resultant of the above two forces make an angle $\theta$ with horizontal line BC. So,
$tan\theta =\frac{F_{1}}{F_{2}} \\ \\tan\theta =\frac{0.04^{2}}{0.03^{2}} \\ \\\theta =tan^{-1}\frac{16}{9}$
Hence, correct answer is option C.