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Electrostatics

This is electric charges and fields chapter.
Answer is 3rd option
Explain with diagram sir.

Question image for This is electric charges and fields chapter. Answ

raju nagula

8 Years agoGrade

1 Answer

Venkat

8 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.

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