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

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

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1 Answers

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.

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