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Electrostatics

This is electric charges and fields
Answer is 3rd option tan^-1(16/7)
Please explain it

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Profile image of raju nagula
8 Years agoGrade
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1 Answer

Profile image of Eshan
8 Years ago
Dear student,

F_1=\dfrac{kq^2}{3^2}
F_2=\dfrac{kq^2}{4^2}
Hence the tan of angle between the resultant andF_2is

tan\theta=\dfrac{F_1}{F_2}=\dfrac{4^2}{3^2}=\dfrac{16}{9}
\implies \theta=tan^{-1}(\dfrac{16}{9})
Probably, the answer given is wrong.