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

raju nagula , 6 Years ago

Grade

1 Answers

Eshan

Last Activity: 6 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 and $F_2$ is

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

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