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Two point changes q₁= 2µC and q₂ = 1µC are placed at distances b = 1cm and a = 2cm from the origin on the y and x axes. The electric field vector at point P (a, b) will subtend an angle ‘ $θ$ ’ with the x - axis, that can be given by

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$\tan θ = 2$

$\tan θ = 4$

$\tan θ = 1$

$\tan θ = 3$

answer is B.

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Detailed Solution

$E_{x} = k \frac{q_{1}}{a^{2}} a n d E_{y} = k \frac{q_{2}}{b^{2}} \Rightarrow \tan θ = \frac{E_{y}}{E_{x}} = \frac{a^{2}}{b^{2}} \times \frac{q_{2}}{q_{1}} = {(\frac{2}{1})}^{2} \times \frac{1}{2} = 2$

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