Two-point charges $q_{1} = 2 μC and q_{2} = 1 μC$ are placed at distances b = 1 cm and a = 2 cm 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 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

$q_{1} = 2 μC {and q}_{2} = 1 μC$

We know that,

$E \propto \frac{q}{d^{2}}$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{2 μC}{(2 cm)^{2}} \times \frac{(1 cm)^{2}}{1 μC} = \frac{1}{2}$

$\Rightarrow E_{2} = 2 E_{1}$

As ${\vec{E}}_{1}$ is directed along +ve x-axis and ${\vec{E}}_{2}$ along +ve y-axis, angle made by their resultant $\vec{E}$

with +ve x-axis is given by $\tan θ = \frac{E_{2}}{E_{1}} = 2 .$

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