Two point charges q1 (√(10)) and q2( - 25 μ C) are placed on the… | JEE Main 2019 (9 Jan, Shift 2)

Q39JEE Main 2019 · 9 JAN · SHIFT 2Medium

Electric Field & Field Lines

Two point charges q1 $\left(\sqrt{10}\,\mu\text{C}\right)$ and q2( $-$ 25 $\mu$ C) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,

[take $\frac{1}{4\pi\varepsilon_0}$ = 9 $\times$ 109 Nm2C ${}^{-2}$ ]

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Answer: (a)

$q_1$ at $(1, 0)$ : $\vec{r}_1 = (-1, 3)$ , $r_1 = \sqrt{10}$ ; $\vec{E}_1 = \dfrac{kq_1}{r_1^3}\vec{r}_1 = \dfrac{9\times10^9 \times \sqrt{10}\times10^{-6}}{10\sqrt{10}}(-1, 3) = (-9\hat{i} + 27\hat{j})\times10^2$
$q_2$ at $(4, 0)$ : $\vec{r}_2 = (-4, 3)$ , $r_2 = 5$ ; $\vec{E}_2 = \dfrac{9\times10^9 \times (-25\times10^{-6})}{125}(-4, 3) = (72\hat{i} - 54\hat{j})\times10^2$
$\vec{E} = (63\hat{i} - 27\hat{j})\times 10^2\text{ V/m}$
So the answer is (a).

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