JEE PYQ: Electrostatics Question 52
Question 52 - 2019 (09 Jan Shift 2)
Two point charges $q_1$ ($\sqrt{10};\mu$C) and $q_2$ ($-25;\mu$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 the $y$-axis is,
$\left[\text{take } \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9;\text{Nm}^2\text{C}^{-2}\right]$
(1) $(63\hat{i} - 27\hat{j}) \times 10^2$
(2) $(-63\hat{i} + 27\hat{j}) \times 10^2$
(3) $(81\hat{i} - 81\hat{j}) \times 10^2$
(4) $(-81\hat{i} + 81\hat{j}) \times 10^2$
Show Answer
Answer: (1)
Solution
Distance from $q_1$ to point: $\sqrt{1+9} = \sqrt{10}$. $E_1 = \frac{9 \times 10^9 \times \sqrt{10} \times 10^{-6}}{10} = 9\sqrt{10} \times 10^2$. Resolving: $\vec{E_1} = 9 \times 10^2(-\hat{i} + 3\hat{j}) \times \frac{1}{\sqrt{10}}$… Similarly for $q_2$. After computation: $\vec{E} = (63\hat{i} - 27\hat{j}) \times 10^2$ V/m.
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