Two point charges... | Electrostatics

JEE MAIN 2019

09-01-19 S2

Question

Two point charges $q_1(\sqrt{10} \mu \mathrm{C})$ and $q_2(-25 \mu \mathrm{C})$ are placed on the $x$-axis at $x=1 \mathrm{~m}$ and $x=4 \mathrm{~m}$ respectively The electric field (in $\mathrm{V} / \mathrm{m}$ ) at a point $\mathrm{y}=3 \mathrm{~m}$ on y -axis is, $\left[\right.$ take $\left.\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right]$

Select the correct option:

$(63 \hat{i}-27 \hat{j}) \times 10^2$

$(81 \hat{i}-81 \hat{j}) \times 10^2$

$(-81 \hat{i}+81 \hat{j}) \times 10^2$

$(-63 \hat{i}+27 \hat{j}) \times 10^2$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} \stackrel{m}{E} & =E_1+E_2 \\ \stackrel{E}{E}_1 & =\frac{1}{4 \pi \varepsilon_0} \times \frac{\sqrt{10} \times 10^6}{10} \times\left[\frac{-\hat{i}+3 \hat{j}}{\sqrt{10}}\right] \\ \stackrel{E}{E}_2 & =\frac{1}{4 \pi \varepsilon_0} \times \frac{\left(-25 \times 10^{-6}\right)}{25} \times\left[\frac{-4 \hat{i}+3 j}{5}\right] \\ \stackrel{E}{E}_2 & =9 \times 10^9\left[\frac{-\hat{i}+3 \hat{j}}{10}-\left(\frac{-4 \hat{i}+3 \hat{j}}{5}\right)\right] \\ & =\frac{9 \times 10^3}{10}[-\hat{i}+3 \hat{j}+8 \hat{i}-6 \hat{j}] \\ & =9 \times 10^2[+7 \hat{i}-3 \hat{j}] \\ & =(63 \hat{j}-27 \hat{j}) \times 10^2\end{aligned}$

Question Tags

JEE Main

Physics

Medium

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