# Q. Two point charges $q_1( \sqrt{10} \mu C)$ and $q^2(-25 \; \mu C)$ are placed on the x-axis at x = l m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is, $\left[\text{take} \frac{1}{4\pi\varepsilon_{0}} = 9\times10^{9} Nm^{2} C^{-2}\right]$

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Solution:

## Let $\vec{E}_1$ & $\vec{E}_2$ are the vaues of electric field due to $q_1$ & $q_2$ respectively magnitude of $E_2 = \frac{1}{4 \pi \in_0} \frac{q_2}{r^2}$ $E_{2} = \frac{9\times10^{9} \times\left(25\right) \times10^{-6}}{\left(4^{2} +3^{2}\right)} V/m$ $E_{2} = 9 \times10^{3} V/m$ $\therefore \vec{E}_{2} = 9 \times10^{3} \left(\cos\theta_{2} \hat{i } -\sin\theta_{2} \hat{j}\right)$ $\because \tan \theta_{2} = \frac{3}{4}$ $\therefore \vec{E}_{2} = 9 \times10^{3} \left(\frac{4}{5} \hat{i} - \frac{3}{5} \hat{j}\right) = \left(72 \hat{i} -54\hat{j}\right) \times10^{2}$ $E_{1} = \frac{1}{4\pi\in_{0}} \frac{\sqrt{10} \times10^{-6}}{\left(1^{2} + 3^{2}\right)}$ $=\left(9 \times10^{9}\right) \times\sqrt{10} \times10^{-7}$ $= 9 \sqrt{10} \times10^{2}$ $\therefore \vec{E}_{1} = 9 \sqrt{10} \times10^{2} \left[\cos\theta_{1} \left(-\hat{i}\right) + \sin \theta_{1} \hat{j}\right]$ $\therefore \tan\theta_{1}= 3$ $E_{1} =9 \times\sqrt{10} \times10^{2} \left[\frac{1}{\sqrt{10} } \left(-\hat{i}\right) + \frac{3}{\sqrt{10}} \hat{j}\right]$ $E_{1} = 9\times10^{2} \left[-\hat{i} +3\hat{j}\right] = \left[-9 \hat{i} + 27\hat{j}\right] 10^{2}$ $\therefore \vec{E} = \vec{E}_{1} + \vec{E}_{2} = \left(63 \hat{i} - 27 \hat{j}\right) \times10^{2} V/m$ $\therefore$ correct answer is (3)

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