JEE MAIN - Physics (2025 - 2nd April Morning Shift - No. 11)
A point charge $+q$ is placed at the origin. A second point charge $+9 q$ is placed at ($\mathrm{d}, 0,0$) in Cartesian coordinate system. The point in between them where the electric field vanishes is:
$(3 \mathrm{d} / 4,0,0)$
$(\mathrm{d} / 4,0,0)$
$(4 \mathrm{d} / 3,0,0)$
$(\mathrm{d} / 3,0,0)$
Explanation
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Let $E_p=0$
$$\begin{aligned} & \therefore \frac{k q}{x^2}=\frac{k 9 q}{(d-x)^2} \\ & \Rightarrow \frac{d-x}{x}=3 \Rightarrow x=\frac{d}{4} \end{aligned}$$
$\therefore$ co-ordinate of P is $\left(\frac{\mathrm{d}}{4}, 0,0\right)$
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