JEE MAIN - Physics (2023 - 25th January Evening Shift - No. 11)
A point charge of 10 $$\mu$$C is placed at the origin. At what location on the X-axis should a point charge of 40 $$\mu$$C be placed so that the net electric field is zero at $$x=2$$cm on the X-axis?
$$x=6$$ cm
$$x=8$$ cm
$$x=4$$ cm
$$x=-4$$ cm
Explanation
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$$ \begin{aligned} & \frac{1}{4 \pi \varepsilon_{0}} \frac{(10 \mu \mathrm{C})}{(2 \mathrm{~cm})^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(40 \mu \mathrm{C})}{[(\mathrm{a}-2) \mathrm{cm}]^{2}} \end{aligned} $$
$\Rightarrow\left(\frac{a-2}{2}\right)^{2}=4$
$\Rightarrow \frac{a-2}{2}=2$
$$ a=6 \mathrm{~cm} $$
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