JEE MAIN - Physics (2024 - 27th January Morning Shift - No. 13)
An electric charge $$10^{-6} \mu \mathrm{C}$$ is placed at origin $$(0,0)$$ $$\mathrm{m}$$ of $$\mathrm{X}-\mathrm{Y}$$ co-ordinate system. Two points $$\mathrm{P}$$ and $$\mathrm{Q}$$ are situated at $$(\sqrt{3}, \sqrt{3}) \mathrm{m}$$ and $$(\sqrt{6}, 0) \mathrm{m}$$ respectively. The potential difference between the points $\mathrm{P}$ and $\mathrm{Q}$ will be :
$$\sqrt{3} \mathrm{~V}$$
$$\sqrt{6} \mathrm{~V}$$
$$0 \mathrm{~V}$$
$$3 \mathrm{~V}$$
Explanation
Potential difference $$=\frac{K Q}{r_1}-\frac{K Q}{r_2}$$
$$\begin{aligned} & r_1=\sqrt{(\sqrt{3})^2+(\sqrt{3})^2} \\ & r_2=\sqrt{(\sqrt{6})^2+0} \end{aligned}$$
As $$r_1=r_2=\sqrt{6} \mathrm{~m}$$
So potential difference $$=0$$
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