2. Electric Potential and Capacitance
hard

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 $P$ and $Q$ will be :

A

 $\sqrt{3} \mathrm{~V}$

B

 $\sqrt{6} \mathrm{~V}$

C

 $0 \mathrm{~V}$

D

$3 \mathrm{~V}$

(JEE MAIN-2024)

Solution

Potential difference $=\frac{K Q}{r_1}-\frac{K Q}{r_2}$

$ \mathrm{r}_1=\sqrt{(\sqrt{3})^2+(\sqrt{3})^2}$

$ \mathrm{r}_2=\sqrt{(\sqrt{6})^2+0}$

As $r_1=r_2=\sqrt{6} \mathrm{~m}$

So potential difference $=0$

Standard 12
Physics

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