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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