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 :
Solution
<p>Potential difference $=\frac{K Q}{r_1}-\frac{K Q}{r_2}$</p>
<p>$$\begin{aligned}
& r_1=\sqrt{(\sqrt{3})^2+(\sqrt{3})^2} \\
& r_2=\sqrt{(\sqrt{6})^2+0}
\end{aligned}$$</p>
<p>As $r_1=r_2=\sqrt{6} \mathrm{~m}$</p>
<p>So potential difference $=0$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
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