Verified Solution • 2025–26 Session moderate

QJEEM24Pha542: JEE Main 2024 Physics - Class 12 JEEM24Pha542 Solution: An electric charge is placed at origin of co-ordinate system.…

An electric charge 10−6μC10^{-6} \mu \mathrm{C} is placed at origin (0,0)(0,0) m\mathrm{m} of X−Y\mathrm{X}-\mathrm{Y} co-ordinate system. Two points P\mathrm{P} and Q\mathrm{Q} are situated at (3,3)m(\sqrt{3}, \sqrt{3}) \mathrm{m} and (6,0)m(\sqrt{6}, 0) \mathrm{m} respectively. The potential difference between the points P\mathrm{P} and Q\mathrm{Q} will be :

Multiple Choice Options

  • A 3 V\sqrt{3} \mathrm{~V}
  • B 6 V\sqrt{6} \mathrm{~V}
  • C 0 V0 \mathrm{~V}
  • D 3 V3 \mathrm{~V}

Step-by-Step Verified Solution

Potential difference =KQr1−KQr2=\frac{K Q}{r_1}-\frac{K Q}{r_2}

amp;r1=(3)2+(3)2amp;r2=(6)2+0\begin{aligned}& r_1=\sqrt{(\sqrt{3})^2+(\sqrt{3})^2} \\& r_2=\sqrt{(\sqrt{6})^2+0}\end{aligned}

As r1=r2=6 mr_1=r_2=\sqrt{6} \mathrm{~m}

So potential difference =0=0

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