Electrostatics

JEE Main level, single correct, medium

176 of 311

Question 176 of 311

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

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Answer: (C) 0V0V

The formula to calculate the potential difference between PP and QQ is given by
∆V=kq(1r1−1r2)...(1)∆V=kq\left(\dfrac{1}{{r}_{1}}-\dfrac{1}{{r}_{2}}\right)...\left(1\right)
Now,
r1=(3)2+(3)2m=6m\begin{matrix}{r}_{1} & = & \sqrt{(\sqrt{3}{)}^{2}+(\sqrt{3}{)}^{2}}m \\ & = & \sqrt{6}m\end{matrix}
And,
r2=(6)2+0m=6m\begin{matrix}{r}_{2} & = & \sqrt{(\sqrt{6}{)}^{2}+0}m \\ & = & \sqrt{6}m\end{matrix}
As r1=r2=6m{r}_{1}={r}_{2}=\sqrt{6}m and the value of the charge remains constant, the potential difference between the points is zero.

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