Q. Point charges $q_{1} = 2\mu C$ and $q_{2} = - 1\mu C$ are kept at points $x = 0$ and $x = 6$ respectively. Electrical potential will be zero at points
A) $x = 2\ and\ x = 9$
B) $x = 1\ and\ x = 5$
C) $x = 4\ and\ x = 12$
D) $x = - 2\ and\ x = 2$
Explanation
(c) Potential will be zero at two points
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At internal point $(M):\frac{1}{4\pi\varepsilon_{0}} \times \left\lbrack \frac{2 \times 10^{- 6}}{(6 - l)} + \frac{\left( - 1 \times 10^{- 6} \right)}{l} \right\rbrack = 0$
$\Rightarrow l = 2$
So distance of $M$ from origin; $x = 6 - 2 = 4$
At exterior point $(N)$
$\frac{1}{4\pi\varepsilon_{0}} \times \left\lbrack \frac{2 \times 10^{- 6}}{(6 + l'')} + \frac{\left( - 1 \times 10^{- 6} \right)}{l''} \right\rbrack = 0 \Rightarrow l^{''} = 6$
So distance of $N$ from origin, $x = 6 + 6 = 12$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.