Electrostatics 4 Question 13
13. A few electric field lines for a system of two charges $Q_{1}$ and $Q_{2}$ fixed at two different points on the $x$-axis are shown in the figure. These lines suggest that
(2010)
(a) $\left|Q_{1}\right|>\left|Q_{2}\right|$
(b) $\left|Q_{1}\right|<\left|Q_{2}\right|$
(c) at a finite distance to the left of $Q_{1}$ the electric field is zero
(d) at a finite distance to the right of $Q_{2}$ the electric field is zero
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Answer:
Correct Answer: 13. (a,d)
Solution:
- From the behaviour of electric lines, we can say that $Q_{1}$ is positive and $Q_{2}$ is negative. Further, $\left|Q_{1}\right|>\left|Q_{2}\right|$
At some finite distance to the right of $Q_{2}$, electric field will be zero. Because electric field due to $Q_{1}$ is towards right (away from $Q_{1}$ ) and due to $Q_{2}$ is towards left (towards $Q_{2}$ ). But since magnitude of $Q_{1}$ is more, the two fields may cancel each other because distance of that point from $Q_{1}$ will also be more.
$\therefore$ The correct options are (a) and (d).
