Q66 / 84 · JEE Advanced 2010 (Paper 1) · Physics

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 Figure

  • Correct answer
    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
  • Correct answer
    D
    at a finite distance to the right of $Q_2$ the electric field is zero

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).

Hence, the correct answer is options: 'A', 'D'.

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