JEE Advanced · Physics · 17. Electrostatics
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

- 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
Answer & Solution
Correct Answer
(D) at a finite distance to the right of \(Q_2\) the electric field is zero
Step-by-step Solution
Detailed explanation
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).
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).
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