question_answer

Two point charges \[(+Q)\] and \[(-2Q)\] are fixed on the X-axis at positions \[a\] and \[2a\] from origin respectively. At what positions on the axis, the resultant electric field is zero                    [MP PET 2001]

A)            Only \[x=\sqrt{2}a\]         

B)            Only \[x=-\sqrt{2}a\]

C)            Both \[x=\pm \sqrt{2}a\]

D)            \[x=\frac{3a}{2}\] only

Correct Answer: B

Solution :

           Suppose electric field is zero at a point P lies at a distance d from the charge + Q. At P     \[\frac{kQ}{{{d}^{2}}}=\frac{k(2Q)}{{{(a+d)}^{2}}}\] Þ \[\frac{1}{{{d}^{2}}}=\frac{2}{{{(a+d)}^{2}}}\] Þ \[d=\frac{a}{(\sqrt{2}-1)}\] Since d > a i.e. point P must lies on negative x-axis as shown at a distance x from origin hence \[x=d-a\] \[=\frac{a}{(\sqrt{2}-1)}-a=\sqrt{2}\,a.\] Actually P lies on negative x-axis so \[x=-\sqrt{2}\,a\]