Electrostatics Question 262

Question: Four charges equal to -q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is [AIEEE 2004]

Options:

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

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Answer:

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}}} $

therefore $ \frac{1}{d^{2}}=\frac{2}{{{(a+d)}^{2}}} $

therefore $ 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 $



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