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
Show Answer
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 $
