Q. Two unit negative charges are placed on straight line. A positive charge q is placed exactly at the mid-point between these unit charges. If the system of these three charges is in equilibrium, the value of q (in C) is
A) 1.0
B) 0.75
C) 0.5
D) 0.25
Explanation
(d) For equilibrium, we have
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$F_{AB} + F_{AC} = 0$
or $\frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q}{(d/2)^{2}} + \frac{1}{4\pi\varepsilon_{0}} \times \frac{q_{1}q_{2}}{d^{2}} = 0$
Given, $q_{1} = q_{2} = - 1\mu\complement$
So, $- \frac{q}{(d/2)^{2}} + \frac{1}{d^{2}}$ =0
or $q = \frac{1}{4}$ =0.25 C
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.