Electrostatics Question 321
Question: A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium, if $ q $ is equal to [IIT 1987; CBSE PMT 1995; Bihar MEE 1995; CPMT 1999; MP PET 1999; MP PMT 1999, 2000; RPET 1999; KCET 2001; AIEEE 2002; AFMC 2002; Kerala PMT 2004; J & K CET 2004]
Options:
A) $ -\frac{Q}{2} $
B) $ -\frac{Q}{4} $
C) $ +\frac{Q}{4} $
D) $ +\frac{Q}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Equilibrium of charge B is considered.
Hence for it’s equilibrium $ |F _{A}|=|F _{C}| $
therefore $ \frac{1}{4\pi {\varepsilon _{0}}}\frac{Q^{2}}{4x^{2}}= $
$ \frac{1}{4\pi {\varepsilon _{0}}}\frac{qQ}{x^{2}} $
therefore $ q=\frac{-Q}{4} $
Short Trick: For such type of problem the magnitude of middle charge can be determined if either of the extreme charge is in equilibrium by using the following formula.
If charge A is in equilibrium then q = ? $ Q _{B}{{( \frac{x _{1}}{x} )}^{2}} $
If charge B is in equilibrium then $ q=-Q _{A}{{( \frac{x _{2}}{x} )}^{2}} $
If the whole system is in equilibrium then use either of the above formula.
