Question

If the charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of q is?
A- Q/2
B- -Q/2
C- Q/4
D- -Q/4

Answer 1

Hint: We know that charges attract or repel each other with a force which is directly proportional to the product of the magnitude of their charges and inversely proportional to the square of distance between them. This is from Coulomb’s law. Initially two charges of the same polarity were held at some distance, so there must be a force of repulsion between them. Now another charge q is placed at the middle point of the line and the system is in equilibrium, that means net force on charge q is 0.

Complete step by step answer:
If we have two charges q and Q and they are separated by a distance r, then force exerted by one charge on another is given by: \[F=\dfrac{kqQ}{{{r}^{2}}}\]
Where k is the constant of proportionality.
Now as per the question after placing the charge q the system is in equilibrium. This is only possible when the force exerted by the charge at the left on the charge q is equal to the force exerted by the charge on the right.
Let us assume the initial distance between the two charges is 2x, then the new charge q, is placed at distance x from each charge.
${{F}_{QQ}}+{{F}_{Qq}}=0 \\ $
$\Rightarrow \dfrac{kQQ}{4{{x}^{2}}}+\dfrac{kqQ}{{{x}^{2}}}=0 \\ $
$\Rightarrow \dfrac{Q}{4}=-q \\ $
$ \therefore q=\dfrac{-Q}{4} \\ $

So, the correct answer is “Option D”.

Note:
Since, the force is a vector quantity, it has both the magnitude and the direction. When we talk about the electrostatic force, the same concept applies to it. Like charges repel each other and unlike charges attract each other. If there is a system of charges and we need to find the net force on any one of the charges then we have to add all the forces acting on it vectorially.