Question

Two point electric charges of values q and 2q are kept at a distance d apart from each other in air. A third charge Q is to be kept along the same line in such a way that the net force acting on q and 2q is zero. Find the location of the third charge from charge q.
(a). $\dfrac{d}{1+ \sqrt{2}}$
(b). $\dfrac{d}{1- \sqrt{2}}$
(c). $\dfrac{d}{1+ \sqrt{6}}$
(d). $\dfrac{d}{1- \sqrt{6}}$

Answer 1

Hint: Net force on the charge q and 2q should be zero according to the question. So, charge Q should be negative and should be placed in between the charges q and 2q. Then Coulomb’s law can be used to determine the forces on charge q and 2q due to other charges. Then we can compare the equations given to determine the distance of charge Q from the charge q.
Formulae used:
Electrostatic force between two charges ${Q}_{1}$ and ${Q}_{2}$ placed at a distance r using the formula, $F=\dfrac{k{Q}_{1}{Q}_{2}}{r^2}$, where k is the electrostatic constant.

Complete step-by-step answer:
We have been given that the two charges q and 2q are placed at a distance d from each other. And a third charge Q has been placed along their line of joining, such that net force on charges q and 2q is zero.

For the given conditions to prevail, the charge Q should be negative and should be placed in between the charges q and 2q. Assume charge Q is placed at a distance x from the charge q.
Using Coulomb’s law, we can express the electrostatic force of attraction or repulsion between two charges ${Q}_{1}$ and ${Q}_{2}$ placed at a distance r using the formula, $F=\dfrac{k{Q}_{1}{Q}_{2}}{r^2}$, where k is the electrostatic constant.
Now, force acting on charge q due to charge Q should be balanced by the force due to the charge 2q.
Therefore, ${F}_{Q}={F}_{2q}$
$\implies \dfrac{kqQ}{x^2}=\dfrac{2kq^2}{d^2}$ ………. (i)
Similarly, forces acting on charge 2q due to charges Q and q should balance each other
Therefore, ${F}_{q}={F}_{Q}$
$\implies \dfrac{2kq^2}{d^2}=\dfrac{2kQq}{(d-x)^2}$ ………. (ii)
Now, from equation (i) and (ii), we can write,
$\dfrac{kqQ}{x^2}=\dfrac{2kQq}{(d-x)^2}$
$\implies \dfrac{1}{x^2}=\dfrac{2}{(d-x)^2}$
$\implies (\dfrac{d-x}{x})^2=2$
$\implies x=\dfrac{d}{\sqrt{2}+1}$
Hence, option a is the correct answer.

Note: One should be careful while placing the charge Q. If we assume the charge to be negative and place it on the line joining the charges q and 2q either on the left of q or in between q and 2q or in the right of 2q, there will only be the repulsion forces, that will disbalance the forces. So we need to consider charge Q as negative.