Question

Two charges $4q\,and\,q$ are placed at a distance $l$ apart. A third charged particle $Q$ is placed at the middle of them. If resultant force on $q$ is zero then the value of $Q$ is:-
A. $q$
B. $ - q$
C. $2q$
D. $ - 2q$

Answer 1

Hint:In order to this question, to find the value of $Q$ , we will first write the formulas of both the forces, ${F_1}\,and\,{F_2}$ . And then we will equate both the equations of forces as their magnitudes are the same. Then we can find the value of $Q$ easily.

Complete step by step answer:
According to Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It acts along the line joining the two charges considered to be point charges. The force due to $ + 4q$ on $ + q$ is given as,
${F_1} = \dfrac{1}{{4\pi { \in _0}}} \times \dfrac{{4q \times q}}{{{l^2}}}$
The force due to $Q$ on $ + q$ is given as:
${F_2} = \dfrac{1}{{4\pi { \in _0}}} \times \dfrac{{Q \times q}}{{{{(\dfrac{l}{2})}^2}}}$

If the resultant force is zero, then the force due to $Q$ should be opposite to the force due to $ + 4q$ . Therefore the charge $Q$ should be negative. The magnitude of the two forces should be the same and it can be written as,
${F_{net}} = F_1 + F_2 = 0 $
So, now we will equate the magnitude of both the forces :-
$F_1 + F_2 = 0 $
$\dfrac{1}{{4\pi { \in _0}}} \times \dfrac{{4q \times q}}{{{l^2}}} = -\dfrac{1}{{4\pi { \in _0}}} \times \dfrac{{Q \times q}}{{{{(\dfrac{l}{2})}^2}}} \\$
$\therefore Q = -q$
Thus, the value of $Q$ is $ - q$.

Hence, the correct option is B.

Note: The overall effective force operating on a body, as well as its directions, is known as the resultant force. In addition, when the item is at rest or travelling at the same speed, the resultant force must be zero.The law is applicable only for the point charges at rest. Coulomb’s Law can be only applied in those cases where the inverse square law is obeyed.