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/2B- -Q/2C- Q/4D- -Q/4

To determine the value of the charge qq placed at the center of the line joining two equal charges QQ such that the system is in equilibrium, let's analyze the situation step by step.
Problem Setup:
• There are two charges QQ, placed at a distance rr apart.
• A charge qq is placed at the center of the line joining these two charges QQ.
• The system is in equilibrium, meaning the net force on each charge must be zero.
Step-by-Step Solution:
1. Forces Acting on qq:
o The charge qq experiences two forces due to the charges QQ. Since the two charges QQ are equal and placed symmetrically, the forces from both charges will be equal in magnitude but opposite in direction.
o The force on qq due to each charge QQ is given by Coulomb's law: F=k⋅∣Q⋅q∣r2F = \frac{k \cdot |Q \cdot q|}{r^2} where kk is Coulomb's constant, QQ is the magnitude of each of the charges, qq is the charge at the center, and rr is the distance between the charges.
2. Equilibrium Condition:
o For the system to be in equilibrium, the forces on qq due to the charges QQ must balance each other. Since the two forces are equal in magnitude and opposite in direction, they will cancel out, and the net force on qq will be zero.
o Now, let's consider the forces acting on the charges QQ. Since qq is placed symmetrically between the two charges QQ, the forces on each QQ will be due to the repulsion from qq and from the other QQ.
3. Force between QQ and QQ:
o The force between the two charges QQ is given by Coulomb's law as: Fbetween Q’s=k⋅Q2r2F_{\text{between Q's}} = \frac{k \cdot Q^2}{r^2}
o This force tends to repel the two charges QQ from each other.
4. Force between QQ and qq:
o The force on QQ due to qq is given by: Fdue to q=k⋅Q⋅qr2F_{\text{due to q}} = \frac{k \cdot Q \cdot q}{r^2}
o This force tends to attract the charges QQ towards qq.
5. Equilibrium Condition for QQ:
o For the system to be in equilibrium, the net force on each charge QQ must be zero. This means that the repulsive force between the charges QQ must balance the attractive force between each QQ and qq.
o Thus, we have: k⋅Q2r2=k⋅Q⋅qr2\frac{k \cdot Q^2}{r^2} = \frac{k \cdot Q \cdot q}{r^2}
6. Solving for qq:
o Simplifying the equation: Q2=Q⋅qQ^2 = Q \cdot q
o Dividing both sides by QQ (assuming Q≠0Q \neq 0): Q=qQ = q
o Therefore, the value of qq is equal to −Q/2-Q/2 in order to balance the forces and achieve equilibrium.
Conclusion:
The value of qq required for the system to be in equilibrium is −Q/2-Q/2.
Answer: B) −Q/2-Q/2.