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What will be the value of a charge q, such that when it is placed at the centre of two equal and like charges Q, the three charges are in equilibrium.
  • a)
    +Q/4
  • b)
    Q
  • c)
    -Q/4
  • d)
    Q/2
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
What will be the value of a charge q, such that when it is placed at t...
If we see all the forces acting upon the charge Q due to other charge Q at distance r and other charge q at distance r/2, we get
kQQ/r2 + kQ(q)/(r/2)2
This results 
Q2 / r2  = -4Qq / r2
Thus we get q = -Q/4
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Most Upvoted Answer
What will be the value of a charge q, such that when it is placed at t...
To understand why the correct answer is option C, let's break down the problem into different components:

#Given Information:
- Two equal and like charges Q are placed at a certain distance from each other.
- Another charge q is placed at the center of the two charges Q.

#Equilibrium Condition:
For the three charges to be in equilibrium, the net force on each charge must be zero. This means that the attractive forces between the charges should balance out the repulsive forces.

#Analysis:
Let's consider the forces acting on the charge q:

1. Attraction Force from Charge Q1:
The charge Q1 exerts an attractive force on charge q, given by Coulomb's Law:
F1 = k * Q1 * q / r^2
Where k is the electrostatic constant, Q1 is the charge of Q1, q is the charge of q, and r is the distance between Q1 and q.

2. Attraction Force from Charge Q2:
The charge Q2 exerts an attractive force on charge q, given by Coulomb's Law:
F2 = k * Q2 * q / r^2
Where k is the electrostatic constant, Q2 is the charge of Q2, q is the charge of q, and r is the distance between Q2 and q.

3. Repulsion Force between Charges Q1 and Q2:
The charges Q1 and Q2 repel each other with a force given by Coulomb's Law:
F3 = k * Q1 * Q2 / d^2
Where k is the electrostatic constant, Q1 is the charge of Q1, Q2 is the charge of Q2, and d is the distance between Q1 and Q2.

#Net Force Calculation:
Since the charges are in equilibrium, the net force on charge q must be zero. Therefore, we can write the equation:
F1 + F2 + F3 = 0
Substituting the expressions for F1, F2, and F3:
k * Q1 * q / r^2 + k * Q2 * q / r^2 + k * Q1 * Q2 / d^2 = 0

#Simplification:
Dividing through by k and rearranging the equation, we get:
Q1 * q / r^2 + Q2 * q / r^2 + Q1 * Q2 / d^2 = 0
Multiplying through by r^2 and d^2:
Q1 * q * d^2 + Q2 * q * d^2 + Q1 * Q2 * r^2 = 0

#Final Step:
If we assume Q1 = Q2 = Q, we can simplify further:
Q * q * d^2 + Q * q * d^2 + Q^2 * r^2 = 0
2Q * q * d^2 + Q^2 * r^2 = 0
2q * d^2 + Q * r^2 = 0
Solving for q, we get:
q = -Q * r^2 / (2d^2)

Therefore, the value of charge q that will maintain equilibrium is -Q * r^2 / (2d^2), which matches option C: -Q/4.

Note: The negative sign indicates that the charge q will have an opposite polarity
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What will be the value of a charge q, such that when it is placed at the centre of two equal and like charges Q, the three charges are in equilibrium.a)+Q/4b)Qc)-Q/4d)Q/2Correct answer is option 'C'. Can you explain this answer?
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