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Two charges, each equal to q, are kept at x=−a and x=a on the x-axis. A particle of mass m and charge q 0 = q/2 is placed at the origin. If charge qo is given a small displacement (y
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Two charges, each equal to q, are kept at x=−a and x=a on the x-axis. ...
Introduction:
In this scenario, we have two charges, each equal to q, placed at x = -a and x = a on the x-axis. A particle of mass m and charge q0 = q/2 is placed at the origin. We need to determine the motion of the particle when it is given a small displacement y.

Explanation:
To analyze the motion of the particle, we need to consider the forces acting on it.

Electric Force:
The electric force between two charges is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

In this case, the particle at the origin experiences electric forces from the charges at x = -a and x = a. The force exerted by the charge at x = -a will be attractive, while the force exerted by the charge at x = a will be repulsive.

Net Electric Force:
To find the net electric force on the particle, we need to consider the vector sum of the forces due to the two charges. Since the charges are located on the x-axis, the forces will act along the x-axis as well.

The force due to the charge at x = -a can be calculated using Coulomb's law:

F1 = k * (q * q0) / (-a)^2

The force due to the charge at x = a can also be calculated using Coulomb's law:

F2 = k * (q * q0) / a^2

Since the charges have opposite signs, the forces will have opposite directions. Therefore, the net force on the particle will be the vector difference of the two forces:

Fnet = F1 - F2

Equation of Motion:
Using Newton's second law, we can relate the net force to the acceleration of the particle:

Fnet = m * a

Substituting the expression for Fnet, we get:

F1 - F2 = m * a

Simplifying the equation, we find:

k * (q * q0) / (-a)^2 - k * (q * q0) / a^2 = m * a

Multiplying both sides by a^2, we get:

k * (q * q0) / a - k * (q * q0) / (-a) = m * a^3

Simplifying further, we have:

2 * k * (q * q0) / a = m * a^3

Conclusion:
By solving the equation of motion, we can determine the acceleration of the particle when it is given a small displacement y. The direction of the acceleration will depend on the relative magnitudes of the charges and the mass of the particle.
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Two charges, each equal to q, are kept at x=−a and x=a on the x-axis. A particle of mass m and charge q 0 = q/2 is placed at the origin. If charge qo is given a small displacement (y
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Two charges, each equal to q, are kept at x=−a and x=a on the x-axis. A particle of mass m and charge q 0 = q/2 is placed at the origin. If charge qo is given a small displacement (y for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two charges, each equal to q, are kept at x=−a and x=a on the x-axis. A particle of mass m and charge q 0 = q/2 is placed at the origin. If charge qo is given a small displacement (y covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two charges, each equal to q, are kept at x=−a and x=a on the x-axis. A particle of mass m and charge q 0 = q/2 is placed at the origin. If charge qo is given a small displacement (y.
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