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Two spherical conductors A and B having equal radii and carrying equal charges on them, repel each other with a force F, when kept apart at some distance. A third spherical conductor C having same radius as that of A, but uncharged, is brought in contact with A, then brought in contact with B and finally removed away from both. The new force of repulsion between the conductors A and B is
  • a)
    F/4
  • b)
    F/8
  • c)
    3F/4
  • d)
    3F/8
Correct answer is option 'D'. Can you explain this answer?
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Two spherical conductors A and B having equal radii and carrying equal...
Given: Two spherical conductors A and B having equal radii and carrying equal charges on them. They repel each other with a force F, when kept apart at some distance.

To find: The new force of repulsion between the conductors A and B when a third spherical conductor C having the same radius as that of A, but uncharged, is brought in contact with A, then brought in contact with B and finally removed away from both.

Solution:

Step 1: When conductor C is brought in contact with conductor A, the charges on A will distribute uniformly over both the conductors A and C due to electrostatic induction. The final charge on both A and C will be Q/2, where Q is the initial charge on A.

Step 2: When conductor C is brought in contact with conductor B, the charges on B will also distribute uniformly over both the conductors B and C due to electrostatic induction. The final charge on both B and C will be Q/2, where Q is the initial charge on A and B.

Step 3: Now, when C is removed away from both A and B, the charges on A and B will redistribute themselves due to electrostatic repulsion. Let the final charges on A and B be Q1 and Q2 respectively.

Step 4: By the principle of conservation of charge, we have:

Initial charge on A + Initial charge on B = Final charge on A + Final charge on B

Q + Q = Q1 + Q2

2Q = Q1 + Q2

Step 5: Using Coulomb's law, the force of repulsion between A and B can be given as:

F = kQ^2/d^2, where k is the Coulomb constant and d is the distance between A and B.

Step 6: Now, the new charges on A and B can be calculated using the principle of superposition of charges. The charges on A and B due to Q/2 charge on C can be considered as a separate system of charges. The force of repulsion between them can be calculated using Coulomb's law as:

F' = k(Q/2)^2/d^2 = kQ^2/4d^2

Step 7: The total force of repulsion between A and B can be obtained by adding the forces F and F':

F_total = F + F' = kQ^2/d^2 + kQ^2/4d^2

Step 8: Simplifying the above equation, we get:

F_total = (5/4)kQ^2/d^2

Step 9: Hence, the new force of repulsion between A and B is 5/4 times the initial force of repulsion between them. Therefore, the correct answer is option D, i.e., 3F/8.
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Two spherical conductors A and B having equal radii and carrying equal charges on them, repel each other with a force F, when kept apart at some distance. A third spherical conductor C having same radius as that of A, but uncharged, is brought in contact with A, then brought in contact with B and finally removed away from both. The new force of repulsion between the conductors A and B isa)F/4b)F/8c)3F/4d)3F/8Correct answer is option 'D'. Can you explain this answer?
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Two spherical conductors A and B having equal radii and carrying equal charges on them, repel each other with a force F, when kept apart at some distance. A third spherical conductor C having same radius as that of A, but uncharged, is brought in contact with A, then brought in contact with B and finally removed away from both. The new force of repulsion between the conductors A and B isa)F/4b)F/8c)3F/4d)3F/8Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two spherical conductors A and B having equal radii and carrying equal charges on them, repel each other with a force F, when kept apart at some distance. A third spherical conductor C having same radius as that of A, but uncharged, is brought in contact with A, then brought in contact with B and finally removed away from both. The new force of repulsion between the conductors A and B isa)F/4b)F/8c)3F/4d)3F/8Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two spherical conductors A and B having equal radii and carrying equal charges on them, repel each other with a force F, when kept apart at some distance. A third spherical conductor C having same radius as that of A, but uncharged, is brought in contact with A, then brought in contact with B and finally removed away from both. The new force of repulsion between the conductors A and B isa)F/4b)F/8c)3F/4d)3F/8Correct answer is option 'D'. Can you explain this answer?.
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