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Point charges +4q,−q and +4q are kept on the X -axis at point x=0,x=a and x=2a respectively. Then
  • a)
    only −q is in stable equilibrium
  • b)
    all the charges are in stable equilibrium
  • c)
    all of the charges are in unstable equilibrium
  • d)
    none of the charges is in equilibrium
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Point charges +4q,−q and +4q are kept on the X -axis at point x=0,x=a...
Explanation:

Given Charges:
- +4q at x=0
- -q at x=a
- +4q at x=2a

Analysis:
- The charge at x=a will experience a repulsive force from the charge at x=0 and an attractive force from the charge at x=2a.
- The net force on the charge at x=a will not be zero. It will be unstable equilibrium as any small displacement will result in the charge moving further away from its initial position.
- Similarly, the other charges at x=0 and x=2a will also be in unstable equilibrium due to the repulsive forces between them.

Conclusion:
- Therefore, all of the charges are in unstable equilibrium as any small displacement will lead to a net force acting on them, causing them to move further away from their initial positions.
This explanation helps in understanding the concept of equilibrium of point charges on the X-axis and how the given configuration results in unstable equilibrium for all the charges.
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Community Answer
Point charges +4q,−q and +4q are kept on the X -axis at point x=0,x=a...
Net force on each of the charge due to the other charges is zero. However, disturbance in any direction other than along the line on which the charges lie, will not make the charges return.
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Point charges +4q,−q and +4q are kept on the X -axis at point x=0,x=a and x=2a respectively. Thena)only −q is in stable equilibriumb)all the charges are in stable equilibriumc)all of the charges are in unstable equilibriumd)none of the charges is in equilibriumCorrect answer is option 'C'. Can you explain this answer?
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