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Two equal negative charges -q are fixed at points (0, -a) and (0, a) on y-axis. A positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge Q will be
- a)Execute simple harmonic motion about the origin
- b)Move to the origin and remain at rest
- c)Move to infinity
- d)Execute oscillatory but not simple harmonic motion
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Two equal negative charges -q are fixed at points (0, -a) and (0, a) o...
Here Q is stay at point (2a, 0) which is very for in x-axis (2a).
Motion is simple harmonic only if Q is released from a point not very far from the origin on x-axis. (for small oscillations) other wise motion is periodic but not simple harmonic. So here such type of motion is not simple harmonic motion.
Motion is simple harmonic only if Q is released from a point not very far from the origin on x-axis. (for small oscillations) other wise motion is periodic but not simple harmonic. So here such type of motion is not simple harmonic motion.
Most Upvoted Answer
Two equal negative charges -q are fixed at points (0, -a) and (0, a) o...
Explanation:
When a positive charge Q is released from rest at the point (2a, 0) on the x-axis, it will experience forces due to the negative charges -q fixed at points (0, -a) and (0, a) on the y-axis. Let's analyze the forces acting on the charge Q.
Force due to the charge at (0, -a):
The force between two charges can be calculated using Coulomb's law:
F1 = k * Q * (-q) / r1^2
The direction of this force will be towards the charge -q at (0, -a).
Force due to the charge at (0, a):
Similarly, the force due to the charge at (0, a) can be calculated:
F2 = k * Q * (-q) / r2^2
The direction of this force will be towards the charge -q at (0, a).
Since charge Q is positive, the forces F1 and F2 will be directed towards the origin.
Resultant force:
The resultant force on charge Q can be obtained by vector addition of F1 and F2:
F = F1 + F2
The direction of this resultant force will depend on the magnitudes and directions of F1 and F2.
Analysis of forces:
Let's consider the case where Q is slightly displaced towards the origin. In this case, the force F1 will increase in magnitude, while the force F2 will decrease in magnitude. Therefore, the resultant force F will be directed towards the origin.
Similarly, if Q is slightly displaced away from the origin, the force F1 will decrease in magnitude, while the force F2 will increase in magnitude. Again, the resultant force F will be directed towards the origin.
Oscillatory motion:
From the above analysis, we can conclude that the force acting on charge Q will always be directed towards the origin, regardless of its position. This means that charge Q will execute oscillatory motion. However, it will not be simple harmonic motion because the force acting on Q is not directly proportional to its displacement from the equilibrium position.
Therefore, the correct answer is option D: Charge Q will execute oscillatory but not simple harmonic motion.
When a positive charge Q is released from rest at the point (2a, 0) on the x-axis, it will experience forces due to the negative charges -q fixed at points (0, -a) and (0, a) on the y-axis. Let's analyze the forces acting on the charge Q.
Force due to the charge at (0, -a):
The force between two charges can be calculated using Coulomb's law:
F1 = k * Q * (-q) / r1^2
The direction of this force will be towards the charge -q at (0, -a).
Force due to the charge at (0, a):
Similarly, the force due to the charge at (0, a) can be calculated:
F2 = k * Q * (-q) / r2^2
The direction of this force will be towards the charge -q at (0, a).
Since charge Q is positive, the forces F1 and F2 will be directed towards the origin.
Resultant force:
The resultant force on charge Q can be obtained by vector addition of F1 and F2:
F = F1 + F2
The direction of this resultant force will depend on the magnitudes and directions of F1 and F2.
Analysis of forces:
Let's consider the case where Q is slightly displaced towards the origin. In this case, the force F1 will increase in magnitude, while the force F2 will decrease in magnitude. Therefore, the resultant force F will be directed towards the origin.
Similarly, if Q is slightly displaced away from the origin, the force F1 will decrease in magnitude, while the force F2 will increase in magnitude. Again, the resultant force F will be directed towards the origin.
Oscillatory motion:
From the above analysis, we can conclude that the force acting on charge Q will always be directed towards the origin, regardless of its position. This means that charge Q will execute oscillatory motion. However, it will not be simple harmonic motion because the force acting on Q is not directly proportional to its displacement from the equilibrium position.
Therefore, the correct answer is option D: Charge Q will execute oscillatory but not simple harmonic motion.
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Two equal negative charges -q are fixed at points (0, -a) and (0, a) on y-axis. A positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge Q will bea)Execute simple harmonic motion about the originb)Move to the origin and remain at restc)Move to infinityd)Execute oscillatory but not simple harmonic motionCorrect answer is option 'D'. Can you explain this answer?
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ample number of questions to practice Two equal negative charges -q are fixed at points (0, -a) and (0, a) on y-axis. A positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge Q will bea)Execute simple harmonic motion about the originb)Move to the origin and remain at restc)Move to infinityd)Execute oscillatory but not simple harmonic motionCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice GATE tests.
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