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A charge +q is placed (a, 0, 0) and another +q is placed (-a, 0, 0). A charge –q is placed at the origin. If it is slightly displaced along y-axis. Then
- a)It will move away
- b)It will execute SHM
- c)It will stand at the displaced position
- d)It will oscillate but not SHM
Correct answer is option 'B'. Can you explain this answer?
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A charge+qis placed(a, 0, 0)and another+qis placed(-a, 0, 0).A charge&...
The correct answer is: It will execute SHMMost Upvoted Answer
A charge+qis placed(a, 0, 0)and another+qis placed(-a, 0, 0).A charge&...
Q is placed at (0, a, 0).
We can use the concept of the electric field to find the force experienced by the charge q at (0, a, 0) due to the charges at (a, 0, 0) and (-a, 0, 0).
First, we need to find the electric field at the position of charge q due to the charge at (a, 0, 0). The electric field at a point due to a charge is given by the equation:
E = k * q1 / r^2
Where E is the electric field, k is the electrostatic constant, q1 is the charge creating the field, and r is the distance between the point and the charge.
In this case, the electric field at the position (0, a, 0) due to the charge at (a, 0, 0) can be calculated as:
E1 = k * q / a^2
Similarly, the electric field at the position (0, a, 0) due to the charge at (-a, 0, 0) can be calculated as:
E2 = k * q / (2a)^2 = k * q / 4a^2
Now, the net electric field at the position (0, a, 0) due to the two charges can be calculated by adding the individual electric fields:
E_net = E1 + E2 = k * q / a^2 + k * q / 4a^2
Simplifying the equation further:
E_net = (5k * q) / 4a^2
Finally, we can calculate the force experienced by the charge q at (0, a, 0) due to the net electric field:
F = q * E_net
F = q * (5k * q) / 4a^2
F = (5k * q^2) / 4a^2
Therefore, the force experienced by the charge q at (0, a, 0) due to the charges at (a, 0, 0) and (-a, 0, 0) is (5k * q^2) / 4a^2.
We can use the concept of the electric field to find the force experienced by the charge q at (0, a, 0) due to the charges at (a, 0, 0) and (-a, 0, 0).
First, we need to find the electric field at the position of charge q due to the charge at (a, 0, 0). The electric field at a point due to a charge is given by the equation:
E = k * q1 / r^2
Where E is the electric field, k is the electrostatic constant, q1 is the charge creating the field, and r is the distance between the point and the charge.
In this case, the electric field at the position (0, a, 0) due to the charge at (a, 0, 0) can be calculated as:
E1 = k * q / a^2
Similarly, the electric field at the position (0, a, 0) due to the charge at (-a, 0, 0) can be calculated as:
E2 = k * q / (2a)^2 = k * q / 4a^2
Now, the net electric field at the position (0, a, 0) due to the two charges can be calculated by adding the individual electric fields:
E_net = E1 + E2 = k * q / a^2 + k * q / 4a^2
Simplifying the equation further:
E_net = (5k * q) / 4a^2
Finally, we can calculate the force experienced by the charge q at (0, a, 0) due to the net electric field:
F = q * E_net
F = q * (5k * q) / 4a^2
F = (5k * q^2) / 4a^2
Therefore, the force experienced by the charge q at (0, a, 0) due to the charges at (a, 0, 0) and (-a, 0, 0) is (5k * q^2) / 4a^2.
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A charge+qis placed(a, 0, 0)and another+qis placed(-a, 0, 0).A charge&...
What is the condition for a particle to oscillate but then also do shm ? how can we differentiate between the two ?
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A charge+qis placed(a, 0, 0)and another+qis placed(-a, 0, 0).A charge–qis placed at the origin. If it is slightly displaced alongy-axis. Thena)It will move awayb)It will executeSHMc)It will stand at the displaced positiond)It will oscillate but notSHMCorrect answer is option 'B'. Can you explain this answer?
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