A student pulls a box across a horizontal floor at a constant speed of 4.0 meters per second by exerting a constant horizontal f
2 answers:
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Answer:
The magnitude of the force on the dipole due to the charge Q =
The magnitude of the torque on the dipole =
Explanation:
Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.
The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by
<em>where</em>,
k is the Coulomb's constant, having value =
is the electrical permittivity of free space.
Also, r>>s, therefore,
The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as
Therefore, the electric force on the charge Q due to the dipole is given by
According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.
Thus, the magnitude of the force on the dipole due to the charge Q =
The magnitude of the torque on the dipole is given by
Since the charge Q is placed in the plane that bisects the dipole, therefore, .