Respuesta :
Answer:
The magnitude of the force on the dipole due to the charge Q = [tex]\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.[/tex]
The magnitude of the torque on the dipole = [tex]\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.[/tex]
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
[tex]\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.[/tex]
where,
k is the Coulomb's constant, having value = [tex]\dfrac{1}{4\pi \epsilon_o}[/tex]
[tex]\epsilon_o[/tex] is the electrical permittivity of free space.
Also, r>>s, therefore, [tex]\rm r^2+s^2\approx r^2.[/tex]
[tex]\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.[/tex]
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
[tex]\rm F=qE[/tex]
Therefore, the electric force on the charge Q due to the dipole is given by
[tex]\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.[/tex]
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 = [tex]\dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.[/tex]
The magnitude of the torque on the dipole is given by
[tex]\rm \tau = Fs\ \sin\theta[/tex]
Since the charge Q is placed in the plane that bisects the dipole, therefore, [tex]\theta = 90^\circ[/tex].
[tex]\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.[/tex]
