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melisa1 [442]
3 years ago
11

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.

The dipole is initially oriented so that the charge Q is located in the plane that bisects the dipole. Assume that r>>s
Immediately after the dipole is released:
a. What is the magnitude of the force on the dipole?
b. What is the magnitude of the torque on the dipole?
Physics
1 answer:
marishachu [46]3 years ago
8 0

Answer:

a) the magnitude of the force is

F= Q(\frac{kqs}{r^3}) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b)  the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = \frac{kq}{r^{2} }

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = \frac{kp}{r^{3} } , where p = q × s

E(r) = \frac{kqs}{r^{3} } where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q(\frac{kqs}{r^3})

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall  F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

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