Apoint charge q is held at a distance r from the center of a dipole that consists of two charges ±qseparated 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,
what is the magnitude of the force on the dipole?
in the space provided, enter the factor that multiplies 1ϵ0 in your answer. express this factor in terms of q, q, s, π, and r.
what is the magnitude of the torque on the dipole?
in the space provided, enter the factor that multiplies 1ϵ0 in your answer. express this factor in terms of q, q, s, π, and r.

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

where,

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, .

the total distance traveled by the car is

30km + 49km = 79 km .