Question

In a certain region of space, electric field is along the z-directionthroughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of105 NC–1 per metre. What are the force and torque experienced by asystem having a total dipole moment equal to 10–7 Cm in the negativez-direction ?

Answer 1

Given: The rate of increment of the magnitude of the electric field is 10 5  N/C and the total dipole moment is 10 −7  C-m in negative z direction.

The force experienced by the system in electric field is given by,

F=qE =q× dE dl ×dl =( q×dl )× dE dl F=p× dE dl

Where, the force is F, the dipole moment is p, the magnitude of point charge is q, the distance of the charge field is l and the rate of change of electric field intensity is dE dl .

By substituting the given values in the above equation, we get

F=− 10 −7 × 10 5 =− 10 −2 =−0.01 N

The negative sign shows that the direction of force is in the opposite to the electric field that is in the negative z direction.

Thus, the force experienced by the system is 0.01 N.

Since, the electric field and the dipole moment are in opposite direction therefore, the angle between them is 180°.

The torque acting on the system is given as,

τ=pEsinθ

Where, the net torque is τ, the electric field is E, the dipole moment is p and the angle between the dipole moment and electric field is θ.

By substituting the given values in the above equation, we get

τ=pEsin180° =0

Thus, the net torque experienced by the system is zero.