In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10⁵ NC^–1 per meter. What are the force and torque experienced by a system having a total dipole moment equal to 10^–7 Cm in the negative z-direction?

Question

In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10⁵ NC^–1 per meter. What are the force and torque experienced by a system having a total dipole moment equal to 10^–7 Cm in the negative z-direction?

Philoid · Accepted Answer

A dipole is a system consisting of two charges equal in magnitude and opposite in nature i.e. a positive charge + q and a negative charge –q separated by some distance d

Electric dipole moment of a dipole is given by,

P = q × d

Where, ‘q’ is magnitude of either of charge (in column)

And ‘d’ is separation between the pair of charges (in metres)

Dipole moment is a vector quantity and its direction is from negative charge to positive charge

We are given dipole moment of the system,

P = 10^-7Cm , along negative Z axis

Here Electric Field is varying at the rate of 10⁵ NC^–1 per metre in positive Z direction

i.e. dE/dl = 10⁵ NC^–1m^-1 along Z direction

so the electric field at any point on z axis at distance of a metres is

Now the total dipole moment equal to 10^–7 Cm in the negative z-direction.

The system and the Forces on it are as shown in the figure

Force on a charged particle in an electric field is given by

F = q × E

Where q is the magnitude of charge and E is the magnitude of Electric Field and the force is same as the direction of electric field in case of a positively charged particle and opposite to the direction of field in case of negatively charged particle

so Force on the + q charge located at a distance l from origin will be

directed towards positive Z axis

(since the electric field on z axis at distance of l metres from origin )

Force on the -q charge located at a distance l + d from origin will be

directed towards negative Z axis

(since the electric field on z axis at distance of l + d metres from origin)

So net force on the system would be

F = F _{+ q} – F_-q

(since F _{+ q} is directed towards positive Z axis and F_-q is directed towards negative Z axis )

So net force will be

Solving we get

We know,

q × d = P which is the dipole moment of the system so we get

Putting the values of P = 10^–7 Cm and = 10⁵ NC^–1m^-1 We get

F = - (10^-7C × 10⁵NC^-1m^-1)

i.e. F = -10^-2 N

Negative sign states that force is along negative Z axis , so the net force on the system is 10^-2 N in negative Z direction.

We know Torque on an Electric Dipole is Vector or cross product of dipole moment P and electric field E and is given as

= P E

which can be further evaluated as

= PE.sin(θ)

is the Torque of the dipole

Where P is the magnitude of dipole moment

E is the magnitude of electric field

is the angle between Dipole moment and electric field

is a unit vector perpendicular to plane containing P and E

Here dipole moment P is along negative z axis and electric field E along positive Z axis so angle between them

So, Sin = Sin180 = 0

So whatever be the magnitude of P and E putting in the equation = PESin

We get the torque τ = 0 Nm

So the torque of the system is zero