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?

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


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