Question

In a certain of space , electric field is along the $z -$ direction throughout the magnitude of electric field is however is not constant , but increases uniformly along the positive $z -$ direction at the rate of ${10^5}N{C^{ - 1}}{M^{ - 1}}$. The force experienced by the system having a total dipole moment equal to ${10^{ - 7}}CM$ in the negative$z -$direction is
A) $- {10^{ - 2}}N$
B) ${10^{ - 2}}N$
C) ${10^{ - 4}}N$
D) $- {10^{ - 4}}N$

Answer 1

This is based on dipole moment so here we will use dipole moment in term of derivative of electric field then use that in the expression of force experienced due to electric field
We have used the formula
Force ( F) experienced by the system
$F = qE$

Complete step by step solution:
Given , Dipole moment of the system in the negative $z - direction$ is ,

p = q \times dl \\\ p = - {10^{ - 7}}CM \\\ \ $$ And rate of increase of electric field per unit length along the positive $$z - direction$$ is $$\dfrac{{dE}}{{dl}} = {10^5}N{C^{ - 1}}$$ Now , we have calculate the force experienced by the system Therefore , Force ( F ) experienced by the system is given by the relation We have $$F = qE$$ Where ,$$F = $$ Force of the experienced by the system $$q = $$ electric charge $$E = $$ electric field So ,$$F = qE$$ Amount change of the electric field is $$F = qdE$$ Multiply and divided by $$dl$$of right hand side , we get $$F = q\dfrac{{dE}}{{dl}} \times dl$$ $$ \Rightarrow F = q\dfrac{{dE}}{{dl}} \times dl$$ $$ \Rightarrow F = qdl \times \dfrac{{dE}}{{dl}}$$ Since , electric dipole moment $$P = q \times dl$$ So ,$$F = P \times \dfrac{{dE}}{{dl}}$$ Substituting the value of electric dipole moment $$P$$, we get $$F = {10^{ - 7}} \times \dfrac{{dE}}{{dl}}$$ Again , substituting the value rate of electric field per unit length $$\dfrac{{dE}}{{dl}}$$,we get $$\ F = - {10^{ - 7}} \times {10^5} \\\ \Rightarrow F = - {10^{ - 7 + 5}} \\\ \Rightarrow F = - {10^{ - 2}}N \\\ \ $$ **The force is $$ - {10^{ - 2}}N$$ in the negative $$z - direction$$ $$ie,$$opposite to the direction of the electric field. Hence, the angle between electric field and dipole moment is $${180^0}$$.** **Notes:** The electric dipole moment for a pair of charges of magnitude q is defined as the magnitude of the charge times the distance between them and the defined direction is toward the positive charge. Applications involve the electric field of a dipole and the energy of a dipole when placed in an electric field.