The electric field intensity at a point on th | vedclass

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Question

(D) The electric field intensity on the axis of a dipole in air is given by $E_{	ext{axis}} = \frac{2kP}{r^3}$.Given $E_{	ext{axis}} = 4 	ext{ NC}^

Vedclass · Accepted Answer

$\frac{1}{16} 	ext{ NC}^{-1}$

Vedclass · Answer

(D) The electric field intensity on the axis of a dipole in air is given by $E_{	ext{axis}} = \frac{2kP}{r^3}$.Given $E_{	ext{axis}} = 4 	ext{ NC}^{-1}$,we have $4 = \frac{2kP}{r^3}$,which implies $\frac{kP}{r^3} = 2$ (Equation $i$).The electric field intensity on the equatorial line of a dipole in a medium with dielectric constant $K$ is given by $E_{	ext{eq}} = \frac{1}{K} \frac{kP}{r_1^3}$.Here,$r_1 = 2r$ and $K = 4$.Substituting these values,we get $E_{	ext{eq}} = \frac{1}{4} \cdot \frac{kP}{(2r)^3} = \frac{1}{4} \cdot \frac{kP}{8r^3} = \frac{1}{32} \cdot \frac{kP}{r^3}$.Using Equation $(i)$,$\frac{kP}{r^3} = 2$,so $E_{	ext{eq}} = \frac{1}{32} 	imes 2 = \frac{1}{16} 	ext{ NC}^{-1}$.Thus,the correct option is $(d)$.

Similar Questions

Two charges each of magnitude $0.01\,C$ and separated by a distance of $0.4\,mm$ constitute an electric dipole. If the dipole is placed in a uniform electric field $\vec{E}$ of $10\,dyne/C$ making a $30^{\circ}$ angle with $\vec{E}$,the magnitude of torque acting on the dipole is:

Two charges $+q$ and $-q$ are kept at a certain distance apart. Then, at any point on the perpendicular bisector of the line joining the two charges:

In the situation shown in the diagram,if $q \ll |Q|$ and $r \gg a$,find the net force on the free charge $-q$ and the net torque on it about $O$ at the instant shown. ($p = 2aQ$ is the dipole moment).

Assertion $:-$ An electric dipole experiences maximum force in a uniform electric field,when it is placed with its axis at right angles to the field direction.
Reason $:-$ When the axis of a dipole is perpendicular to a uniform external electric field,then torque acting on it will be zero.

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