The electric field intensity at a point on th | vedclass

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(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 $(d)$.

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