The electric field intensity at a point on the axis of an electric dipole in air is $4 \mathrm{NC}^{-1}$.…
The electric field intensity at a point on the axis of an electric dipole in air is $4 \mathrm{NC}^{-1}$. Then the electric field intensity at a point on the equatorial line which is at a distance equal to twice the distance on the axial line and if the dipole is in a medium of dielectric constant 4 is
$1 \mathrm{NC}^{-1}$
$\frac{1}{8} \mathrm{NC}^{-1}$
$16 \mathrm{NC}^{-1}$
$\frac{1}{16} \mathrm{NC}^{-1}$
Solution
As, electric field intensity on the axis,
$
\begin{aligned}
E_{\text {axis }}=\frac{2 k P}{r^3} \Rightarrow 4=\frac{2 k P}{r^3} \text { for } r & >a \\
& \left(\because \text { given, } E_{\text {axis }}=4\right)
\end{aligned}
$
Electric field intensity on equatorial line,
$
E_{\mathrm{eq}}=\frac{k^{\prime} P}{r_1^3}
$
where, $\quad r_1=2 r$ and $k^{\prime}=\frac{k}{4}$
So,
$
E_{\text {eq }}=\frac{k P}{4 \times 8 r^3}
$
Now, from the Eq. (i), we get
$
E_{\mathrm{eq}}=\frac{2}{4 \times 8}=\frac{1}{16} \mathrm{NC}^{-1}
$
Hence, the correct option is (d)
