Question

The electric field intensity at a point on the axis of an electric dipole in ah is $4 \; NC^{-1}$. Then the electric field intensity at a point on the equitorial hue 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

• A. $1 \; NC^{-1}$
• B. $\frac{1}{8} \; NC^{-1}$
• C. $16 \; \; NC^{-1}$
• D. $\frac{1}{16} \; NC^{-1}$

Answer 1

Answer: (D)

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

Answer 2

As, electric field intensity on the axis,
$E_{\text {axis }}=\frac{2 k P}{r^{3}}$
$\Rightarrow 4=\frac{2 k P}{r^{3}}$ for $r > >a$
$\left(\because \text { given, } E_{\text {axis }}=4\right)$
$\Rightarrow \frac{k P}{r^{3}}=2 \ldots$ (i)
Electric field intensity on equatorial line,
$E_{ eq }=\frac{k^{\prime} P}{r_{1}^{3}}$
where, $r_{1}=2 r$ and $k'=\frac{k}{4}$
So, $E_{ eq }=\frac{k P}{4 \times 8 r^{3}}$
Now, from the E (i), we get
$E_{ eq }=\frac{2}{4 \times 8}=\frac{1}{16} NC ^{-1}$