An electric dipole coincides on $Z$-axis and its mid point is on origin of the co-ordinate system. The electric field at an axial point at a distance $z$ from origin is ${\overrightarrow{E}}_{(z)}$ and electric field at an equatorial point at a distance $y$ from origin is ${\overrightarrow{E}}_{(y)}$ Here $z = y \gg a,\ so\ \left| \frac{{\overrightarrow{E}}_{(z)}}{{\overrightarrow{E}}_{(y)}} \right| = \ldots$

Question

Dr. Abubacker · Accepted Answer

(d) The magnitude of electric field at a axial point $P$ at a distance $z$ from the origin is given by
[image:1_image301.emf]
$\left| {\overrightarrow{E}}_{(z)} ight|\frac{4qaz}{4\pi\varepsilon_{0}\left( z^{2} - a^{2} ight)^{2}} = \frac{2pz}{4\pi\varepsilon_{0}\left( z^{2} - a^{2} ight)^{2}}$
Where $p = 2qa$ is the electric dipole moment
For $z > > a,\left| {\overrightarrow{E}}_{(z)} ight| = \frac{2p}{4\pi\varepsilon_{0}z^{3}}$
The magnitude of electric field at a equatorial point $Q$ at a distance $y$ from the origin is given by
$\left| {\overrightarrow{E}}_{(y)} ight| = \frac{1}{4\pi\varepsilon_{0}}\frac{2qa}{\left( y^{2} + a^{2} ight)^{3/2}} = \frac{p}{4\pi\varepsilon_{0}\left( y^{2} + a^{2} ight)^{3/2}}$
For $y > > a,\left| {\overrightarrow{E}}_{(y)} ight| = \frac{p}{4\pi\varepsilon_{0}y^{3}}$
For $z = y > > a,\ 	herefore\frac{\left| {\overrightarrow{E}}_{(z)} ight|}{\left| E_{(y)} ight|} = 2$

Explanation

Reviewed by Dr. Abubacker

Want more practice?