Obtain the ratio of Electric field formed by an ideal dipole at points lying on axial and equatorial axis.

### Asked by avaneesh1162 | 12th Jun, 2021, 11:06: AM

Expert Answer:

###
Let a dipole consists of charges +q and -q that are separated by a distance d is placed at origin of
X-Y coordinate system so that mid point of dipole coinicides with origin.
Electric field at a point x = r on x-axis is given as

..................................(1)

Where K = 1/ (4πε_{o} ) is Coulomb's constant
if we consider dipole moment p = ( q × d ) and r >> d/2 , then eqn.(1) is written as
E_{x} = K × [ ( 2 p ) / r^{3} ] ............................ (2)
Electric field at a point y = r on y-axis is given as
E_{y} = 2 E cosθ .................................(3)
Where E is given as
Hence we get E_{y }as , .........................(4)

In above expression , we used the approximation r >> d and dipole moment p = ( q × d )
From eqn.(2) and eqn.(4) , we get ( E_{x} / E_{y} ) = 2

Let a dipole consists of charges +q and -q that are separated by a distance d is placed at origin of

X-Y coordinate system so that mid point of dipole coinicides with origin.

Electric field at a point x = r on x-axis is given as

..................................(1)

Where K = 1/ (4πε

_{o}) is Coulomb's constantif we consider dipole moment p = ( q × d ) and r >> d/2 , then eqn.(1) is written as

E

_{x}= K × [ ( 2 p ) / r^{3}] ............................ (2)Electric field at a point y = r on y-axis is given as

E

_{y}= 2 E cosθ .................................(3)Where E is given as

Hence we get E

_{y }as , .........................(4)In above expression , we used the approximation r >> d and dipole moment p = ( q × d )

From eqn.(2) and eqn.(4) , we get ( E

_{x}/ E_{y}) = 2### Answered by Thiyagarajan K | 12th Jun, 2021, 02:59: PM

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