Question

why electric dipole field deceasing by 1/r3??

Answer 1

because, electric dipole = distance * charge
E = \[\frac{ k \times q \times d }{ r ^{3} }\]
Therefore, distance on numerator cancels in unit distance^3 to distance^2

Answer 2

At long distances, the two charges of the dipole look to be very close together so that the field each creates would be expected to partially cancel each other. The appearance of closeness of the two opposite charges makes it look like a near net charge of zero is generating the field. So you would expect the field to drop off more quickly than 1/r^2. How much more quickly? Well that's easy to calculate. Just add together the fields due to each charge of the dipole separately to find out.



Now calculate the total field at the point in space:

\[E=kQ \left[ \frac{ 1 }{ r^{2} }-\frac{ 1 }{ r'^{2} } \right]=kQ \left[ \frac{ 1 }{ r^{2} }-\frac{ 1 }{ r^{2}+d^{2}-2rd \cos(A) } \right]=kQ \left[ \frac{ d^{2}-2rd \cos(A) }{ r^{2} \left\{ r^{2}+d^{2}-2rd \cos(A) \right\} } \right] \]

Divide the top and bottom of the fraction by r, then note in the numerator that d^2/r is very small for large r and in the denominator that r^2 >> d^2 and r^2 >> 2rd:

\[E=kQ \left[ \frac{ \frac{ d^{2} }{ r } - 2d \cos(A)}{ r \left\{ r^{2}+d^{2}-2rd \cos(a) \right\} } \right] \approx kQ \left[ \frac{ -2d \cos(A) }{ r^{3} } \right] \]

And there you see at large distance r, the field falls off as 1/r^3.

Answer 3

It needs to be mentioned that I did not do a vector addition of the two fields because I assumed a very far distance where the two E fields are almost parallel and their magnitudes can therefore be added together as if they were scalars.

Answer 4

Another way to show that the electric field of a dipole indeed falls off as 1/r3 is to find the individual potentials at the distant point due to the two charges independently. Since potential is a scalar, we can add them merely as two numbers by the principle of superposition. Therefore the net potential at the point P can be quickly observed to be

\[V = kQ/Ra - kQ/Rb\]

Now if we assume large distance , i.e the far field, we can observe that in the term Ra-Rb/Ra*Rb, Ra*Rb can be approximated as r2 and Ra-Rb can be considered to be \[dcos \theta \] where the angle theta is the angle made by Ra or Rb with the Radial vector R

Now that we have the potential equation, we can simply find the electric field by using the definition as negative gradient of potential is the electric field. If we do this using proper calculus, we will find that the field indeed falls of as 1/r3