Question

For a dipole, why is the electric potential zero directly in the middle?

Answer 1

Please note that the potential \[\varphi \]
of an electric dipole, is given by the subsequent formula:
\[\varphi = \frac{{p\;z}}{{{{\left( {{x^2} + {z^2}} \right)}^{3/2}}}}\]
we are in the x,z-plane P is the electric dipole moment and our reference frame is located in the mid point of our electric dipole

Answer 2

https://www.physicsforums.com/threads/electric-potential-at-the-center-of-a-dipole.658004/

Answer 3

ohhhhhhhhhh... that's what I needed. Thank you @TheDream

Answer 4

No problem :D

Answer 5

@Michele_Laino @physical.yogaling \[ \varphi = \frac{{p\;z}}{{{{\left( {{x^2} + {z^2}} \right)}^{3/2}}}} \] is the equation for electric field.

The potential can be written simply as\[V = \frac{ Q }{ 4\pi \epsilon _{0} }\left( \frac{ 1 }{r _{+} } -\frac{ 1 }{ r _{-}}\right)\] where r+ is the distance from the positive charge and r- is the distance from the negative charge so at the midpoint between r+= r- you get V=0

Answer 6

Please note that my formula expresses the potential function in the Gauss system, at all distances from the electric dipole, furthermore, your formula is an approximation, which is valid only for large distance from the electric dipole.

Answer 7

@Michele_Laino

Excuse me but your formula is dimensionally incorrect to start and the distance dependence is incorrect for the potential and my formula for potential is exact (not an approximation) for any point in space.

Answer 8

Please note that the unit of measure of p is esu*cm
so the unit measure of phi is:
(esu*cm^2)/cm^3 = esu/cm

Answer 9

please note that p is the electric dipole moment, namely charge*distance

Answer 10

What is your point?

Answer 11

@Michele_Laino @physical.yogalingB
OK, your formula is correct, please accept my apology. but mine is also correct and more easily shows why the potential is zero at the center of the dipole and on the axis perpendicular to the dipole through the center of the dipole.

Answer 12

no problem you are welcome! :)