Equipotential surfaces for an electric dipole

[vcsharp2003]

Homework Statement

Question:
Equipotential surfaces associated with an electric dipole are

a) spheres centered on the dipole

b) cylinders with axes along the dipole moment.

c) planes parallel to the dipole moment.

d) planes perpendicular to the dipole moment.

e) None of the above

Relevant Equations

None

The answer is given as (a), but I think it's not correct based on the equipotential surfaces diagram given in our book for an electric dipole as below.

The red dashed lines, which are supposed to be the equipotential surfaces, are surely not representing a sphere centred at the dipole center. So, in my view, the correct answer is "(e) None of the above".

 

[PhDeezNutz]

I would agree with you. To me arguing that the equipotential surfaces of a dipole are spheres centered at the dipole is essentially saying that a dipole and a monopole are the same thing……which is insane.

Physical dipole or ideal dipole.

 

[vcsharp2003]

I would agree with you.

Ok. Thanks for your answer. One more question regarding this example. Are the equipotential surfaces for a dipole still spherical or are they some form of a conic?

 

[BvU]

That you can answer yourself using the expressions for the potential field...


$\$

 

[PhDeezNutz]

Ok. Thanks for your answer. One more question regarding this example. Are the equipotential surfaces for a dipole still spherical or are they some form of a conic?

Are we talking about a physical dipole or an ideal dipole?

for an ideal dipole definitely not because the spacing is so small the equipotential lines will smoosh together.

I'd have to work it out for a physical dipole.

 

[vcsharp2003]

Are we talking about a physical dipole or an ideal dipole?

for an ideal dipole definitely not because the spacing is so small the equipotential lines will smoosh together.

I'd have to work it out for a physical dipole.

As a guess, just by looking at the diagram, each dashed red line is definitely not a circle, so probably it would be some sort of an ellipse. And therefore, each equipotential surface would not be spherical.

 

[vanhees71]

Simply write down the potential $\Phi(\vec{r})$ for a dipole and check, what the surfaces $\Phi(\vec{r})=\text{const}$ are.

 

[BvU]

Don't look like no spheres to me though...
https://www.theochem.ru.nl/~pwormer/Knowino/knowino.org/wiki/File_Dipole_potential_field.html

 

[hutchphd]

One knows by (anti)symmetry that the plane between them is equipotential usually chosen =0

 

[BvU]

Well, that plane can be considered a (rather big ) sphere ...

$\$

 

[hutchphd]

centered very far away

 

[kuruman]

This is an elaboration of post #7 by @vanhees71.

Say it is true that "spheres centered on the dipole" are equipotentials. Call the direction of the dipole $\mathbf{\hat z}.~$ You could go to distance $r$ at angle $\theta$ relative to $z$ and construct such a sphere of radius $r$ centered at the dipole. Now go to the diametrically opposing point on the sphere and see whether the potential has the same value as the starting point. It does not. In fact it is the negative of that value. That's because $$V_i=\frac{k~p\cos(\theta)}{r^2}~~\text{and}~~V_f=\frac{k~p\cos(\theta + \pi)}{r^2}=-\frac{k~p\cos(\theta)}{r^2}=-V_i.$$The two points on the surface of the sphere are not at the same potential. One can see that the only equipotential is the $xy$-plane where $\theta = \frac{\pi}{2}$ and $V_i=V_f=0.$

 

[vcsharp2003]

Well, that plane can be considered a (rather big ) sphere ...

$\$

Why would that plane be a sphere? Or maybe you were just joking.

The zero potential points will lie only on the equatorial plane that bisects the dipole axis and is perpendicular to the dipole axis, since all points on such a plane will be equidistant from either +q or -q charge resulting in algebraic sum of the potentials due to +q and -q being zero.

 

[haruspex]

Why would that plane be a sphere?

A plane is the limit of a sphere radius r, centre at r from O, as r tends to infinity.
It is treated as a special case of a sphere in some branches of geometry, such as https://en.wikipedia.org/wiki/Inversive_geometry

 

[vcsharp2003]

A plane is the limit of a sphere radius r, centre at r from O, as r tends to infinity.
It is treated as a special case of a sphere in some branches of geometry, such as https://en.wikipedia.org/wiki/Inversive_geometry

I see. I never knew that. Thankyou.

 

[haruspex]

each dashed red line is definitely not a circle

Don't look like no spheres to me though

Quite…

$\frac 1{\sqrt{y^2+(x-a)^2}}-\frac 1{\sqrt{y^2+(x+a)^2}}=c$
$y^2+(x+a)^2+y^2+(x-a)^2-2\sqrt{(y^2+(x+a)^2)(y^2+(x-a)^2)}=c^2(y^2+(x+a)^2)(y^2+(x-a)^2)$
$2(y^2+x^2+a^2)-2\sqrt{(y^2+x^2+a^2+2xa)(y^2+x^2+a^2-2xa)}=c^2(y^2+x^2+a^2+2xa)(y^2+x^2+a^2-2xa)$
Writing $z^2=(x^2+y^2+a^2)$ and $w^4=z^4-4a^2x^2$:
$2z^2=2w^2+c^2w^4$

 

[vcsharp2003]

Quite…

$\frac 1{\sqrt{y^2+(x-a)^2}}-\frac 1{\sqrt{y^2+(x+a)^2}}=c$
$y^2+(x+a)^2+y^2+(x-a)^2-2\sqrt{(y^2+(x+a)^2)(y^2+(x-a)^2)}=c^2(y^2+(x+a)^2)(y^2+(x-a)^2)$
$2(y^2+x^2+a^2)-2\sqrt{(y^2+x^2+a^2+2xa)(y^2+x^2+a^2-2xa)}=c^2(y^2+x^2+a^2+2xa)(y^2+x^2+a^2-2xa)$
$2(y^2+x^2+a^2)-2\sqrt{(y^2+x^2+a^2+2xa)(y^2+x^2+a^2-2xa)}=c^2(y^2+x^2+a^2+2xa)(y^2+x^2+a^2-2xa)$
Writing $z^2=(x^2+y^2+a^2)$ and $w^4=z^4-4a^2x^2$:
$2z^2=2w^2+c^2w4$

If I wanted to know the equation of the red dashed line i.e. equation of red dashed curve in two dimensions like in terms of x and y, then would I plugin z = 0 in the final form of equation of the surface you have got? If I do that then the first term on the right side would be an imaginary term.

 

[haruspex]

If I wanted to know the equation of the red dashed line i.e. equation of red dashed in two dimensions like in terms of x and y, then would I plugin z = 0 in the final form of equation of the surface you have got? If I do that then the first term on the right side would be an imaginary term.

My z has nothing to do with x,y,z coordinates.

 

[vcsharp2003]

My z has nothing to do with x,y,z coordinates.

I see.

 

[vcsharp2003]

My z has nothing to do with x,y,z coordinates.

I did my own analysis, as shown below, to get the equation of an equipotential line that is shown in my original post ( i.e. not an equipotential surface). I end up with an equation that's impossible to interpret as being an equation of some standard curve like circle or ellipse.

 

[haruspex]

I did my own analysis, as shown below, to get the equation of an equipotential line that is shown in my original post ( i.e. not an equipotential surface). I end up with an equation that's impossible to interpret as being an equation of some standard curve like circle or ellipse.

… as the image @BvU linked illustrates.

 

[vcsharp2003]

… as the image @BvU linked illustrates.

I see. Then, one can conclude that the red dashed line is surely neither a circle nor an ellipse. It's just some non-standard closed curve.

 

[haruspex]

I see. Then, one can conclude that the red dashed line is surely neither a circle nor an ellipse. It's just some non-standard closed curve.

Right.

 

[kuruman]

I did my own analysis, as shown below, to get the equation of an equipotential line that is shown in my original post ( i.e. not an equipotential surface). I end up with an equation that's impossible to interpret as being an equation of some standard curve like circle or ellipse.

The point dipole potential in polar coordinates shown in post #12 is easier to handle. It is an approximation in the limit $r=\sqrt{x^2+y^2}>>a$, i.e. for distances much larger than the separation between charges. You can get this result using a Taylor expansion of your expression for small $a/r.$

You can make a polar plot of equipotentials, quite easily, if you have a plotting program. The relevant equation is obtained as follows $$V=\frac{k~p~\cos(\theta)}{r^2}=\text{const.}\implies r=\pm C\sqrt{\cos(\theta)}~.$$ Below are shown equipotential plots for $C=1, \frac{3}{2},2$. Positive values in blue, negative values in red. Note that there is no separation between equipotentials at the origin as expected for a pure point dipole.

 

[vcsharp2003]

The point dipole potential in polar coordinates shown in post #12 is easier to handle. It is an approximation in the limit r=x2+y2>>a, i.e. for distances much larger than the separation between charges.

I used to think that a point dipole is one in which the positions of +q and
-q charges coincide like in non-polar dielectrics. But, it seems, I was wrong.

 

[kuruman]

I used to think that a point dipole is one in which the positions of +q and
-q charges coincide like in non-polar dielectrics. But, it seems, I was wrong.

It's just a model for finding the electric potential of, for example, polarized atoms or molecules in which the positive charge density $\rho_+$ and the negative charge density $\rho_-$ come into play. The monopole term is $$Q=\int_V\left(\rho_+(r)+\rho_-(r)\right)dV$$ and is zero if the atom or molecule is neutral. The dipole term of the atom is the charge density weighted by the position of the charge, $$\vec{p}=\int_V\left[\rho_+ (r) +\rho_-(r)\right]~\vec{r}~dV =\int_V\rho_+(r)~\vec{r}~dV+ \int_V\rho_-(r)~\vec{r}~dV.$$ This term may or may not be zero depending on how the positive and negative charges are distributed within the atom or molecule. Notably, the water molecule has a dipole moment which can be considered a point dipole but the picture is not that of a positive and a negative point charge separated by some distance.