# Electric dipole moment when charges are not symmetrically opposite each other [closed]

In the above figure, I want to calculate the net dipole moment.

I know that dipole moment is a collection of equal and opposite charges. But, according to that concept, the above figure cannot be a dipole.

Could someone help me with the concept of dipoles in detail.

• is this spherical shell or circular arc
– user182868
Aug 16, 2018 at 5:23
• 4 Arcs, forming a ring Aug 16, 2018 at 5:25
• There's currently no specific question here. Could you please edit your question to indicate what aspect of dipoles you don't understand and need help with? You say "the above figure cannot be a dipole", but there's no reason given that anyone would expect it to be a dipole. Please be as specific as possible so answers can be as helpful as possible. Aug 20, 2018 at 8:20
• I have specified in the first line itself that I want to calculate the net dipole moment for the given figure. Aug 21, 2018 at 0:11
• @David Z : the question had been post on Aug 16. On this same day two answers were given. On Aug 17 I gave my own answer. On Aug 28 0r 29 (10 days later) you put it on hold as off-topic. Question : When the question would be closed as off-topic the moderation team will call me not to answer off-topic questions and I must delete my answer ??? Aug 30, 2018 at 23:44

$\boldsymbol{\S\:}$ 1. Electric dipole moment of an arc with uniform linear charge density $\;\lambda$

The electric dipole moment of a charge distribution in a volume $\;V\;$ with volume charge density $\;\rho\left(\mathbf{r}\right)\;$ is defined with respect to the origin as

$$\mathbf{p}=\iiint\limits_{V}\rho\left(\mathbf{r}\right)\mathbf{r}\mathrm d V \tag{01}\label{01}$$ (don't confuse the term dipole with the existence necessarily of a dipole $\;q\boldsymbol{d}$, that is of two point charges $\;q,-q\;$ separated by a vector $\;\boldsymbol{d}\;$ from $\;-q\;$ to $\;q\;$).

Now, suppose we want to find the electric dipole moment of an arc $\;\mathbf{AB}\;$ of radius $\;R\;$ and angle $\;\theta\;$ with uniform linear charge density $\;\lambda$, see Figure-01 below

In equation \eqref{01} replacing $\;\rho\left(\mathbf{r}\right)\mathrm d V\;$ by $\;\lambda \mathrm ds=\lambda R\mathrm d\omega \;$ and $\;\mathbf{r}=R\left(\cos\omega,\sin\omega\right)\;$ we have for the infinitesimal dipole moment $\;\mathrm d\mathbf{p}\;$ of the infinitesimal arc $\;\mathrm ds$ $$\mathrm d\mathbf{p}=\lambda R^{2}\left(\cos\omega,\sin\omega\right)\mathrm d\omega \tag{02}\label{02}$$ and so for the moment of the arc $\mathbf{AB}$ $$\mathbf{p}=\int\limits_{\mathbf{A}}^{\mathbf{B}}\mathrm d\mathbf{p}=\int\limits_{0}^{\theta}\lambda R^{2}\left(\cos\omega,\sin\omega\right)\mathrm d\omega=\lambda R^{2}\left(\sin\theta,1-\cos\theta\right) \tag{03}\label{03}$$ that is $$\mathbf{p}=2\lambda R^{2}\sin\left(\frac{\theta}{2}\right)\left[\cos\left(\frac{\theta}{2}\right),\sin\left(\frac{\theta}{2}\right)\right] \tag{04}\label{04}$$ a vector along the bisector of the angle $\;\theta\;$ as expected by symmetry, see Figure-02 below

$\boldsymbol{\S\:}$ 2. Electric dipole moment of the question

According to the aforementioned in $\boldsymbol{\S\, 1\:}$ the electric dipole moments of the quadrants are \begin{align} \mathbf{p}_1 & = \dfrac{\boldsymbol{+}2q}{\pi R/2}R^2 \begin{bmatrix} \hphantom{\boldsymbol{-}}1\vphantom{\dfrac{a}{b}}\\ \hphantom{\boldsymbol{-}}1\vphantom{\dfrac{a}{b}} \end{bmatrix} =\dfrac{4qR}{\pi}\left(\mathbf{i}\boldsymbol{+}\mathbf{j}\right) \tag{05.1}\label{eq05.1}\\ \mathbf{p}_2 & = \dfrac{\boldsymbol{+}3q}{\pi R/2}R^2 \begin{bmatrix} \boldsymbol{-}1\vphantom{\dfrac{a}{b}}\\ \hphantom{\boldsymbol{-}} 1\vphantom{\dfrac{a}{b}} \end{bmatrix} =\dfrac{6qR}{\pi}\left(\boldsymbol{-}\mathbf{i}\boldsymbol{+}\mathbf{j}\right) \tag{05.2}\label{eq05.2}\\ \mathbf{p}_3 & = \dfrac{\boldsymbol{-}6q}{\pi R/2}R^2 \begin{bmatrix} \boldsymbol{-}1\vphantom{\dfrac{a}{b}}\\ \boldsymbol{-} 1\vphantom{\dfrac{a}{b}} \end{bmatrix} =\dfrac{12qR}{\pi}\left(\mathbf{i}\boldsymbol{+}\mathbf{j}\right) \tag{05.3}\label{eq05.3}\\ \mathbf{p}_4 & = \dfrac{\boldsymbol{+}q}{\pi R/2}R^2 \begin{bmatrix} \hphantom{\boldsymbol{-}} 1\vphantom{\dfrac{a}{b}}\\ \boldsymbol{-} 1\vphantom{\dfrac{a}{b}} \end{bmatrix} =\dfrac{2qR}{\pi}\left(\mathbf{i}\boldsymbol{-}\mathbf{j}\right) \tag{05.4}\label{eq05.4} \end{align} with sum $$\mathbf{p}=\mathbf{p}_1\boldsymbol{+}\mathbf{p}_2\boldsymbol{+}\mathbf{p}_3\boldsymbol{+}\mathbf{p}_4=\dfrac{4qR}{\pi} \begin{bmatrix} \:\:3\:\:\vphantom{\dfrac{a}{b}}\\ 5\vphantom{\dfrac{a}{b}} \end{bmatrix} =\dfrac{4qR}{\pi}\left(3\mathbf{i}\boldsymbol{+}5\mathbf{j}\right)=\dfrac{4\sqrt{34}qR}{\pi}\left(\dfrac{3\mathbf{i}\boldsymbol{+}5\mathbf{j}}{\sqrt{34}}\right) \tag{06}\label{eq06}$$ So for the magnitude of the total electric dipole moment we have $$\Vert \mathbf{p} \Vert=\dfrac{4\sqrt{34}qR}{\pi} \tag{07}\label{eq07}$$ In Figure-03 below we see all electric dipole moments under scale

A choice of an equivalent electric dipole with the same moment is shown in Figure-04 below. We suppose that $\;q>0$.

I added substantial edits to the end after the OP's comment.

A simple point dipole is composed of one positive charge $q$ and one negative charge $-q$ separated by a distance, $d$, so that the dipole moment is

$$\vec{p} = qd .$$

More generally, the (static) electric dipole moment is the first term of the electrostatic multipole expansion. In your example, since we have a discrete distribution of charges, we only need to calculate the components of the dipole moment via the sum

$$\vec{p}(\vec{r}) = \sum_{i = 1}^{N} q_{i} (r_{i} - r)$$

where N is the number of point charges, $i \in \{1,...N\}$, $q_{i}$ is the electric charge of the $i^{th}$ point charge, $r$ is the distance from the origin to the observation/test point, and $r_{i}$ is the distance from the origin to the $i^{th}$ point charge. When doing this, take extra care to to have a term for each charge in the sum with its corresponding distance. Also notice that the dipole moment is origin dependent, so choose your origin wisely to take advantage of symmetries.

Essentially, any configuration of positive and negative charges will have a dipole contribution to the multipole moment, but the dipole moment is typically the dominant term and its also the easiest multipole term to calculate which is why it's a decent exercise. If you're interested, try calculating the next term in the multipole expansion - the quadrupole moment - for your example system!

EDIT: Apologies, I did not interpret the image correctly at first. Since the charge is continuous, you must find an expression for the charge distribution and then integrate over it. Perhaps you can try writing down a charge distribution using diract delta function and the heaviside function, and then integrate over it to obtain the dipole moment. Using polar coordinates $(r, \phi)$, with the origin at the ring's center, phi is measured from the horizontal, and supposing the ring has radius $a$, perhaps try something like,

$$\rho(r,\phi) = \delta(r-a) [2q\Theta(\frac{\pi}{2} - \phi) + 3q(\Theta(\pi - \phi) - \Theta(\frac{\pi}{2} - \phi)) - 6q(\Theta(\frac{3\pi}{2} - \phi) - \Theta(\pi - \phi)) + q(\Theta(2\pi - \phi) - \Theta(\frac{3\pi}{2} - \phi))]$$

where $\Theta(\psi - \phi)$ is the Heaviside step function (1 for $\psi > \phi$, and zero otherwise), and $\delta$ is the Dirac delta function.

Then integrate over the surface of the circle:

$$\vec{p}(\vec{r}) = \int \rho(r',\phi') \delta(r - r') dr'd\phi'$$

EDIT: for help in how to integrate the step function. For help in how to integrate the dirac delta function see equation 5 of this and this might be helpful

$\vec p=\vec p_{1}+\vec p_{2}+\vec p_{3}+\vec p_{4}$

$\vec p_{1}=$ electric dipole moment(EDM) vector at origin of ring due to quarter arc having linear charge density $\left(\lambda_{1}=\dfrac{2q}{\pi r/2}\right)$

$\vec p_{2}=$E.D.M vector due to $\left(\lambda_{2}=\dfrac{3q}{\pi r/2}\right)$

$\vec p_{3}=$E.D.M vector due to $\left(\lambda_{3}=\dfrac{-6q}{\pi r/2}\right)$

$\vec p_{4}=$E.D.M vector due to $\left(\lambda_{4}=\dfrac{q}{\pi r/2}\right)$

we know , for continuous linear charge distribution $\vec p=\displaystyle\int \vec r dq$

where $\vec r$ is position vector of elemental charge $dq$

$\vec p=\lambda_{1}R^2\displaystyle\int_{0}^{\dfrac{\pi}{2}}\left[cos\theta \hat x+sin\theta\hat y\right]d\theta+\lambda_{2}R^2\displaystyle\int_{\dfrac{\pi}{2}}^{\pi}\left[cos(\pi-\theta )(\hat{-x})+sin(\pi-\theta)\hat y\right]d\theta$+$\lambda_{3}R^2\displaystyle\int_{\pi}^{\dfrac{3\pi}{2}}\left[cos(\pi+\theta) ({-\hat x})+sin(\pi+\theta)({-\hat y})\right]d\theta+ \lambda_{4}R^2\displaystyle\int_{0}^{\dfrac{-\pi}{2}}\left[cos(-\theta) \hat x+sin(-\theta)({-\hat y})\right]d\theta$

$\vec p=R^2\left[\lambda_{1}(\hat x+ \hat y)+\lambda_{2}(\hat{ -x}+ \hat y)+\lambda_{3}(\hat {-x}+ \hat {-y})+\lambda_{4}(\hat {x}+ \hat {-y})\right]$

if $\lambda_{1}=\lambda(say)\implies \lambda_{2}=1.5\lambda\ \ ;\lambda_{3}=-3\lambda \ \ ;\lambda_{4}=0.5\lambda;$

$\vec p=R^2\lambda[(3\hat x+5\hat y)]=\dfrac{4qR}{\pi}(3\hat x+5\hat y)$

• Ty for clarifying. Why should it be zero? The charge is not symmetric about the origin? Aug 17, 2018 at 14:42