What is Electrostatic Potential?

What is Electrostatic Potential?

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Last updated on September 6th, 2019 at 04:47 pm

What is Electrostatic Potential?

Electrostatic potential energy is the amount of work required to move a charge against an electric field from a point of reference to a specific point of interest. However, an electrostatic potential is the amount of work needed to move a unit positive charge from a reference point to a specific point inside the field without producing an acceleration. In electrostatics, this point of reference is often conveniently chosen as infinity. Thus the electrostatic potential is also often defined as the amount of work done to move a unit positive charge from infinity to a point within the electric field. In all these descriptions, please note the differences between electrostatic potential energy and electrostatic potential. While the former one is defined for any charge, the latter is defined for unit positive charge only.

    \[ V(\vec{r})=-\int_{\infty}^{\vec{r}}\vec{E}(\vec{r'}).d\vec{r'} \]

Thus the electrostatic potential due to a point charge Q at a distance \vec{r} is given by

    \[ \boxed{V(\vec{r})=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r}}} \]

As already mentioned in our earlier article that an electrostatic force is conservative in nature. This means that the work done depends only on the initial and the final position of the particle and not on the path it follows to reach the final position. Consequently curl associated with any conservative force field becomes zero. Therefore we can always associate a scalar potential with an electric field as


    \[ \boxed{\vec{E}(\vec{r})=-\vec{\nabla}V(\vec{r})} \]

This scalar potential is known as electrostatic potential.

The introduction of a scalar potential is very useful since it allows us to apply the conservation of mechanical energy, which simplifies the solution of a large number of problems. Moreover, being a scalar quantity we need not take care of the direction which comes hand in hand with an electric field. This description of potential is also helpful to describe the potential energy associated with a charge distribution as the electrostatic potential is defined as the potential energy of a test particle divided by the charge of that particle. Now suppose we have the most general charge distribution due to point charges q, volume charge density \rho, surface charge density \sigma and line charge density \lambda. In that case, we can simply apply the superposition principle to get the total electrostatic potential as.

    \[ \boxed{V(\vec{r})=\frac{1}{4\pi\epsilon_0}\left[\sum_{i=1}^{N}\frac{q_i}{|\vec{r}-\vec{r}_i|}+\int_{V}\frac{\rho(\vec{r_0})dV_0}{|\vec{r}-\vec{r_0}|}+ \int_{S}\frac{\sigma(\vec{r_0})dS_0}{|\vec{r}-\vec{r_0}|}+\int_{L}\frac{\lambda(\vec{r_0})dl_0}{|\vec{r}-\vec{r_0}|}\right]} \]

The S.I. unit of electrostatic potential is volt (V). 1 V = 1 Joule/Coulomb = 1 Newton-meter/Coulomb.

Multipole Expansion of Electrostatic Potential

A multipole expansion is a series of functions that depend on angles. They are useful, as they can be truncated according to the approximation requirement of the original function. This useful methodology gives us the edge to deal with many difficult problems in electrostatics. Here we will show the multipole expansion for electrostatic potential due to a volume charge distribution.

Let us consider a volume charge distribution in volume V as shown in the diagram below. The volume charge density is \rho(\vec{r'}). Any coordinate with a superscript ‘ denotes the source coordinate. The test point is P(\vec{r}). We further assume that the test point is at a large distance from the source.

multipole expansion of electrostatic potential

The electrostatic potential at \vec{r} due to this charge distribution is

    \[ V(\vec{r})=\frac{1}{4\pi\epsilon_0}\left\int_{V}\frac{\rho(\vec{r'})}{|\vec{r}-\vec{r'}|}dV'\right \]

Here dV' is the volume element at position \vec{r'}. The main trick of multipole expansion is shifiting the origin from chrge element dV' to the point O and for thet we will express the distance |\vec{r}-\vec{r'}| in terms of \vec{r} and \vec{r}'.

    \[ \frac{1}{\left|\vec{r}-\vec{r'}\right|}=\left(r^2-2\vec{r}\cdot\vec{r'}+r'^2\right)^{-\frac{1}{2}} \]

    \[ =\frac{1}{r}\left[1-\frac{1}{2}\left(-\frac{2\vec{r}\cdot\vec{r'}}{r^2}+\frac{r'^2}{r^2}\right)+\frac{3}{8}\left(....\right)^2+ ... \right] \]

We can neglect terms involving higher orders of  \frac{r'}{r} and approximate the potential as

    \[ V(\vec{r})=\frac{1}{4\pi\epsilon_0}\left\int_{V}\left[ \frac{1}{r}+\frac{\vec{r}\cdot\vec{r'}}{r^3}+ H.O. \right]\rho(\vec{r'})dV' \right \]

    \[ V(\vec{r})=V_{mono}+V_{di}+H.O. \]

Here V_{mono} is the electrostatic potential if the total charge were concentrated at the origin. V_{di} is the potential for a dipole of dipole moment \int_V \vec{r'}\rho(\vec{r'})dV'. H.O. constitutes of potential due to quadrupole and other higher order poles.

Alternative Derivation

    \[ \left|\vec{r}-\vec{r'} \right|^2=r^2+r'^2-2rr'Cos\alpha \]

    \[ =r^2\left[1+\frac{r'^2}{r^2}-2\frac{r'}{r}Cos\alpha \right] \]

    \[ =r^2\left[ 1+\left(\frac{r'}{r} \right)\left(\frac{r'}{r}-2Cos\alpha \right)\right] \]

    \[ \left|\vec{r}-\vec{r'} \right|=r\sqrt{1+\left(\frac{r'}{r}\right)\left(\frac{r'}{r}-2Cos\alpha\right)} \]

    \[ =r\sqrt{1+\epsilon} \]

    \[ \frac{1}{|\vec{r}-\vec{r'}|}=\frac{1}{r}\left(1-\frac{\epsilon}{2}+\frac{3}{8}\epsilon^2-\frac{5}{16}\epsilon^3+..........\right) \]

Here we have used Binomial expansion formula,

    \[ (a+b)^n=\sum_{k=0}^{n}\left(^nC_k\right)a^{n-k}b^k \]

    \[ \frac{1}{|\vec{r}-\vec{r'}|}=\frac{1}{r}\left[1+\left(\frac{r'}{r}\right)Cos\alpha+\left(\frac{r'}{r}\right)^2\frac{(3Cos^2\alpha-1)}{2}+......\right] \]

    \[ \frac{1}{|\vec{r}-\vec{r'}|}=\frac{1}{r}\sum_{n=0}^{\infty}\left(\frac{r'}{r}\right)^nP_n(Cos\alpha) \]

P_n is Legendre polynomial.

    \[ V(\vec{r})=\frac{1}{4\pi\epsilon_0}\int_V \left[\frac{1}{r}\sum_{n=0}^{\infty}\left(\frac{r'}{r}\right)^nP_n(Cos\alpha)\right]\rho(\vec{r'})dV' \]

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