Electric Field, Electrostatic Potential, Work done in moving a charge, Electrostatic Potential Energy

Electric Field:

The region or space sorrounding any charge or group of charges where electric force can be experienced is is called electric field.

Electric field intensity:

Electric field intensity at a point can be defined as the force experienced by unit positive charge in the electric field.

Let us consider a point charge q and \(\vec{F}\) be the force acting on the test charge in the electric ield then electric field intensity is defined as

\begin{align*}\vec{E}=\lim_{q\rightarrow0}\frac{\vec{F}}{q} \end{align*}\begin{align*} \vec{F}=q\vec{E}\rightarrow 1\end{align*}

Unit of \(\vec{E}\) is Newton/Coulomb or Volt/metre.

From equation 1,\begin{align*}\vec{E}=\frac{\vec{F}}{q}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}\rightarrow 2 \end{align*}

If point charge is not at origin i.e. \(\vec{r^{'}}\) from the origin then electric field intensity \(\vec{E}\) at a point P at a distance \(\vec{r}\) from the origin is given by \begin{align*}\vec{E}=\frac{q}{4\pi\epsilon_0}\frac{\vec{r}-\vec{r^{'}}}{|\vec{r}-\vec{r^{'}}|^3} \end{align*}If we consider charge distribution consisting N point charges \(q_1, q_2,...q_N\) located at points \(r_1,r_2,...,r_N\) respectively then

Electric field at a distance \(\vec{r}\) is \begin{align*}\vec{E}(r)=\frac{1}{4\pi\epsilon_0}\sum_{i=1}^{N}q_i\frac{\vec{r}-\vec{r^{'}}}{|\vec{r}-\vec{r^{'}}|^3}= \frac{1}{4\pi\epsilon_0}\int_{v}\frac{\vec{r}-\vec{r^{'}}}{|\vec{r}-\vec{r^{'}}|^3}\rho(r^{'})dv^{'}= \frac{1}{4\pi\epsilon_0}\int_{s}\frac{\vec{r}-\vec{r^{'}}}{|\vec{r}-\vec{r^{'}}|^3}\sigma(r^{'})da^{'}\end{align*}

Electrostatic Potential:

The potential at a point is defined as the amount of work done in bringing a unit positive test charge from infinity to that point without acceleration in direction opposite to the electric field.

\begin{align*}w=\int_{-\infty}^{r}(-\vec{F})dr=-\int_{-\infty}^{r}\frac{Q}{4\pi\epsilon_0 r^2}dr=\frac{Q}{4\pi\epsilon_0 r} \end{align*}

Then the electric potential\((V)=\frac{w}{q}=\frac{Q}{4\pi\epsilon_0 r}(if\space q=1\space coulomb)\)

Potential difference:

The potential difference between two points in electric field may be defined as the amount of work done in moving a unit positive test charge from one point to another point without acceleration in direction opposite to the electric force.

Letus consider a charge +Q at a point O and a unit test charge should bring from point B to point A i.e. from low potential to high potential(i.e. r1 to r2) then the work done gives p.d. between A and B.

\begin{align*}V_A-V_B=W_{AB}=-\int_{r_1}^{r_2}\frac{Q}{4\pi\epsilon_0 r^2}dr \end{align*}\begin{align*}=\frac{Q}{4\pi\epsilon_0} \left [ \frac{1}{r_2}-\frac{1}{r_1} \right ]\end{align*}

Unit of Potential

SI unit, 1 volt=1joule/1coulomb

CGS unit, 1 statvolt=1erg/1 stat coulomb

1 volt=300 stat volt

Potential due to a group of point charges:

Let us consider there are N charges \(q_1,q_2,...q_N\) are at a distance of \(r_1,r_2...r_N\) from the test charge \(q_i\) at a point A.

Let V1 be the potential at point A due to charge q1,V2 be the potential at point A due to charge q2,V3 be the potential at point A due to charge q3 and so on. Then the total potential at point A due to all charges is given by the sum of potentials given by each charges.

\begin{align*}V=V_1+V_2+...+V_N \end{align*}\begin{align*}=\frac{q_1}{4\pi\epsilon_0 r_1}+\frac{q_2}{4\pi\epsilon_0 r_2}+\frac{q_3}{4\pi\epsilon_0 r_3}+...+\frac{q_N}{4\pi\epsilon_0 r_N} \end{align*}\begin{align*}\therefore V=\frac{1}{4\pi\epsilon_0}\sum_{i=1}^{N}\frac{q_{i}}{r_{i}}\end{align*}

Relation between Potential Gradient and Electric Field Intensity

\begin{align*}V=-\int_{\infty}^{P}\vec{E}.d\vec{r} \end{align*}\begin{align*}or,dV=-\vec{E}.d\vec{r}\rightarrow 1 \end{align*}

If co-ordinates of point P are in terms of x, y & z as shown in Fig., then electric potential, V is also the function of x, y &z and rate of change of V along these axes (x, y & z) is represented by its partial derivatives\(\frac{\partial V}{\partial x},\frac{\partial V}{\partial y}\space and \space \frac{\partial V}{\partial z}\)respectively. That is,\begin{align*}dV=\frac{\partial V}{\partial x}dx+\frac{\partial V}{\partial y}dy+\frac{\partial V}{\partial z}dz \end{align*}\begin{align*}=(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}).(dx\hat{i}+dy\hat{j}+dz\hat{k}) \end{align*}\begin{align*}But,\space d\vec{r}=dx\hat{i}+dy\hat{j}+dz\hat{k}\end{align*}where dx, dy and dz are components of displacement vector \(d\vec{r}\) along x, y and z axes. \begin{align*}Also,\space \frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}=(\frac{\partial }{\partial x}\hat{i}+\frac{\partial }{\partial y}\hat{j}+\frac{\partial }{\partial z}\hat{k})V =\bigtriangledown V\end{align*}\begin{align*}That\space is, dV=\bigtriangledown V.d\vec{r}\rightarrow 2 \end{align*}From 1 and 2, we get\begin{align*}E=-\bigtriangledown V=-grad\space V\rightarrow 3 \end{align*}

Eq. (3) represents the electric field intensity in terms of potential gradient.

Electric Potentials due to Different types of Charge Distribution

Electric potentials due to more than one point charges are obtained by scalar addition because electric potential by a point charge is the linear function of the value of charge.
So, this addition of potentials of individual charge is known as the “Principle of superposition of electric potentials”.
This principle states that the potential at a point is the sum of the potentials at that point by an individual charge. That is,

Total potential = potential due to group of point charge + potential due to surface charge + potential due to linear charge distribution + potential due to volume charge distribution.

\begin{align*}V=V_1+V_2+V_3 +V_4\rightarrow 1 \end{align*}

(i) Electric Potential due to group of point charges:

Assume there are ‘n’ number of charges \(q_1, q_2, q_3…..q_{n}\)situated at distances \(r_1, r_2, r_3 …. r_{n}\) respectively from the test charge qiat point P as shown in Fig. -1.

The net electric potential at point P is given by the algebraic sum of electric potentials by individual charge. That is,

\begin{align*}v_1=\frac{1}{4\pi \epsilon_0}\sum \frac{q_{i}}{r_{i}}\rightarrow 2 \end{align*}

In case, if charge \(q_i\)is at \(\vec{r^{'}}\)and point of consideration P is at \(\vec{r}\) from the origin then Eq. (2) can be written as \begin{align*}V_1=\frac{1}{4\pi \epsilon_0}\sum \frac{q_{i}}{|\vec{r}-\vec{r^{'}}|}\rightarrow 3 \end{align*}

(ii) Electric Potential due to surface charge distribution:

If \(\sigma\) be the surface charge distribution over the surface and \(d\vec{s}\)be the surface element as shown in Fig.

The electric potential at point P is given as,\begin{align*}V_2=\frac{1}{4\pi \epsilon_0}\int_{s}\frac{\sigma d\vec{s}}{|\vec{r}-\vec{r^{'}}|} \rightarrow 4\end{align*}

(iii) Electric Potential due to linear charge distribution:

In case, if charge is not concentrated at a point but is distributed along a line with \(\lambda\) as linear charge density then electric potential due to this linear charge distribution is given by

\begin{align*}V_3=\frac{1}{4\pi \epsilon_0}\int_{l}\frac{\lambda d\vec{l}}{|\vec{r}-\vec{r^{'}}|}\rightarrow 5 \end{align*}

(iv) Electric Potential due to volume charge distribution:

If q charge is uniformly distributed over the volume V as shown in Fig, then electric potential at point P is given as

\begin{align*}V_4=\frac{1}{4\pi \epsilon_0}\int_{v}\frac{\rho d\vec{V}}{|\vec{r}-\vec{r^{'}}|}\rightarrow 6 \end{align*}

where \(\rho \) be the volume charge density and \(d\vec{V}\) be the volume element.

By substituting the values of V1, V2, V3 and V4 in Eq. (1), we get,

\begin{align*}V=\frac{1}{4\pi \epsilon_0}\sum \frac{q_{i}}{|\vec{r}-\vec{r^{'}}|}+\frac{1}{4\pi \epsilon_0}\int_{s}\frac{\sigma d\vec{s}}{|\vec{r}-\vec{r^{'}}|}+\frac{1}{4\pi \epsilon_0}\int_{l}\frac{\lambda d\vec{l}}{|\vec{r}-\vec{r^{'}}|}+\frac{1}{4\pi \epsilon_0}\int_{v}\frac{\rho d\vec{V}}{|\vec{r}-\vec{r^{'}}|}\rightarrow 7 \end{align*}

Eq. (7) represents total electric potential at a point due to different types of charges.

Work done in moving a Charge:

Let us consider a point test charge q moving from one point A to another point B in the presence of electric field \(\vec{E}(x)\0. Then the force acting on the charge at any point is

\begin{align*}\vec{F}=q\vec{E}\rightarrow 1 \end{align*}

So small work done due to force exists on the charge is

\begin{align*}dw=-\vec{F}.d\vec{l}\rightarrow 2 \end{align*}

the -ve sign appears because we are calculating the work done on the charge against the action of field.

Now the total work done on moving charge from point A to B is

\begin{align*}w=-\int_{A}^{B}\vec{F}.d\vec{l}=-q\int_{A}^{B}\vec{E}.d\vec{l}\rightarrow 3 \end{align*}

But we can define electric field as negative gradient of scalar potential due to the field irrotational i.e.

\begin{align*}E=-\bigtriangledown \phi \rightarrow 4 \end{align*}

From equation 3 and 4, we get

\begin{align*}w=q\int_{A}^{B}\bigtriangledown \phi .d\vec{l}=q\int_{A}^{B} d\phi \end{align*}\begin{align*}\therefore w=q(\phi_{B}-\phi_{A}) \rightarrow 5\end{align*}

which shows that q\(\phi\) can be interpreted as the potential energy of the test charge in the electrostatic field. From eqn 3 and 5, it can be obsetrved that the line integral of the electric field between twopointsis independent of the path and is negative of the potential difference between the points. So if the path is closed, the line integral is zero.

\begin{align*}\int_{A}^{B}\vec{E}.d\vec{l}=-(\phi_{B}-\phi_{A}) =0\end{align*}

Electrostatic potential energy:

As we know the product of the scalar potential and a charge of point object can be interpreted as potential energy i.e. if a point charge \(q_{i}\) is brought from infinity to a point \(\vec{r_{i}}\) in a region of localized electric field described by the scalr potential \(\phi\vec{(r_{i})}\) which vanishes at infinity, the work done on the change (i.e. its potential energy) is given by

\begin{align*}w_{i}=q_{i}\phi\vec{(r_{i})}\rightarrow 1 \end{align*}

where \(\phi(r_{i})\) is potential at \(r_{i}\) from the origin due to (n-1) charges \(q_{j}(j=1,2...n-1)\) located at \(r_{}\) distance from origin then \begin{align*}\phi\vec{(r_{i})}=\frac{1}{4\pi \epsilon_0}\sum_{j=1}^{n-1}\frac{q_{j}}{|\vec{(r_{i})}-\vec{(r_{j})}|}\rightarrow 2 \end{align*}

where \(r_{j}\) be the distance of the charge \(q_{j}\) as a source coordinate and \(r_{i}\) be the distance of the charge \(q_{i}\) as a field coordinate.

If the charge \(q_{j}\) is at origin then \(\vec{r_{j}}=0\)

\begin{align*}Then\space \phi\vec{(r_{i})}=\frac{1}{4\pi \epsilon_0}\sum_{j=1}^{n-1}\frac{q_{j}}{|\vec{r_{i}}|}\end{align*}

From equaion 1,\begin{align*}W_i=q_i\frac{1}{4\pi \epsilon_0}\sum_{j=1}^{n-1}\frac{q_{j}}{|\vec{(r_{i})}-\vec{(r_{j})}|} \end{align*}

Now the total potential energy of all charges due to all the forces actiong between them is \begin{align*}W=\frac{1}{4\pi \epsilon_0}\sum_{i=1}^{n}\sum_{j<i}\frac{q_{j}}{|\vec{(r_{i})}-\vec{(r_{j})}|} \end{align*}

Now a more symmetric form can be written by summing over i and j unrestricted, and then by dividing by 2\begin{align*} \frac{1}{8\pi \epsilon_0}\sum_{i=1}^{n}\sum_{j}\frac{q_{j}}{|\vec{(r_{i})}-\vec{(r_{j})}|};i\neq j\rightarrow 4\end{align*}

Now in case of continuous charge distribution i.e. volume charge density can be defined as \begin{align*}\rho (\vec{r})=\frac{dq}{dv}\end{align*}\begin{align*} dq=\rho (\vec{r}) dv\end{align*}\begin{align*}\therefore q=\int \rho(r')dv' \end{align*}

So potential energy takes the form of \begin{align*}W= \frac{1}{8\pi \epsilon_0}\int \int \frac{\rho (r) \rho (r')}{|\vec{r}-\vec{r'} |} d^3rd^3r' \rightarrow 5\end{align*}

which is the required expression for electrostatic potential energy due to a group of point charges and also in continuous system under the consideration of source coordinate (\(r_j=r'\)) system.

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Griffiths, D.J., Introduction to Electrodynamics, PHI Learning Private Limited, 2013