Coloumb's law, Electric lines of Force, Electric Flux, Gauss law, Application of Gauss law, Differential form of Gauss law

Coulomb’s Law

History:

In 1784 AD Charles-Augustin de Coulomb found (through experiment) the law which governs the force that one charge exerts upon another.

(Interestingly, Coulomb used a torsion balance similar in form to the one Cavendish used to verify Newton’s law of gravity thirteen years later. Cavendish’s balance, however, was much larger.)

Coulomb’s law applies to point charges, i.e., charges whose physical dimensions are small compared to the distance separating them. Ideally, a point charge would have all its charge confined to a mathematical point. (There actually seem to be such objects: the electron, positron and a quark, for example, appear to be point charges.) Also, spherically symmetric charges can be treated as point charges (with all their charge concentrated at their centers) so long as their separation is large enough and they do not touch one another.

Formula:

Consider two point charges q1 and q2 separated by a distance r. Coulomb’s law states that the force that any charge exerts on the other charge is directed along the line connecting them and has magnitude given as,

\(F=k\frac{\left |q_1 \right |\left |q_2 \right |}{r^2}=\frac{1}{4\pi \epsilon_0}\frac{\left |q_1 \right |\left |q_2 \right |}{r^2} \rightarrow 1\)

where k is a constant know as the Coulomb’s law constant. In vacuum or air, \(k=\frac{1}{4\pi \epsilon_0}=8.99\times 10^9 Nm^2/C^2\)

If q1 and q2 have the same sign the force is repulsive; if q1 and q2 have opposite signs the force is attractive. See the Figure 1. Since, force is a vector quantity, but relation for force in Eqn. (1) just represents the magnitude of force.

We can also express Coulomb’s formula in vectorial form as expressed below,

\(\vec{F}=\frac{1}{4\pi \epsilon_0}\frac{\left |q_1 \right |\left |q_2 \right |}{r^2}\hat{r}=\frac{1}{4\pi \epsilon_0}\frac{\left |q_1 \right |\left |q_2 \right |}{r^3}\vec{r}\rightarrow 2\)

Properties:

(1)Coulomb’s law applies only between pairs of charged particles; and Coulomb’s Law is only accurate if the charges involved are point like charges.

(2) the forces implied by Coulomb’s Law are always along the direct line joining between the two charges.

(3)If they are like charges, the force is repulsive, where as if they are unlike charges, the force is attractive.

(4)the force between any pair of charges is independent of the possible presence of any other charges nearby.

Electric lines of Force

Electric lines of force is an imaginary curve drawn in an electric field at which tangent at any point of curve gives the direction of electric field intensity at that point.Lines of force originate at a positive charge and terminate at a negative charge.Tangent at any point to a line of force shows the direction of the electric field at that point.Lines of force contract longitudinally in case of two unlike charges and exert a lateral pressure between two like charges.

Lines of force do not intersect and are continuous curves. They do not form any closed loop. Number density of lines of force at point gives the magnitude of electric field. Electric lines of force do not pass through a charged conductor, as the electric field inside a charged conductor is zero.

Electric field lines are produced for different charges

Point charge

Two like Charges

(1)Equal charges: same number density, N lies mid-way between charges

(2)Unequal charges: N shifts towards the smaller charge.

Two opposite charges

Uniform Electric Field

Electric Flux

Electric flux, \(\Phi\) through a surface is the measure of the number of electric field lines crossing the surface. Consider an imaginary surface which cuts across some E-field lines. We say that there is some electric flux through this surface. To make the notion of flux precise, we must first define a surface vector.

Definition:

surface vector A=\(\vec{A}=A\hat{n}\) , associated with a flat surface of area A. Magnitude of vector A = area A of surface. Direction of vector A = direction perpendicular (normal) to surface = direction of unit normal \(\hat{n}\) . The electric flux \(\Phi\)through a surface A is defined as \(\Phi=\vec{E}.\vec{A}=E.Acos\theta\)

Gauss Law

Gauss's law relates charges and electric fields in a subtle and powerful way. It states that “The electric flux through any closed surface S is a constant \(\frac{1}{\epsilon_0}\) times the total charge inside S”. Symbolically,

\(\Phi=\vec{E}.\vec{A}=\frac{q}{\epsilon_0} \rightarrow 1\)

Gauss' Law can be derived from Coulomb's Law if the charges are stationary, but Gauss' Law is more general than Coulomb's Law. Coulomb's Law is only true if the charges are stationary. Gauss' Law is always true, whether or not the charges are moving.

Derivation of Gauss’ Law using Coulombs law:

Assume a point charge +Q as shown in figure. Electric field at a distance r from it can be obtained from Coulomb's Law, the E-field of a point charge is

\begin{align*}E=\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2} \end{align*}

We get the same result by applying Gauss's Law: Here, electric flux linked with this spherical gaussian surface is given as,\(\vec{E}.\vec{A}=EA\), as E is radially directed outward from the centerandis perpendicular to thesurface area of the sphere.

i.e. \(\Phi=E\times 4\pi r^2\)

According to Gauss’ Law, electric flux is related with electric charge as,

\begin{align*}E\times 4\pi r^2=\frac{Q}{\epsilon_0} \end{align*}

Solving it for E, we get,\begin{align*}E=\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2} \end{align*}

Thus, we see that relation for electric field can be obtained from Gauss’ Law and hence Gauss’ law is in consistent with the Coulomb’s Law.

Application of Gauss law:

Electric field due to uniformly charged spherical shell:

Let us consider any sphere of radius ‘r’, on which Q charges are spread uniformly. In order to find electric field due to charged sphere, we can use Gauss law. For this purpose, we need to consider suitable gaussian surface (dottted line) at different distances from the center of sphere as r > R, r = R & r < R as shown in figure.

Case (i), r > R, The electric lines of force is directed radially outward, so electric flux linked with the gaussian surface (Figure a) is expressed as,

\(\Phi=E\times 4\pi r^2\)

Now, total charge enclosed by the dotted gaussian surface in Fig. a = Q .

So, using Gauss law, we can write,

\begin{align*}\Phi=\frac{Q}{\epsilon_0} \end{align*}\begin{align*}or,\space E\times 4\pi r^2= \frac{Q}{\epsilon_0}\end{align*}\begin{align*}E=\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2} \rightarrow 1\end{align*}

From Eq. (1), electric filed is same as that of all the charges are concentrated at the center of sphere.

Case (ii), r = R, The electric lines of force is directed radially outward.In this case gaussian surface as shown in Figure (b) just encloses the charged sphere. so electric flux linked with the gaussian surface is expressed as,

\(\Phi=E\times 4\pi R^2\)

Now, total charge enclosed by the dotted gaussian surface in Fig. b = Q .

So, using Gauss law, we can write,

\begin{align*}\Phi=\frac{Q}{\epsilon_0} \end{align*}\begin{align*}or,\space E\times 4\pi R^2= \frac{Q}{\epsilon_0}\end{align*}\begin{align*}E=\frac{1}{4\pi \epsilon_0}\frac{Q}{R^2} \rightarrow 2\end{align*}

From Eq. (2), electric filed is same as that of all the charges are concentrated at the center of sphere.

Case (iii), r <R,
In this case gaussian surface is inside the charged sphere as shown in Figure (c). So, the electric fluxlinked with the gaussian surface is expressed as,

\(\Phi=E\times 4\pi r^2\)

Now, total charge enclosed by the gaussian surface (dotted line in Fig. c) = 0 ( zero) . So, using Gauss law, we can write,

\begin{align*}\Phi=\frac{0}{\epsilon_0} \end{align*}\begin{align*}or,\space E\times 4\pi r^2=0 \end{align*}\begin{align*}E=0\rightarrow 3 \end{align*}

From Eq.(3), we can say that inside the charged sphere, ther is no electric field. The plot for the variation of electric field with respect to r is shown in Figure 3.

Electric field due to infinitely long charged sheet

Let us consider infinitely long sheet of charge having surface charge density,\(\sigma\).The electric lines of force is directed vertical direction. P be the any point at a distance ‘r’ from the surface, where we need to calculate electric field due to sheet of charge. In order to use Gauss’ law, we need to consider arbitrary shaped gaussian surface but for simplicity here, we have assumed cylindrical shaped gaussian surface, in which point ‘P’ lies in the center of top circular surface of radius ‘x’ and it is extend down to a same distance ‘r’ below the sheet of charge by piercing into it as shown in figure.

The total electric flux linked with the Gaussian surface is given by

\begin{align*}\Phi=EA_{top}Cos0^{\circ}+EA_{bottom}Cos0^{\circ}+EA_{side}Cos90^{\circ} =2(E\times \pi r^2)\end{align*}

Now, total charge enclosed by the gaussian surface,\begin{align*}Q=\sigma\times A_{top}=\sigma \pi r^2 \end{align*}

So, using Gauss law, we can write, \begin{align*}\Phi=\frac{Q}{\epsilon_0} \end{align*}\begin{align*}or,\space2(E\times \pi r^2)=\frac{\sigma \pi r^2 }{\epsilon_0 }\end{align*}\begin{align*}E=\frac{\sigma }{2\epsilon_0}\rightarrow 1\end{align*}From Eq.(1), we observe that electric filed due to infinitely long sheet of charge is uniform and does not depend on the distance from the sheet of charge.

Electric field due to long charged conducting plate

Let us consider long charged conducting plate having surface charge density,\(\sigma\). The electric lines of force is directed vertically upward. Let ‘P’ be the any point at a distance ‘r’ from the surface, where we need to calculate electric field due to charged conducting plate. In order to use Gauss’ law, we need to consider arbitrary shaped gaussian surface but for simplicity here, we have assumed cylindrical shaped gaussian surface, in which point ‘P’ lies in the center of top circular surface of radius ‘x’ and it is extend by same distance (r) into the surface of the charged conducting plate as shown in Figure 1.

The electric field inside the conductor is always zero. So, the electric flux linked with the gaussian surface inside the conductor is zero. The total electric flux linked with the Gaussian surface is given by

\begin{align*}\Phi=EA_{top}Cos0^{\circ}+EA_{curve}Cos90^{\circ}=E\times \pi x^2\end{align*}

Now, total charge enclosed by the gaussian surface,\begin{align*}Q=\sigma\times A_{top}=\sigma \pi x^2 \end{align*}

So, using Gauss law, we can write, \begin{align*}\Phi=\frac{Q}{\epsilon_0} \end{align*}\begin{align*}or,\space (E\times \pi x^2)=\frac{\sigma \pi x^2 }{\epsilon_0 }\end{align*}\begin{align*}E=\frac{\sigma }{\epsilon_0}\rightarrow 1\end{align*}

From Eq.(1), we observe that electric filed due to long charged conducting plate is uniform and does not depend on the distance from the charged plate. Variation of electric field from the charged conducting plate is presented in Figure 2.

Differential form of Gauss’ Law

Let \(\rho\) be the volume charge density and dV be the volume element as shown in Figure 1. Now the total charge enclosed by dV is given as

\begin{align*}q=\int_V\rho(r)dV \end{align*}

So, total electric flux \begin{align*}\int_s \vec{E}.d\vec{S}=\frac{\int_V\rho(r)dV }{\epsilon_0} \end{align*}

Gauss divergence theorem for \(\vec{F}\) states that\begin{align*}\oint_s\vec{F}.\hat{n}dS=\oint_s\vec{F}.d\vec{S}=\int_v\bigtriangledown.\vec{F}dV \end{align*}

Divergence theorem applied to electric field \(\vec{E}\) which becomes as,\begin{align*} \oint_s\vec{E}.d\vec{S}=\int_v\bigtriangledown.\vec{E}dV\end{align*}\begin{align*}\implies \int_v\bigtriangledown.\vec{E}dV= \frac{\int_V\rho(r)dV }{\epsilon_0} \end{align*}\begin{align*} \bigtriangledown.\vec{E}=\frac{\rho(r) }{\epsilon_0}\rightarrow 1\end{align*}

Equation (1) represents the differential form of Gauss’ theorem which is also called Poisson’s Equation.

Poisson's Equation and Laplace Equation

We have Gauss law in differential form as

\begin{align*} \bigtriangledown.\vec{E}=\frac{\rho }{\epsilon_0}\rightarrow 1\end{align*}

\begin{align*}Since\space \vec{E}=-\bigtriangledown V \end{align*}

Now taking divergence on both sides, we get

\begin{align*}\bigtriangledown.\vec{E}=-\bigtriangledown^2 V\rightarrow 2 \end{align*}

From 1 and 2, we get

\begin{align*}\bigtriangledown^2 V=-\frac{\rho }{\epsilon_0} \end{align*}

which is called Poisson's equation. Where V is scalar potential and \(\rho \) be volume charge density of material.

In free space,\(\rho =0\) then equation 3 reduces to

\(\bigtriangledown^2 V=0\) which is called Laplace equation.

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