Gauss's Law and it's Application

Gauss's Law

Statement

The total dielectric flux passing through a closed surface in vacuum enclosing a charge is \(\frac{1}{\epsilon _o}\) times the charge enclosed by the closed surface.

Mathematically,

$$\phi = \frac{1}{\epsilon _o} \times \text{charge enclosed (q)} $$

$$\phi = \frac{q}{\epsilon _o}$$

where,\(\epsilon _o\) is permittivity of free space.

In medium,$$\phi = \frac{q}{\epsilon}$$

To verify Gauss theorem suppose a point charge placed at O in vacuum show that the electric field intensity at point 'P' that lies at distance'r' from the charge 'q' is

$$E = \frac{1}{4\pi \epsilon _o}. \frac{q}{r^2}$$

When a spherical surface passing through point'P' is constructed that has the centre 'O' and radius 'r' at every pointon its surface electric field intensity has the same value. If \('\phi '\) represents the electric flux passing through the sphere and 'A' its area then

$$\phi = EA$$

$$= \frac{1}{4\pi \epsilon _o}. \frac{q}{r^2} 4\pi r^2$$

$$=\frac{q}{\epsilon _o}$$

$$\therefore \phi=\frac{q}{\epsilon _o}$$

Application of Gauss's Theorem

Gauss's theorem is useful for determining the electric field intensity produced by a charged conductor as follows:

Electric Field Intensity due to a Charged Sphere

Consider a sphericalConductor having radius 'R' and center 'O'. It is carrying charge 'q' on it's surface. Due to this distribution of charged electric field intensity around it can be defined. Following three distinct points can be considered.

1. At a point outside the surface of sphere(r>R)

Consider 'P' is the concerned point where electric field intensity due to the distribution of charge on the spherical conductor is to be calculated. For this, we construct a Gauss' sphere concentric with the charged sphere and passing through point 'P'. So, that due to symmetry at every point on the surface of the sphere electric field intensity has the same magnitude.

It \(\phi\) represents the electric flux passing through Gauss' surface and 'A' be area of Gauss' surface and 'A' be area of Gauss's Sphere. Then

$$\phi = EA$$

$$\phi= E4\pi r^2 \dots(i)$$

Also from Gauss' theorem

$$\phi =\frac{ \text{charge enclosed by Gaussian Surface}}{\epsilon _o}$$

$$\phi =\frac{q}{\epsilon _o} \dots (ii)$$

Equating (i) and (ii), we get

$$E. 4\pi r^2 = \frac{q}{\epsilon _o}$$

$$\therefore E = \frac{1}{4\pi r^2}.\frac{q}{\epsilon _o}$$

This is the electric field intensity produced distanced 'r' from the center of a spherical conductor carrying charge 'q'.

2. At a point on the surface of sphere(r=R)

When the concerned point lies on the surface the gauss' surface also becomes the sphere having same radius as that of the charged surface.

Then,

$$\phi = EA$$

$$\phi= E4\pi r^2 \dots(i)$$

Also from Gauss' theorem

$$\phi =\frac{ \text{charge enclosed by Gaussian Surface}}{\epsilon _o}$$

$$\phi =\frac{q}{\epsilon _o} \dots (ii)$$

Equating (i) and (ii), we get

$$E. 4\pi R^2 = \frac{q}{\epsilon _o}$$

$$\therefore E = \frac{1}{4\pi \epsilon}.{\frac{q}{R^2}}$$

Hence, electric field intensity on the surface of charged sphere has constant magnitude.

$$E = \frac{1}{4\pi r^2}. \frac{q}{\epsilon _o}$$

3. At a point inside the surface of sphere(r<R)

Now, the Gaussian surface lies inside the charged sphere. As the charge resides only on the surface of the sphere, electric charge enclosed by Gaussian surface is O. If 'E ' represents electric field Intensity at point 'P' then

$$\phi = EA$$

$$\phi= E4\pi r^2 \dots(i)$$

Also from Gauss' theorem

$$\phi =\frac{ \text{charge enclosed by Gaussian Surface}}{\epsilon _o}$$

$$\phi = 0\dots (ii)$$

Equating (i) and (ii) we get

$$E. 4\pi R^2 = 0 $$

$$E = 0$$

Hence, electric field intensity on the surface of charged sphere is zero.

4. Electric Field Intensity due to an Infinite Charged Plane Sheet.

Consider an infinite plane sheet is given positive charge density is \(\sigma \). P is a point at distance 'r' from the sheet where electric field intensity is to be calculated. For this gaussian cylinder is drawn perpendicular to the sheet whose cross-section passes through point P. The cylinder extends to the other sides also.

Since, the lines of force leave the charged surface perpendicularly. Here lines of force pass perpendicularly through the cross-sectional area and do not pass through curved surface area. Therefore, total electric flux through Gaussian surface.

$$\phi = EA + EA \dots (i) $$

$$\text {Charge enclosed by Gaussian surface(q)} = \sigma A \dots (i) $$

Also from Gauss' theorem

$$\phi =\frac{ \text{charge enclosed by Gaussian Surface}}{\epsilon _o}$$

$$\phi =\frac{\sigma A}{\epsilon _o} \dots (iii)$$

Equating (i) and (iii), we get

$$2EA=\frac{\sigma A}{\epsilon _o}$$

$$\therefore E=\frac{\sigma}{2\epsilon _o}$$

It is independent of the distance of the concerned point from the plane.

5. Electric Field Intensity near a Charged Plane.

Consider a charged plane with surface charge density \(\sigma \).We have to calculate an electric field intensity at point P that lies at the distance 'r' from the plane in order to use Gauss' theorem. We are going to construct a Gaussian area of cross-sectional area 'A' of the Gaussian cylinder but not across the curved surface. Therefore, total electric flux through Gaussian surface \((\phi) = E.A\)

$$\phi = E..A\dots (i)$$

Charge enclosed by Gaussian Surface\(q = \sigma A\dots (ii)\)

From Gauss Theorem,

$$\phi =\frac{ \text{charge enclosed by Gaussian Surface}}{\epsilon _o}$$'

$$\phi =\frac{\sigma A}{\epsilon _o} \dots (iii)$$

Equating (i) and (iii), we get

$$EA=\frac{\sigma A}{\epsilon _o}$$

$$\therefore E=\frac{\sigma}{\epsilon _o}$$

This indicates an electric field intensity near a charged plane is uniform which is represented by parallel lines of force.

6. Electric Field Intensity due to Linear Charge Distribution

Suppose a linear distribution of charge with linear charge density \((\lambda)\) as shown in the figure. P is a point at the distance 'r' from the charge distribution where electric field intensity 'E' is to be calculated.

To make the use of Gauss Theorem here we draw the cylindrical Gaussian surface of length 'l' and the radius 'r' whose c.urvedsurface passes through point 'P'. The lines of force pass through the curved surface of the Gaussian surface perpendicularly and no lines of force pass through the cross-section of the cylinder.

If \(\phi \) represents the electric flux passing through the Gaussian surface by the definition of flux,

$$\phi = E.A \text{where} A =2\pi rl, \text{is the curved surface area of Gaussian cylinder}$$

$$\phi = E.2\pi rl\dots (i)$$

Now, charge enclosed by Gaussian surface(q) = charge present in length 'l'

$$\text{or,} q = \lambda l$$

$$\phi =\frac{ \text{charge enclosed by Gaussian Surface}}{\epsilon _o}$$'

$$\phi =\frac{\lambda l}{\epsilon _o} \dots (ii)$$

Equating (i) and (iii), we get

$$E. 2\pi rl=\frac{\lambda l}{\epsilon _o}$$

$$ E=\frac{\lambda }{2\pi \epsilon _o r}$$

$$\therefore E \propto \frac {1}{r} $$

Hence, the electric field intensity produced by a linear charge distribution is non-uniform rather it depends upon the linear charge density and the distance of the concerned point.