## Solid Angle:

Just as we talk about the angles in two dimensions we talk about solid angle in three dimensions. The solid angle is the measure of total opening of cone around its vertex. We measure it by drawing a sphere centered at the vertex of cone ‘a’ of the sphere intercepted by cone and dividing it by the square of distance

The complete solid angle corresponds to the case for which ‘a’ = surface area of sphere i.e.

## Electric Flux:

* Consider a closed surface S placed in uniform electric field , and divide this surface into infinitesimally small parts of each. (The surface area is a vector and its direction is same as the direction perpendicular to the plane of area)
*

Electric flux is thus defined as the sum of dot products of and for all elementary areas constituting the surface. It is denoted by **f** and

where q is the angle between electric field and the area vector.

*Positive Electric Flux:
*

If the angle between **E** and **dS** is *acute* then flux is said to be positive or if number of field lines leaving the surface are more than the number of electric lines of force entering the given surface.

*Negative Electric Flux :
*

If the number of electric lines of force entering the surface are more than the field lines leaving the surface or if the angle between E and dS is *obtuse*, the the flux is said to be negative flux.

## GAUSS LAW:

* Electric field intensity can be calculated from Coulomb’s law for point charges only, but if we have some complex configuration of charges the field intensity can be computed using Gauss Law. It states that
*

** ” for any distribution of charges, the total electric flux linked with a closed surface is times the total charge within the surface “**. Mathematically,

where the first equality applies if the surface enclosed discrete charges and the second applies if the surface encloses continuous charge distribution.

*Proof of Gauss’s Law:
*

* To prove gauss law, consider a single point charge q enclosed in a closed surface of arbitrary shape. For positive charge, the electric field is pointing radially outwards. Imagine any infinitesimally small surface such that,
*

*But
*

where **dW** is the solid angle subtended at O by the surface area **dS**.

If there are number of charges q_{1}, q_{2}, …, q_{n}

then we can write gauss law as . In cgs system, Gauss Law can be stated as .

## Derivation of Coulomb’s Law from Gauss Law

* :
*

* Consider a point charge q and we have to find the electric field at a distance ‘ r‘ from charge. The gaussian surface is the spherical surface of radius ‘r‘ with centre on charge q. From symmetry electric field must have same value at all the points on the surface. Thus angle between and is zero,*

* If a test charge q¢ is located at point where electric field is determined, then,
*

* which is nothing but Coulomb’s law.
*

**Selecting a Gaussian Surface:
**

* In application of gauss’s laws to field calculations, some judgment is required in choosing the surface. Two useful guiding principles are that the point or points at which the field is to be determined must lie on the surface and the surface must have enough symmetry so that it is possible to evaluate the integral. Thus if a problem has spherical or cylindrical symmetry, the gaussian surface will usually be spherical or cylindrical respectively.
*

*Applications of Gauss Theorem :
*

## 1. Electric Field Due to Infinitely Long Wire of Uniform Charge Density (l) :

In order to find electric field due to wire at P, select cylindrical surface of radius r and height l to be gaussian surface. The electric field lines are parallel to upper and lower surface of cylinder and hence makes no contribution to the electric flux. It is the curved surface which contributes to the electric flux.

where r is the radius and l is the length of the cylinder.

**2**

## . Electric Field Due to Infinite Sheet of Charge with Uniform Charge Density ( s ) :

** **Electric field is to be determined at P. To apply gauss theorem, let us consider a cylinder to be a gaussian surface with ends on each side of sheet as shown. Let S be the surface area of the two end surfaces. In this case field lines are parallel to curved surface, hence their contribution to flux is zero. Only end surfaces will contribute.

Thus, we see that magnitude of electric field is independent of the distance from the sheet.

## 3. Electric Field At Any Point Due to Two Charged Conducting Plates

** :
**

Consider two parallel plane conductors P and Q given opposite charges with **s _{1}** and

**s**be the surface charge density for positive and negative conductor respectively.

_{2} Electric field at A due to P and Q will be,

Thus, net electric field,

If **s _{1} = s_{2}**, then,

**E = 0**

*Point lying inside two conductors:
*

If point lies outside the two conductors then direction of electric field due to both the conductors is same and net electric field is,

If **s _{1} = s_{2}** , then

**E = s/e**.

_{0}**4. Electric field At Point Inside A uniformly Charged Sphere:
**

Let us consider a sphere in which charge is uniformly distributed. Let r be the charge density. To find the electric field at any point inside the sphere at a distance of r from the centre. The gaussian surface is thus a sphere of radius r,

Charge inside gaussian sphere = Charge per unit volume ´ volume

i.e

For point lying on the surface of the sphere,

## 5. Electric Field Due To Uniformly Charged Spherical Shell :

Consider a shell of radius R with charge uniformly distributed over its surface.

*Electric Field At Any Outside Point :*

Consider any point lying at a distance r from the centre of the shell of radius R such that ( r > R). The gaussian surface in this case is a spherical shell of radius r. At all points on this sphere electric field is same.

According to Gauss Law,

## Electric Field At A Point On the Surface

* :
*

In case the point P lies on the shell i.e. if r = R, the gaussian surface is assumed to be sphere of radius R. Hence E = k q/r2

*Electric Field At Point Inside The Shell :
*

If r < R, the electric field is zero because charge enclosed inside gaussian surface is zero.

## Relation Between Surface Density Of Charge & Radius of Curvature

* :
*

Consider two spherical conductors with radius **r _{1}** and

**r**and having charges

_{2}**q**and

_{1}**q**. If the two spheres are connected by wire the potential will become same, i.e.

_{2}**V _{1} = V_{2}**

If the electric field at the two surfaces be E_{1} and E_{2}, then

Similarly, the ratio of surface charge densities on the two spheres will be,

or surface density of charge is inversely proportional to the radius of curvature or we can say that charge always tends to concentrate towards the region with low value of radius of curvature or pointed ends.

## Conductor in An Electric Field:

In any conductor or piece of metal electrons are always acts as charge carriers and the number of electrons are always equal to the number of positive ions in it. When conductor is placed in an electric field, the positive charge carriers moves in the direction of the field and negative charge carriers in direction opposite to the field. Therefore, there is an accumulation of positive charges on one side and accumulation of negative charges on the other. These charges are called induced charges and they generate an electric field whose direction is opposite to the applied field. This electric field is called induced electric field. The accumulation of charges will keep on increasing unless applied and induced field totally balance each other. At this point, net field inside conductor becomes zero, hence force (F = qE) acting on the conductor is also reduced to zero. This whole process completes in a fraction of a second. Hence, we assume, E = 0 inside a conductor.