- Introduction
- |
- Electric field due to continous charge distributions
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- Gauss's Law
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- Applications of Gauss's Law
- |
- Derivation of Coulumb's Law
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- Electric field due to line charge
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- Electric field due to charged solid sphere
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- Electric field due to an infinite plane sheet of charge

- Introduction to Gauss's Law
- Field due to continous distibution of charges
- How To Apply Gauss's Law
- Derivation of Coulomb's Law using Gauss's Law
- Electric field due to a line charge (using Gauss's Law)

- Gauss's law was suggested by Karl Frederich Gauss(1777-1855) who was German scientist and mathematician.
- Gauss's law is basically the relation between the charge distribution producing the electrostatic field to the behavior of electrostatic field in space.
- Gauss's law is based on the fact that flux through any closed surface is a measure of total amount of charge inside that surface and any charge outside that surface would not contribute anything to the total flux.
- Before we look further to study Gauss's law in detail let's study electric field due to continuous charge distributions.

- So far as in the previous chapter we have discussed force and field due to discrete charges.
- We now assume that charges on a surface are located very close togather so that such a system of charges can be assumed to have continous distribution of charges.
- In a system of closely spaced charges, total charge could be continously distributed along some line, over a surface or throughout a volume.
- First divide the continous charge distribution into small elements containing Δq amount of charge as shown in fig 1
- Electric field at point A due to element carrying charge Δq is

where r is the distance of element under consideration from point A and**r**ˆis the unit vector in the direction from charge element towards point A. - Total electric field at point A due to all such charge elements in charge distribution is

where index i refers to the i_{th}charge element in the entire charge distribution. - Since the charge is distributed continuously over some region , the sum becomes integral. Hence total field at A within the limit Δq→0 is,

and integration is done over the entire charge distribution. - If a charge q is uniformly distributed along a line of length L, the linear charge density λ is defined by

and the unit of λ is Coulumb/meter(C/m).

- For charge distributed non-uniformly over a line, linear charge density is

where dQ is the amount of charge in a small length element dL.

- For a charge Q uniformly distributed over a surface of area A, the surface charge density σ is

and unit of surface charge density is C/m^{2}. For non uniform charge distributed over a surface charge density is

where dA is a small area element of charge dQ.

- Similarly for uniform charge distributions volume charge density is

and for non uniform distribution of charges

and unit of volume charge distribution is C/m^{3}.

- We already know about electric field lines and electric flux. Electric flux through a closed surface S is

which is the number of field lines passing through surface S.

**Statement of Gauss's Law**

“ELectric flux through any surface enclosing charge is equal to q/ε_{0}, where q is the net charge enclosed by the surface”

mathematically,

where q_{enc}is the net charge enclosed by the surface and**E**is the total electric field at each point on the surface under consideration.

- It is the net charge enclosed in the surface that matters in Gauss's law but the total flux of electric field
**E**depends also on the surface choosen not merely on the charge enclosed.

- So if you have information about distribution of electric charge inside the surface you can find electric flux through that surface using Gauss's Law.

- Again if you have information regarding electric fluxthrough any closed surface then total charge enclosed by that surface can also be easily calculated using Gauss's Law.

- Surface on which Gauss's Law is applied is known as Gaussian surface which need not be a real surface.

- Gaussian surface can be an imaginary geometrical surface which might be empty space or it could be partially or fully embedded in a solid body.

- Again consider equation 11

In left hand side of above equation**E**·**da**is scalar product of two vectors namely electric field vector**E**and area vector**da**. Area vector**da**is defined as the vector of magnitude |da| whose direction is that of outward normal to area element da. So,**da**=**n**ˆda where**n**ˆ is unit vector along outward normal to da.

From above discussion we can conclude that,

**(1)**If both**E**and surface area da at each points are perpendicular to each other and has same magnitude at all points of the surface then vector**E**has same direction as that of area vector as shown below in the figure.

since**E**is perpendicular to the surface

**(2)**If**E**is parallel to the surface as shown below in the figure

and integral is also zero for zero value of electric field vector**E**at all points on the surface.

Class 12 Maths Class 12 Physics

- NCERT Solutions: Physics 12th
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- Concepts of Physics - Vol. 2 HC Verma
- Dinesh New Millennium Physics Class -XII (Set of 2 Vols) (Free Complete Solutions to NCERT Textbook Problems & NCERT Exemplar Problems in Physics-XII)
- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- CBSE Chapterwise Questions-Solutions Physics, Class 12
- CBSE 15 Sample Question Paper: Physics for Class 12th

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