- Introduction
- |
- Electric field due to continous charge distributions
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- Gauss's Law
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- Applications of Gauss's Law
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- 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)

- Coulumb's law can be derived from Gauss's law.

- Consider electric field of a single isolated positive charge of magnitude q as shown below in the figure.

- Field of a positive charge is in radially outward direction everywhere and magnitude of electric field intensity is same for all points at a distance r from the charge.

- We can assume Gaussian surface to be a sphere of radius r enclosing the charge q.

- From Gauss's law

since**E**is constant at all points on the surface therefore,

surface area of the sphere is A=4πr^{2}

thus,

- Now force acting on point charge q' at distance r from point charge q is

This is nothing but the mathematical statement of Coulomb's law.

- Consider a long thin uniformly charged wire and we have to find the electric field intensity due to the wire at any point at perpandicular distance from the wire.

- If the wire is very long and we are at point far away from both its ends then field lines outside the wire are radial and would lie on a plane perpandicular to the wire.

- Electric field intensity have same magnitude at all points which are at same distance from the line charge.

- We can assume Gaussian surface to be a right circular cylinder of radius r and length l with its ends perpandicular to the wire as shown below in the figure.

- λ is the charge per unit length on the wire. Direction of
**E**is perpendicular to the wire and components of**E**normal to end faces of cylinder makes no contribution to electric flux. Thus from Gauss's law

- Now consider left hand side of Gauss's law

Since at all points on the curved surface**E**is constant. Surface area of cylinder of radius r and length l is A=2πrl therefore,

- Charge enclosed in cylinder is q=linear charge density x length l of cylinder,

or, q=λl

From Gauss's law

Thus electric field intensity of a long positively charged wire does not depends on length of the wire but on the radial distance r of points from the wire.

Class 12 Maths Class 12 Physics

- NCERT Solutions: Physics 12th
- NCERT Exemplar Problems: Solutions Physics Class 12
- 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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