Find the intensity of the electric field due to a point charge by Gauss's theorem and the column...

Electric field due to point charge

A : Gauss theorem is applicable on any closed surface. R : In order to find the value of electric field due to a charge distribution, Gauss' theorem should be applid on a symmetrical closed surface.

Assertion: Electric field intensity at a point due to a positive charge is always opposite in direction to electric field intensity due to negative charge, and it is independent to locațions of charges. Reason: The direction of electric field due to a point charge is away from the positive charge and towards the negative charge along the line joining the point charge and location where the electric field is to be calculated.

Electric Field Due To An Isolated Point Charge

Electric Field Due To A Point Charge Vector Notation

It is not convenient to use a spherical Gaussian surface to find the electric field due to an electric dipole using Gauss’s theorem because

The intensity of electric field at a point due to charged conductor of any shape or plane charged sheet is

Which of the following statement(s) is/are correct? If the electric field due to a point charge varies as r^(-2.5) instead of r^(-2),then the Gauss law will still be valid. The Gauss law can be used to calculate the field distribution around an electric dipole. If he electric field between two point charges is zero somewhere,then the sign of the two charges is the same. The work done by the external force is moving a unit positive charge from point A at potential "V_(A)" to point B at potential "V_(B)" is,(V_(B)-V_(A))

Assertion : With the help of Gauss's theorm we can find electric field at any point. Reason : Gauss's theorem can be applied for any type of charge distribution.