Title: Field Due To Plane Charged Conductor in Nepali
Consider a plane charged conductor having uniformly distributed charge density(sigma) . Let P be a point at a distance from the plane charged conductor at which the field strength has to be found. For this purpose a small area A, surrounding P is drawn in such a way that this surface area is parallel to the plane charged conductor. Also a cylindrical surface of base area A is constructed with its walls perpendicular to the plane of charged conductor. A The electric field due to this conductor is E and it is perpendicular to the surface area A. Also the electric field E. is parallel to the wall of the cylinder and no lines of force cross the side walls of the cylinder. Therefore, electric flux crossing the side walls of the cylinder is zero .
So, the flux though the surface area A at P is given as,(phi)= E.A.
The net charge enclosed by the surface Area A = (sigma)*A
from Gauss's theorem, we can write,
Total flux= (total charge enclosed)/(epsilon)
So,
E=(sigma)/(epsilon)
where, (epsilon)=permittivity of freespace
Links:
Application of Gauss's Law:
• Application of Gauss's...
Permittivity:
• Permittivity ||P-07||#...
action of Point:
• Revealing the Action o...
Quantization of Charge:
• Quantization of Charge...
Conservation of Charge :
• Conservation of Charge...
Surface Charge Density:
• Surface Charge Density...
Van De Graff Generator:
• Van De Graff Generator...
Coulumb's law :
• Coulumb's Law Class11 ...
Electric Field:
• Electric Field Class11...
Electric Field Strength:
• Electric Field Strengt...
Gauss Law;
• Gauss's Law in Nepali ...
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