Use gauss law to derive expression for electric field intensity at any point
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Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge. It is one of the four equations of Maxwell’s laws of electromagnetism. It was initially formulated by Carl Friedrich Gauss in the year 1835 and relates the electric fields at the points on a closed surface and the net charge enclosed by that surface.
The electric flux is defined as the electric field passing through a given area multiplied by the area of the surface in a plane perpendicular to the field. Yet another statement of Gauss’s law states that the net flux of a given electric field through a given surface, divided by the enclosed charge should be equal to a constant.
Usually, a positive electric charge is supposed to generate a positive electric field. The law was released in 1867 as part of a collection of work by the famous German mathematician, Carl Friedrich Gauss.
Gauss Law Equation
Let us now study Gauss’s law through an integral equation. Gauss’s law in integral form is given below:
∫E⋅dA=Q/ε0 ….. (1)
Where,
E is the electric field vector
Q is the enclosed electric charge
ε0 is the electric permittivity of free space
A is the outward pointing normal area vector
Flux is a measure of the strength of a field passing through a surface. Electric flux is defined as
Φ=∫E⋅dA …. (2)
We can understand the electric field as flux density. Gauss’s law implies that the net electric flux through any given closed surface is zero unless the volume bounded by that surface contains a net charge.
Gauss’s law for electric fields is most easily understood by neglecting electric displacement (d). In matters, the dielectric permittivity may not be equal to the permittivity of free-space (i.e. ε≠ε0). In the matter, the density of electric charges can be separated into a “free” charge density (ρf) and a “bounded” charge density (ρb), such that:
Ρ = ρf + ρb
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Plz mark me brainliest
Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge. It is one of the four equations of Maxwell’s laws of electromagnetism. It was initially formulated by Carl Friedrich Gauss in the year 1835 and relates the electric fields at the points on a closed surface and the net charge enclosed by that surface.
The electric flux is defined as the electric field passing through a given area multiplied by the area of the surface in a plane perpendicular to the field. Yet another statement of Gauss’s law states that the net flux of a given electric field through a given surface, divided by the enclosed charge should be equal to a constant.
Usually, a positive electric charge is supposed to generate a positive electric field. The law was released in 1867 as part of a collection of work by the famous German mathematician, Carl Friedrich Gauss.
Gauss Law Equation
Let us now study Gauss’s law through an integral equation. Gauss’s law in integral form is given below:
∫E⋅dA=Q/ε0 ….. (1)
Where,
E is the electric field vector
Q is the enclosed electric charge
ε0 is the electric permittivity of free space
A is the outward pointing normal area vector
Flux is a measure of the strength of a field passing through a surface. Electric flux is defined as
Φ=∫E⋅dA …. (2)
We can understand the electric field as flux density. Gauss’s law implies that the net electric flux through any given closed surface is zero unless the volume bounded by that surface contains a net charge.
Gauss’s law for electric fields is most easily understood by neglecting electric displacement (d). In matters, the dielectric permittivity may not be equal to the permittivity of free-space (i.e. ε≠ε0). In the matter, the density of electric charges can be separated into a “free” charge density (ρf) and a “bounded” charge density (ρb), such that:
Ρ = ρf + ρb
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hey mate here is the proof
a)Electric field intensity at any point outside a uniformly charged spherical shell:
Consider a thin spherical shell of radius R and with centre O. Let charge + q be uniformly distributed over the surface of the shell.
Let P be any point on the Gaussian sphere S1 with centre O and radius r, as shown in the following figure.
According to Gauss's law, we can write the flux through ds as:
At any point on the surface of the shell, r = R
a)Electric field intensity at any point outside a uniformly charged spherical shell:
Consider a thin spherical shell of radius R and with centre O. Let charge + q be uniformly distributed over the surface of the shell.
Let P be any point on the Gaussian sphere S1 with centre O and radius r, as shown in the following figure.
According to Gauss's law, we can write the flux through ds as:
At any point on the surface of the shell, r = R
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