let us consider the total flux through a sphere of radius `r`, which encloses a point charge `q` at its centre. Divide the sphere into small area elements, as shown in Fig. below
Electric flux through a closed surface `S`
= `q/ε_0`
`q =` total charge enclosed by `S`.
Let us consider the total flux through a sphere of radius r, which encloses a point charge q at its centre. Divide the sphere into small area elements, as shown in
The flux through an area element `ΔS` is
`Δφ=E.ΔS=q/(4πε_0r^2)r. Δ S`
The unit vector `hat r` is along the radius vector from the centre to the area element. Now, since the normal to a sphere at every point is along the radius vector at that point, the area element `ΔS` and `hat r` have the same direction. Therefore,
`Δφ=q/(4πε_0r^2)ΔS`
The total flux through the sphere is obtained by adding up flux through all the different area elements:
`φ=sum_(allΔs )q/(4πε_0r^2 )ΔS`
Since each area element of the sphere is at the same distance r from the charge,
`φ=q/(4πε_0r^2 )sum_(allΔs )ΔS=q/(4πε_0r^2 )S`
Now `S`, the total area of the sphere, equals `4πr^ 2` . Thus,
`φ=(q/(4πε_0r^2 ))*(4πr^2) =q/ε_0`
Here the electric field is uniform and we are considering a closed cylindrical surface, with its axis parallel to the uniform field `E`. The total
flux `φ` through the surface is `φ = φ_1 + φ_2 + φ_3,` where `φ_1` and `φ_2` represent the flux through the surfaces `1` and `2` of the cylinder and `φ_3` is the flux through the curved cylindrical part of the closed surface. Now the normal to the surface `3` at every point is perpendicular to `E`, so by definition of flux, `φ_3 = 0`. Further, the outward normal to `2` is along `E` while the outward normal to `1` is opposite to `E`.
Therefore,
`φ_1 = –E S_1, φ_2 = +E S_2`
`S_1 = S_2 = S`
where `S` is the area of circular cross-section. Thus, the total flux is zero, as expected by Gauss’s law. Thus, whenever you find that the net electric flux through a closed surface is zero, we conclude that the total charge contained in the closed surface is zero.
`\color{green} ✍️ \color{green} \mathbf(KEY \ CONCEPT)`
The great significance of Gauss’s law , is that it is true in general, and not only for the simple cases we have considered above. Let
us note some important points regarding this law:
`color{blue}{(i)}` Gauss’s law is true for any closed surface, no matter what its shape or size.
`color{blue}{(ii)}` The term `q` on the right side of Gauss’s law , includes the sum of all charges enclosed by the surface. The charges may be located anywhere inside the surface.
`color{blue}{(iii)}` When the surface is so chosen that there are some charges inside and some outside, the electric field is due to all the charges, both inside and outside `S`. The term `q` on the right side of Gauss’s law, however, represents only the total charge inside `S`.
`color{blue}{(iv)}` The surface that we choose for the application of Gauss’s law is called the Gaussian surface. You may choose any Gaussian surface and apply Gauss’s law. However, take care not to let the Gaussian surface pass through any discrete charge. This is because electric field due to a system of discrete charges is not well defined at the location of any charge .However, the Gaussian surface can pass through a continuous charge distribution.
`color{blue}{(v)}` Gauss’s law is often useful towards a much easier calculation of the electrostatic field when the system has some symmetry. This is facilitated by the choice of a suitable Gaussian surface.
`color{blue}{(vi)}` Finally, Gauss’s law is based on the inverse square dependence on distance contained in the Coulomb’s law. Any violation of Gauss’s law will indicate departure from the inverse square law.
let us consider the total flux through a sphere of radius `r`, which encloses a point charge `q` at its centre. Divide the sphere into small area elements, as shown in Fig. below
Electric flux through a closed surface `S`
= `q/ε_0`
`q =` total charge enclosed by `S`.
Let us consider the total flux through a sphere of radius r, which encloses a point charge q at its centre. Divide the sphere into small area elements, as shown in
The flux through an area element `ΔS` is
`Δφ=E.ΔS=q/(4πε_0r^2)r. Δ S`
The unit vector `hat r` is along the radius vector from the centre to the area element. Now, since the normal to a sphere at every point is along the radius vector at that point, the area element `ΔS` and `hat r` have the same direction. Therefore,
`Δφ=q/(4πε_0r^2)ΔS`
The total flux through the sphere is obtained by adding up flux through all the different area elements:
`φ=sum_(allΔs )q/(4πε_0r^2 )ΔS`
Since each area element of the sphere is at the same distance r from the charge,
`φ=q/(4πε_0r^2 )sum_(allΔs )ΔS=q/(4πε_0r^2 )S`
Now `S`, the total area of the sphere, equals `4πr^ 2` . Thus,
`φ=(q/(4πε_0r^2 ))*(4πr^2) =q/ε_0`
Here the electric field is uniform and we are considering a closed cylindrical surface, with its axis parallel to the uniform field `E`. The total
flux `φ` through the surface is `φ = φ_1 + φ_2 + φ_3,` where `φ_1` and `φ_2` represent the flux through the surfaces `1` and `2` of the cylinder and `φ_3` is the flux through the curved cylindrical part of the closed surface. Now the normal to the surface `3` at every point is perpendicular to `E`, so by definition of flux, `φ_3 = 0`. Further, the outward normal to `2` is along `E` while the outward normal to `1` is opposite to `E`.
Therefore,
`φ_1 = –E S_1, φ_2 = +E S_2`
`S_1 = S_2 = S`
where `S` is the area of circular cross-section. Thus, the total flux is zero, as expected by Gauss’s law. Thus, whenever you find that the net electric flux through a closed surface is zero, we conclude that the total charge contained in the closed surface is zero.
`\color{green} ✍️ \color{green} \mathbf(KEY \ CONCEPT)`
The great significance of Gauss’s law , is that it is true in general, and not only for the simple cases we have considered above. Let
us note some important points regarding this law:
`color{blue}{(i)}` Gauss’s law is true for any closed surface, no matter what its shape or size.
`color{blue}{(ii)}` The term `q` on the right side of Gauss’s law , includes the sum of all charges enclosed by the surface. The charges may be located anywhere inside the surface.
`color{blue}{(iii)}` When the surface is so chosen that there are some charges inside and some outside, the electric field is due to all the charges, both inside and outside `S`. The term `q` on the right side of Gauss’s law, however, represents only the total charge inside `S`.
`color{blue}{(iv)}` The surface that we choose for the application of Gauss’s law is called the Gaussian surface. You may choose any Gaussian surface and apply Gauss’s law. However, take care not to let the Gaussian surface pass through any discrete charge. This is because electric field due to a system of discrete charges is not well defined at the location of any charge .However, the Gaussian surface can pass through a continuous charge distribution.
`color{blue}{(v)}` Gauss’s law is often useful towards a much easier calculation of the electrostatic field when the system has some symmetry. This is facilitated by the choice of a suitable Gaussian surface.
`color{blue}{(vi)}` Finally, Gauss’s law is based on the inverse square dependence on distance contained in the Coulomb’s law. Any violation of Gauss’s law will indicate departure from the inverse square law.