Question

State Gauss's law. Using Gauss law deduce the expression of electric field due o uniformly charged sheet.Also state the direction of electric field. *

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Answer 1

Gauss law states that electric flux through any closed surface is equal to the net charge enclosed inside the surface divided by the permittivity of vacuum.

Gauss law states that electric flux through any closed surface is equal to the net charge enclosed inside the surface divided by the permittivity of vacuum.Electric field due to a uniformly charged thin shell.

In the diagram, O is the centre of the spherical shell of radius R, and charge enclosed by the spherical shell is q.

In the diagram, O is the centre of the spherical shell of radius R, and charge enclosed by the spherical shell is q.Let a point A is outside the spherical shell, where we have to calculate the electric field.The distance between the centre of spherical shell and the point A.

In the diagram, O is the centre of the spherical shell of radius R, and charge enclosed by the spherical shell is q.Let a point A is outside the spherical shell, where we have to calculate the electric field.The distance between the centre of spherical shell and the point A.ϕ=E.Scosθ

In the diagram, O is the centre of the spherical shell of radius R, and charge enclosed by the spherical shell is q.Let a point A is outside the spherical shell, where we have to calculate the electric field.The distance between the centre of spherical shell and the point A.ϕ=E.ScosθSo, ϕ=E.4πr2−−−−−−−−−−−−−−(1)

In the diagram, O is the centre of the spherical shell of radius R, and charge enclosed by the spherical shell is q.Let a point A is outside the spherical shell, where we have to calculate the electric field.The distance between the centre of spherical shell and the point A.ϕ=E.ScosθSo, ϕ=E.4πr2−−−−−−−−−−−−−−(1)As per gauss law,

In the diagram, O is the centre of the spherical shell of radius R, and charge enclosed by the spherical shell is q.Let a point A is outside the spherical shell, where we have to calculate the electric field.The distance between the centre of spherical shell and the point A.ϕ=E.ScosθSo, ϕ=E.4πr2−−−−−−−−−−−−−−(1)As per gauss law,ϕ=ϵoq−−−−−−−−−(2)

In the diagram, O is the centre of the spherical shell of radius R, and charge enclosed by the spherical shell is q.Let a point A is outside the spherical shell, where we have to calculate the electric field.The distance between the centre of spherical shell and the point A.ϕ=E.ScosθSo, ϕ=E.4πr2−−−−−−−−−−−−−−(1)As per gauss law,ϕ=ϵoq−−−−−−−−−(2)From equation (1) and (2)

In the diagram, O is the centre of the spherical shell of radius R, and charge enclosed by the spherical shell is q.Let a point A is outside the spherical shell, where we have to calculate the electric field.The distance between the centre of spherical shell and the point A.ϕ=E.ScosθSo, ϕ=E.4πr2−−−−−−−−−−−−−−(1)As per gauss law,ϕ=ϵoq−−−−−−−−−(2)From equation (1) and (2)E.4πr2=ϵoq

In the diagram, O is the centre of the spherical shell of radius R, and charge enclosed by the spherical shell is q.Let a point A is outside the spherical shell, where we have to calculate the electric field.The distance between the centre of spherical shell and the point A.ϕ=E.ScosθSo, ϕ=E.4πr2−−−−−−−−−−−−−−(1)As per gauss law,ϕ=ϵoq−−−−−−−−−(2)From equation (1) and (2)E.4πr2=ϵoqNow,

In the diagram, O is the centre of the spherical shell of radius R, and charge enclosed by the spherical shell is q.Let a point A is outside the spherical shell, where we have to calculate the electric field.The distance between the centre of spherical shell and the point A.ϕ=E.ScosθSo, ϕ=E.4πr2−−−−−−−−−−−−−−(1)As per gauss law,ϕ=ϵoq−−−−−−−−−(2)From equation (1) and (2)E.4πr2=ϵoqNow,E=4πϵor2q

In the diagram, O is the centre of the spherical shell of radius R, and charge enclosed by the spherical shell is q.Let a point A is outside the spherical shell, where we have to calculate the electric field.The distance between the centre of spherical shell and the point A.ϕ=E.ScosθSo, ϕ=E.4πr2−−−−−−−−−−−−−−(1)As per gauss law,ϕ=ϵoq−−−−−−−−−(2)From equation (1) and (2)E.4πr2=ϵoqNow,E=4πϵor2qThis is the electric field at the distance r, outside from the shell.

Hope this helps.......