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State Gauss's theorem ? Derive the expression for electric field due Spherical hollow conductor and infinitely charged sheet and linearly charged conductors.
Asked by avaneesh1162 | 13 Jun, 2021, 08:56: AM
Gauss law states that electric flux φ through a closed surface equals  ( q / ε) , where q is total charge enclosed inside the closed surface
and εo is permittivity of free space .

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Let us consider a spherical hollow conductor of radius R that has surface charge so that total charge on the sphere is q .

If we need to find electric field at a distance r from centre of hollow sphere, we make a gaussian surface
using a concentric holow sphere of radius r as shown in figure.

Electric flux φ over the gaussian surface = Electric field × area

since electric field is uniform ,  electric flux φ = E × 4π r2

According to Gauss theorem ,  electric flux φ = E × 4π r2 =  ( q / εo )

Hence   E =  q / ( 4πεo r2  )

when r < R , Gaussian surface will be inside the hollow sphere and enclosed charge is zero.

Hence electric field inside hollow sphere is zero.

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Let us consider an infinit plane that has surface charge density σ ( charge per unit area ) .

Electric field near to infinite plane is determined by making gaussian surface using a square box as shown in figure .

Let this square box enclose part of the charge of infinite plane within the area of box .
Also this square box intersects the plane so that top part of box is above the plane and bottom part of box is below the plane.
( part of the box below the plane is not visible in the figure ).
Hence a gaussian closed surface is formed that enclose part of the charge of infinite plane.

Let d be the height of top surface of gaussian surface above plane .

Then electric field at a distance d from plane is determined by gauss theorem .

E × 2A  =  ( σ × A ) / εo

LHS of above expression is total electric flux over the closed surface and RHS is the enclosed charge divided by εo.

Hence Electric field E = σ / ( 2 εo )

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Let us consider a linearly charged thin rod of infinite length that has charge density λ ( charge per unit length ) .

To get electric field at distance r from rod , we form a gaussian closed surface by using a cylinder of length l
and radius r as shown in figure. Let the axis of cylinder coincide withe rod.

By Gauss theorem ,  E × ( 2 π r l ) =  ( λ l ) / εo

LHS of above expression is total electric flux and RHS is enclosed charge divided by εo

Hence electric field E = λ / [ ( 2 π r ) εo ]
Answered by Thiyagarajan K | 13 Jun, 2021, 03:09: PM

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