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 | 13th Jun, 2021, 08:56: AM

Expert Answer:

###
Gauss law states that electric flux φ through a closed surface equals ( q / ε_{o }) , where q is total charge enclosed inside the closed surface
and ε_{o} 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 r^{2}
According to Gauss theorem , electric flux φ = E × 4π r^{2} = ( q / ε_{o} )
Hence E = q / ( 4πε_{o} r^{2} )
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 ) / ε_{o}
LHS of above expression is total electric flux over the closed surface and RHS is the enclosed charge divided by ε_{o}.
Hence Electric field E = σ / ( 2 ε_{o} )
-----------------------------------------------------
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 ) / ε_{o}
LHS of above expression is total electric flux and RHS is enclosed charge divided by ε_{o}
Hence electric field E = λ / [ ( 2 π r ) ε_{o} ]

Gauss law states that electric flux φ through a closed surface equals ( q / ε

_{o }) , where q is total charge enclosed inside the closed surfaceand ε

_{o}is permittivity of free space .------------------------------------------------------

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 r

^{2}According to Gauss theorem , electric flux φ = E × 4π r

^{2}= ( q / ε_{o})Hence E = q / ( 4πε

_{o}r^{2})when r < R , Gaussian surface will be inside the hollow sphere and enclosed charge is zero.

Hence electric field inside hollow sphere is zero.

-----------------------------------------------------------------

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 ) / ε

_{o}LHS of above expression is total electric flux over the closed surface and RHS is the enclosed charge divided by ε

_{o}.Hence Electric field E = σ / ( 2 ε

_{o})-----------------------------------------------------

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 ) / ε

_{o}LHS of above expression is total electric flux and RHS is enclosed charge divided by ε

_{o}Hence electric field E = λ / [ ( 2 π r ) ε

_{o}]### Answered by Thiyagarajan K | 13th Jun, 2021, 03:09: PM

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