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) , where q is total charge enclosed inside the closed surface
and εo 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 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.
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 )
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 | 13th Jun, 2021, 03:09: PM