Question

Verify Gauss theorem and write it's applications

Answer 1

The total dielectric flux passing through a closed surface in vacuum enclosing a charge is 1ϵo times the charge enclosed by the closed surface.Application of Gauss's Theorem

Gauss's theorem is useful for determining the electric field intensity produced by a charged conductor as follows:

Electric Field Intensity due to a Charged Sphere

Consider a sphericalConductor having radius 'R' and center 'O'. It is carrying charge 'q' on it's surface. Due to this distribution of charged electric field intensity around it can be defined. Following three distinct points can be considered.

Point P lying outside the charged sphere

1. At a point outside the surface of sphere(r>R)

Consider 'P' is the concerned point where electric field intensity due to the distribution of charge on the spherical conductor is to be calculated. For this, we construct a Gauss' sphere concentric with the charged sphere and passing through point 'P'. So, that due to symmetry at every point on the surface of the sphere electric field intensity has the same magnitude.

It ϕϕ represents the electric flux passing through Gauss' surface and 'A' be area of Gauss' surface and 'A' be area of Gauss's Sphere. Then

ϕ=EAϕ=EA

ϕ=E4πr2…(i)ϕ=E4πr2…(i)

Also from Gauss' theorem

ϕ=charge enclosed by Gaussian Surfaceϵoϕ=charge enclosed by Gaussian Surfaceϵo

ϕ=qϵo…(ii)ϕ=qϵo…(ii)

Equating (i) and (ii), we get

E.4πr2=qϵoE.4πr2=qϵo

∴E=14πr2.qϵo∴E=14πr2.qϵo

This is the electric field intensity produced distanced 'r' from the center of a spherical conductor carrying charge 'q'.

Point P lying on the surface of the sphere.

2. At a point on the surface of sphere(r=R)

When the concerned point lies on the surface the gauss' surface also becomes the sphere having same radius as that of the charged surface.

Then,

ϕ=EAϕ=EA

ϕ=E4πr2…(i)ϕ=E4πr2…(i)

Also from Gauss' theorem

ϕ=charge enclosed by Gaussian Surfaceϵoϕ=charge enclosed by Gaussian Surfaceϵo

ϕ=qϵo…(ii)ϕ=qϵo…(ii)

Equating (i) and (ii), we get

E.4πR2=qϵoE.4πR2=qϵo

∴E=14πϵ.qR2∴E=14πϵ.qR2

Hence, electric field intensity on the surface of charged sphere has constant magnitude.

E=14πr2.qϵoE=14πr2.qϵo

Point P lying inside the sphere.

3. At a point inside the surface of sphere(r<R)

Now, the Gaussian surface lies inside the charged sphere. As the charge resides only on the surface of the sphere, electric charge enclosed by Gaussian surface is O. If 'E ' represents electric field Intensity at point 'P' then

ϕ=EAϕ=EA

ϕ=E4πr2…(i)ϕ=E4πr2…(i)

Also from Gauss' theorem

ϕ=charge enclosed by Gaussian Surfaceϵoϕ=charge enclosed by Gaussian Surfaceϵo

ϕ=0…(ii)ϕ=0…(ii)

Equating (i) and (ii) we get

E.4πR2=0E.4πR2=0

E=0E=0

Hence, electric field intensity on the surface of charged sphere is zero.

4. Electric Field Intensity due to an Infinite Charged Plane Sheet.

Field of a charged plane conductor.

Consider an infinite plane sheet is given positive charge density is σσ. P is a point at distance 'r' from the sheet where electric field intensity is to be calculated. For this gaussian cylinder is drawn perpendicular to the sheet whose cross-section passes through point P. The cylinder extends to the other sides also.

Since, the lines of force leave the charged surface perpendicularly. Here lines of force pass perpendicularly through the cross-sectional area and do not pass through curved surface area. Therefore, total electric flux through Gaussian surface.

ϕ=EA+EA…(i)ϕ=EA+EA…(i)

Charge enclosed by Gaussian surface(q)=σA…(i)Charge enclosed by Gaussian surface(q)=σA…(i)

Also from Gauss' theorem

ϕ=charge enclosed by Gaussian Surfaceϵoϕ=charge enclosed by Gaussian Surfaceϵo

ϕ=σAϵo…(iii)ϕ=σAϵo…(iii)

Equating (i) and (iii), we get

2EA=σAϵo2EA=σAϵo

∴E=σ2ϵo∴E=σ2ϵo

It is independent of the distance of the concerned point from the plane.

5. Electric Field Intensity near a Charged Plane.

Infinite plane sheet of charge

Consider a charged plane with surface charge density σσ.We have to calculate an electric field intensity at point P that lies at the distance 'r' from the plane in order to use Gauss' theorem. We are going to construct a Gaussian area of cross-sectional area 'A' of the Gaussian cylinder but not across the curved surface. Therefore, total electric flux through Gaussian surface (ϕ)=E.A(ϕ)=E.A

ϕ=E..A…(i)ϕ=E..A…(i)

Charge enclosed by Gaussian Surfaceq=σA…(ii)q=σA…(ii)

From Gauss Theorem,

ϕ=charge enclosed by Gaussian Surfaceϵoϕ=charge enclosed by Gaussian Surfaceϵo

'

ϕ=σAϵo…(iii)ϕ=σAϵo…(iii)

Equating (i) and (iii), we get

EA=σAϵoEA=σAϵo

∴E=σϵo∴E=σϵo

This indicates an electric field intensity near a charged plane is uniform which is represented by parallel







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by anshu jurriya 2003