Question

state gauss's law. derive an expression for electric field due to spherical shell?​

Answer 1

Consider a charged thin spherical shell of radius R. Let the charge +q distribute uniformly on the outer surface of the conductor, so that the surface charge density be ‘σ’. Case 1: Outside the spherical shell: Let ‘P’ be a point outside the conductor at a distance r from the centre of the spherical shell. With ‘r’ as the radius imagine a Gaussian surface. The electric field at every point on the Gaussian surface is same and radial (θ = 0). The flux through the Gaussian surface of radius r is Φ = E.S cosθS – Surface area of the Gaussian surface of radius r. ∴ Φ = E.(4πr2) ……(1) From Gauss’s Law Φ = q ε 0 qε0 …….(2) Equating (1) and (2), we get ^ n n^ = Unit radial vector Case 2: On the surface of the spherical shell: The electric field on the surface of the conductor r = R, then Case 3: Inside the spherical shell: Since the charges enclosed by the Gaussian surface of radius ('r') is zero, the electric field inside the spherical shell is zeroRead more on Sarthaks.com - https://www.sarthaks.com/623688/state-gausss-derive-expression-electric-intensity-point-outside-uniformly-charged-shell

Answer 2

Answer:

Consider a charged thin spherical shell of radius R.

Let the charge +q distribute uniformly on the outer surface of the conductor,

so that the surface charge density be ‘σ’. Case 1: Outside the spherical shell:

 Let ‘P’ be a point outside the conductor at a distance r from the centre of the spherical shell.  

With ‘r’ as the radius imagine a Gaussian surface. The electric field at every point on the Gaussian surface is same and radial (θ = 0).  The flux through the Gaussian surface of radius r is  Φ = E.S cosθS – Surface  area of the Gaussian surface of radius r.  

∴ Φ = E.(4πr2)  ……(1)  

From Gauss’s Law Φ =  q ε 0 qε0 …….(2)

Equating (1) and (2),

we get ^ n n^ = Unit radial vector

Case 2: On the surface of the spherical shell:  The electric field on the surface of the conductor r = R, then

Case 3: Inside the spherical shell: Since the charges enclosed by the Gaussian surface of radius ('r') is zero, the electric field inside the spherical shell is zero