apply gauss law to show that for a charged spherical shell, the electric field outside the shell is , as - Physics - Electric Charges And Fields

Question

apply gauss law to show that for a charged spherical shell, the
electric field outside the shell is , as if the entire charge were
concentrated at the centre

Mousam Maity · Accepted Answer

Dear Student, Let the surface charge density of the spherical shell of radius R be σ C/m2 and the total charge contained in it be q coulombs The charged spherical shell has surface density σ C/m2 which can be written like $σ = \frac{Total charge on the spherical shell}{Spherical area of the shell} = \frac{q}{4 π R^{2}} o r, q = 4 σ π R^{2} (1)$

Let the Gaussian surface through point P which will be a spherical shell of radius r encloses the charged spherical shell Let Vector E be the electric field at a point on gaussian shell Consider a small area element vector dS around the point Then the electric flux through area element, $d Φ = \vec{E} \vec{d S}$ As vector dS is normal along the surface of shell, it is along the direction of vector E Thus, $d Φ = E d S Total electric flux, Φ = \oint_{S} E d S = E \oint_{S} d S Now, \oint_{S} d S = Surface area of spherical shell of radius r=4 π r^{2} ∴ Φ = E \times 4 π r^{2} (1)$ According to Gauss theorem, $Φ = \frac{q}{ϵ_{0}} (2) Form eq(1) and eq(2) E \times 4 π r^{2} = \frac{q}{ϵ_{0}} o r, E = \frac{q}{4 π ϵ_{0} r^{2}} For (r>R)$ Electric field lines due to a positive spherical shell:

Electric field lines due to a negative spherical shell:

Regards,

apply gauss law to show that for a charged spherical shell, the electric field outside the shell is , as if the entire charge were concentrated at the centre