Name: Electric Field of a Charged Plane Sheet
Uploaded: 2019-07-19
Duration: 6 min 58 s
Description: Another distribution of charge. This time on an infinite plane sheet. Let’s see once again how Gauss’s law comes to the rescue in electric field calculations.

Question

Find the expression for electric field intensity

E

due to uniformly charges thin spherical shell sphere at a point outside the shell.

Accepted Answer

Gauss Law for electrostatics states that the total electric flux passing through a closed surface equals the enclosed charge in the surface divided by the permittivity of the medium. Mathematically it is written as:

$\oint Eds = \frac{q}{ε_{0}}$

Electric field due to a uniformly charged thin shell:
In the diagram, $O$ is the centre of the spherical shell of radius $R$ , and charge enclosed by the spherical shell is $q$ with uniformed charge density $σ$ .
Let a point $P$ is outside the spherical shell, where we have to calculate the is an electric field. The distance between the centre of the spherical shell and the point A.is $r$ .
The total charge on the Gaussian surface of radius $R$ is

$q = σ (surface area) = σ 4 π R^{2}$ .

Hence $\oint Eds = \frac{q (4 π R^{2}) σ}{ε_{0}} or E \oint ds = \frac{q (4 π R^{2}) σ}{ε_{0}}$

If the point $P$ lies outside the surface of the shell at a distance $r$ from the centre of the sphere

As $\oint E dS = \oint dS = E . 4 {πr}^{2}$ we have $E (4 {πr}^{2}) = \frac{q (4 {πR}^{2}) σ}{ε_{0}} or E = \frac{σ}{ε_{0}} . \frac{R^{2}}{r^{2}} (for r > R)$

Hence, he required expression is : $E = \frac{{σR}^{2}}{ε_{0} r^{2}}$

Find the expression for electric field intensity $E$ due to uniformly charges thin spherical shell sphere at a point outside the shell.

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