Electric field intensity due to infinite sheet of charge
![](https://mashalscienceacademy.files.wordpress.com/2019/09/p12c1num21-1.png?w=243)
Consider a sheet of infinite extent on which charges are uniformly distributed with charge density σ. We want to calculate electric intensity E at point P at a distance r from the sheet. Imagine a Gaussian surface in the form of a cylinder passing through the sheet. Let the height of the cylinder is 2r and cross-sectional area A. As E is perpendicular to the end faces and parallel (angle of E and area of the curved surface is 90°.) to the curved faces of the cylinder, therefore, the flux through the curved faces is zero and all flux contribution come through the end faces.
![](https://mashalscienceacademy.files.wordpress.com/2019/09/p12c1num22.png?w=424)
Since charge per unit area is σ and area enclosed is ΔA, therefore, total charge enclosed by the Gaussian surface is Q = σ ΔA. Put this value in the above equation.
![](https://mashalscienceacademy.files.wordpress.com/2019/09/p12c1num23.png?w=552)
In vector form
![](https://mashalscienceacademy.files.wordpress.com/2019/09/p12c1num24.png?w=75)
where n is a unit vector normal to the sheet.
Electric Field intensity due to a charge conducting spherical shell
Consider a charged conducting spherical shell (a hollow one) of radius R. Suppose a +ve charge Q is given to the shell which distributes evenly on the surface. We calculate E on point p inside the sphere (rB ˂ R).
![](https://mashalscienceacademy.files.wordpress.com/2019/09/p12c1num26.png?w=283)
Consider a point B situated inside the shell at a distance r from the center of the sphere, hence rB is less than R . Apply Gauss’s law by taking a Gaussian surface of radius rB through the point B. As there is no charge inside the Gaussian surface, therefore, Q=0 and,
![](https://mashalscienceacademy.files.wordpress.com/2019/09/p12c1num25.png?w=146)
The charge is evenly distributed on the surface, therefore,(1) E is constant at all points at radial distance rB from the centre (think over, E will be constant at the surface of any sphere drawn inside this sphere). (2) Vector ΔA̅ is also radial, therefore, both E̅ and ΔA̅ have same direction, therefore,
![](https://mashalscienceacademy.files.wordpress.com/2019/09/p12c1num27.png?w=420)
Sum of ΔA is the surface area and equal to 4πrB2, therefore, E (4πrB2) = 0
Or E = 0 (as 4πrB2≠ 0)
B was chosen any arbitrary point; therefore, we conclude that the field inside a charged spherical shell is zero.
(How can you explain the safety of the passengers of airplanes in thunderstorms).
You can calculate the field intensity when the point p is outside the charge sphere. It will be
![](https://mashalscienceacademy.files.wordpress.com/2019/09/p12c1num28.png?w=101)
This result is identical to Coulomb’s law. Do it yourself.