Electric Field Inside the Spherical Shell
Let’s look at point P inside the spherical shell to see how the electric field there is. We may use symmetry to create a spherical Gaussian surface that passes through P, is centered at O, and has a radius of r. Now, based on Gauss’s Law,
ϕ = q ⁄ ε0 = E × 4πr2
Since surface charge density is spread outside the surface, there is no charge contained inside the shell. Therefore, the electric field inside shell from the above formula is also zero, i.e.,
E = 0 (since q = 0)
Also, Check
Applications of Gauss’s Law
Gauss’s Law states that the total electric flux out of a closed surface equals the charge contained inside the surface divided by the absolute permittivity. The electric flux in an area is defined as the electric field multiplied by the surface area projected in a plane perpendicular to the field. Now that we’ve established what Gauss law is, let’s look at how it’s used. Application of Gauss Law is important for Class 12 students.
In this article, our main focus is on the Application of Gauss Law with a brief discussion of Gauss Law.
Table of Content
- What is Gauss Law?
- Applications of Gauss Law
- Electric Field due to Infinite Wire
- Electric Field due to Infinite Plane Sheet
- Electric Field due to Thin Spherical Shell
Contact Us