Question

A thin spherical shell of radius ‘R’ is

uniformly charged to a surface charge

density  .Using Gauss’s theorem derive the

expression for the electric field produced

outside the shell.​

Answer 1

Answer:

$\frac{kq}{r^2}$

Explanation:

• Guess theorem say that product of surface area and electric field is equal to charge inside Gaussian surface divided by epsilon

Procedure

• Let a Gaussian surface at distance r+dr from centre of sphere
• now accoridng to Gauss law

$e \times 4\pi( {r + dr})^{2} = \frac{\sigma}{\epsilon}$

$e \times 4\pi {r}^{2} = \frac{q}{\epsilon}$

$e = \frac{kq}{ {r}^{2} }$

Note : dr is neglected because it is too small as compared to r

Additional information

• electric potential at surface of sphere is kq/r