Question

the value of electric field at the centre of hollow spherical metallic shell of radius 5m,having positive charge of 10 C is​

Answer 1

Given:

Radius of metallic spiracle shell is 5 metres and a charge of 10 C is present along its surface.

To find:

Value of electrostatic field intensity at the centre of the metallic shell.

Calculation:

Let electrostatic field intensity be denoted by E, area of the Gaussian surface be denoted by s:

Applying Gauss' Theorem:

$\therefore \: \displaystyle \: \oint \vec{E}. \vec{ds} = \dfrac{q_{enclosed}}{\epsilon_{0}}$

$= > \: \displaystyle \: \oint E \times ds \times \cos( \theta) = \dfrac{q_{enclosed}}{\epsilon_{0}}$

$= > \: \displaystyle \: \oint E \times ds \times \cos(\theta ) = \dfrac{q_{enclosed}}{\epsilon_{0}}$

$= > \: \displaystyle \: E \: \oint ds \:\cos(\theta)= \dfrac{q_{enclosed}}{\epsilon_{0}}$

Now , the charge of 10 coulombs is located along the surface of the shell and hence the enclosed charge will be zero coulombs.

$= > \: \displaystyle \: E \: \oint ds \:\cos(\theta)= \dfrac{0}{\epsilon_{0}}$

$= > \: \displaystyle \: E \: \oint ds \:\cos(\theta) = 0$

$= > \: \displaystyle \: E = 0 \: N{C}^{ - 1}$

So, the electrostatic field Intensity at the centre of the metallic shell will be zero.