Question

Define gaussian surface? find relation far electric field intensity due to charged hollow conducting shell a,inside b,outside c,on the surface draw geographically also

Answer 1

Answer:

Gaussian surface:

The surface that we choose for the application of Gauss's law is called the Gaussian surface. The Gaussian surface can pass through a continuous charge distribution. Gauss's law is useful for the calculation of the electrostatic field for a symmetric system.

Explanation:

Relation far electric field intensity due to charged hollow conducting shell :

Consider a thin spherical shell of radius R with a positive charge q distributed uniformly on the surface. As the charge is uniformly distributed, the electric field is symmetrical and directed radially outward.

Inside the shell:

In this case, we select a gaussian surface concentric with the shell of radius r (r>R).

So, ∮E.ds=E(4πr^2)

According to gauss law,

E(4πr^2) = Q end/ε0

Since charge enclosed inside the spherical shell is zero.

So, E=0

Hence, the electric field due to a uniformly charged spherical shell is zero at all points inside the shell.

Electric field outside the shell:

For point r>R; draw a spherical gaussian surface of radius r.

Using gauss law, ∮E.ds= q 0/q end

Since

E is perpendicular to gaussian surface, angle betwee

Eis 0.

Also E being constant, can be taken out of integral.

So, E(4πr^2)= q/q0

So, E=1/4πε0 ×q/r^2

Thus electric field outside a uniformly charged spherical shell is same as if all the charge q were concentrated as a point charge at the center of the shell.

Hope you get it ..

Thank you