Question

using gauss's law , derive an expression for electric field intensity at a point outside a thin uniformly charged sphere . the test charge used to measure the electric field at a point should be negligible small . why ?​

Answer 1

Answer:

IT'S Q/Aε₀

Explanation:

HERE Q= CHARGE

A = AREA

ε₀ IS RELATIVE PERMEATTIVITY

Answer 2

Answer:

Let, R be the radius of the shell and Q be the charge uniformly distributed on the surface.

i) For a point outside the shell :

By Gauss's law, E.4πr

2

=

ϵ

0

Q

en

here r be the distance from centre of shell (r>R) and charge enclosed by surface Q

eq

=Q

So, E

out

=E=

4πϵ

0

r

2

Q

ii) For a point inside the shell :

By Gauss's law, E.4πr

2

=

ϵ

0

Q

en

here r be the distance from centre of shell (r<R) and charge inside the shell, Q

eq

=0

So, E

in

=E=0