how does the electric field intensity vary with the increase of distance of the point from the center of a charged conducting sphere? explain with graph
Answers
Answer:
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.
(i) 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
E
is 0.
Also
E
being constant, can be taken out of integral.
So, E(4πr
2
)=
q
0
q
So, E=
4πε
0
1
r
2
q
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.
(ii) 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
)=
ε
0
Q
end
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.
Find the electric field intensity due to a uniformly charged spherical shell at a point (i) outside the shell and (ii) inside the shell. Plot the graph of electric field with distance from the centre of the shell.
Explanation: