Question

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

Answer 1

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.

Answer 2

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:

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