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How does the electric field intensity vary with the increase of distance of the point from the centre of a charged conducting sphere? Explain with graph.​

User Bluesmonk
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1 Answer

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Answer:

(i) Electric field outside the shell:

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

Using gauss law, ∮E.ds=q0qend

Since E is perpendicular to gaussian surface, angle betwee E is 0.

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

So, E(4πr2)=q0q

So, E=4πε01r2q

User ZengJuchen
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