Question

Find electric field intensity due to uniformly charged spherical shell of radius r at a point are equal to r are greater than or less than air

Answer 1

Explanation:

) Electric field outside the shell:

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

Using gauss law, \oint E.ds=\dfrac{q_{end}}{q_{0}}∮E.ds= q 0

q {E}

E is perpendicular to gaussian surface, angle betwee{E}

E is 0.

Also

E being constant, can be taken out of integral.

So, E(4\pi r^{2})=\dfrac{q}{q_{0}}E(4πr2 )= q 0

q

So, E=\dfrac{1}{4\pi \varepsilon_{0}}\dfrac{q}{r^{2}}E=

4πε 012q

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)(r>R).

So, \oint E.ds=E(4\pi r^{2})∮E.ds=E(4πr

2

)

According to gauss law,

E(4\pi r^{2})=\dfrac{Q_{end}}{\varepsilon _{0}}E(4πr

2

)=

ε

0

Q

end

Since charge enclosed inside the spherical shell is zero.

So, E=0E=0

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