The surface charge density of a sphere of radius ris O. It is placed at a point A. The electric field
intensity at B due to this sphere is E. Another charged sphere of radius 2r is placed at B. If the
intensity at the centre of A and B is E/2 then the surface charge density of B is
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Answer:
Total charge , Q=∫
0
R
ρdV=∫
0
R
R
ρ
0
r
4πr
2
dr=
R
4πρ
0
[R
4
/4]=πρ
0
R
3
b) By Gauss's law, the field inside sphere, E.4πr
2
=
ϵ
0
Q
en
Here, Q
en
=∫
0
r
ρdV=∫
0
r
R
ρ
0
r
4πr
2
dr=
R
4πρ
0
[r
4
/4]=πρ
0
r
4
/R=Qr
4
/R
4
So, E.4πr
2
=
R
4
ϵ
0
Qr
4
or E=
R
4
KQr
2
where K=
4πϵ
0
1
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