11.3 Solutions
501
to be concentrated at the centre. The interaction energy between the shell
and the sphere of radius r will be
dU =
1
4πε 0
(dq)(dq )
r
=
1
4πε 0
(4πr 2 drρ)
4
3
πr 3 ρ
r
=
4πρ 2 r 4 dr
3ε 0
Total interaction energy
U =
dU =
4πρ 2
3ε 0
R
0
r
4 dr =
4πρ 2 R 5
15ε 0
=
3q 2
20πε 0 R
where we have substituted the value of ρ.
(b) U =
3
5
×9 × 10 9 × (92 × 1.6 × 10 −19 ) 2
1.5 × (238) 1/3 × 10 −15
= 1.259 × 10
−11 J
= 78.7 MeV
Fig. 11.37
11.40 (i) E =
Q
4πε 0
2
x 2 −
1
(x + d) 2 −
1
(x − d) 2
= −
2Q d 2 (3x 2 − d 2 )
4πε 0 x 2 (x 2 − d 2 ) 2
(ii) For x >> d, 3x
2
− d
2
3x
2 and x
2
− d
2
x
2
E = −
6Q d 2
4πε 0 x 4
(iii) E = −
6 × 9 × 10 9 × 2 × 10 −6 × (10 −4 ) 2
(0.2) 2
= 0.675 N/C