510
11 Electrostatics
Q
=
4
3
π (r 3 − b 3 )Q
4π
3
(a 3 − b 3 )
=
(r 3 − b 3 )
(a 3 − b 3 )
Q
E =
Q
4πε 0 r 2 =
Q
4πε 0 r 2
(r 3 − b 3 )
(a 3 − b 3 )
(c) Region r > a : E =
Q
4πε 0 r 2
11.54 (a) Construct a Gaussian surface in the form of a cylinder of radius r > R
and height h. By Gauss’ law
(ε 0 E) (2πrh) = Q = ρπ R
2 h
∴
E =
1
2
ρ
ε 0
R 2
r
(b) Construct a Gaussian surface in the form of a cylinder of height h and
radius r < R coaxial with the cylinder of radius R and height h. By
Gauss’ law
(ε 0 E) (2πrh) = Q
= ρπ R
2 h
∴
E =
1
2
ρ
ε 0
R 2
r
(r < R)
11.55 (a) E =
q
4 πε 0 r 2
∴
q = 4 πε 0 r
2 E =
1
9 × 10 9 × (0.5)
2
× 800 = 2.22 × 10
−8 C
(b) E =
σ
ε 0
∴
σ = Eε 0 = 120 × 8.85 × 10
−12
= 1.062 × 10
−9 C/m
2
(c) q = ε 0 φ
φ = E A = 120 × (100 × 75) = 9 × 10
5 N m
2
/C
11.56 (a) E 0
E · d A = q
(Gauss’ law)
(1)
Consider an isolated positive point charge. Construct a Gaussian surface,
a sphere of radius r centred at the charge. At every point on the spherical