524
11 Electrostatics
The charge on C 2 is
q 2 = C 2 V 2 = 6 × 10
−6
× 3000 = 0.018 C
The combined capacitance
C =
C 1 C 2
C 1 + C 2
=
3 × 6
3 + 6
× 10
−6
= 2 × 10
−6 F
Take the lower charge to find the maximum voltage V :
V =
q 1
C
=
0.012
2 × 10 −6 = 6000 V
11.84 If the dielectric is present, Gauss’ law gives
ε 0
E · ds = q − q
=
q
K
(1)
where −q is the induced surface charge, q is the free charge and K is the
dielectric constant. Construct a Gaussian surface in the form of a coaxial
cylinder of radius r and length l, closed by end caps. Applying (1),
ε 0 E (2πrl) =
q
K
or E =
q
2πε 0 rlk
(2)
In (1) the integral is contributed only by the curved surface and not the end
caps. The potential difference between the central rod and the surrounding
tube is given by
V = −
b
a
E · dr =
b
a
E dr =
b
a
q
2πε 0 l K
dr
r
=
q
2πε 0 l K
ln
b
a
The capacitance is given by
C =
q
V
=
2πε 0 l K
ln (b/a)
11.85 The field at point P is caused entirely by the charge Q on the inner sphere,
Fig. 11.45, and has the value
E =
Q
4πε 0 r 2