528
11 Electrostatics
U 3 =
1
2
C 3 V
2
3 =
1
2
× 3 × 10
−6
× 80
2
= 0.0096 J
Note that U 1 + U 2 + U 3 = 0.012 J = U =
1
2
C V 2
11.91 (a) The combination of C 1 and C 2 in series yields C 12 =
8 × 4
8 + 4
= 2.667 ×
10 −6 F.
C = C 12 + C 3 = (2.667 + 3.0) × 10
−6
= 5.667 × 10
−6 F
q = C V = 5.667 × 10
−6
× 100 = 5.667 × 10
−4 C
q 3 = C 3 V = 3 × 10
−6
× 100 = 3 × 10
−4 C
q 1 = q 2 = q − q 3 = (5.667 − 3.0) × 10
−4
= 2.667 × 10
−4 C
(b) V 3 = 100 V
V 1 =
q 1
C 1
=
2.667 × 10 −4
8 × 10 −6 = 33.33 V
V 2 =
q 2
C 2
=
2.667 × 10 −4
4 × 10 −6 = 66.66 V
(c) U 1 =
1
2
C 1 V
2
1 =
1
2
× 8 × 10
−6
× (33.33)
2
= 0.00444 J
U 2 =
1
2
C 2 V
2
2 =
1
2
× 4 × 10
−6
× (66.66)
2
= 0.00889 J
U 3 =
1
2
C 3 V
2
3 =
1
2
× 3 × 10
−6
× 100
2
= 0.015 J
Note that U 1 + U 2 + U 3 = U =
1
2
C V 2 , as it should.
11.92 Let the effective capacitance between points a and b be C. Apply a potential
difference V between a and b and let C be charged to q, Fig. 11.47.
Let the charge across C 1 and C 5 be q 1 and q 5 , respectively; the charges across
various capacitors are shown in Fig. 11.47.
Fig. 11.47