Chapter 24 Capacitance and Dielectrict SV | Bài giảng vật lí đại cương | Đại học Bách Khoa Hà Nội

4/1/2020 Capacitor
A parallel-plate capacitor
Chapter 24: CAPACITANCE AND DIELECTRICS
Exercises: 3, 11, 13, 15, 17, 23, 25, 29, 35, 37, 39, 41, 45, 49,
The electric field between the plates of a parallel-plate capacitor
Problems: 51(51), 57, 59(59), 61(61), 63(63), 65(65), 67(67), 69, 71(71)
Basic configuration of a capacitor.
In the limit where the plates are infinitely large, the system has planar
symmetry and we can calculate the electric field everywhere using
Capacitance is a measure of the capacity of storing electric Gauss’s law given Dang Duc Vuong
charge for a given potential difference V . The SI unit of    capacitance is the farad (F) E    2 2 
Email: vuong.dangduc@hust.edu.vn Q Q 0 0 0 C     V  V V 
 V  V   Eds  Ed  U  V   Ed ab   Q Q Q A  A C      0 V  Ed   d d d   0 0 One Love. One Future. 4/1/2020
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Dang Duc Vuong - SEP - HUST 2 1 2 Cylindrical Capacitor Spherical Capacitor
The electric field is non-vanishing only in the region a< rDue to the cylindrical symmetry of the system, we choose our Gaussian surface to
Using Gauss’s law, we obtain
be a coaxial cylinder with length and radius r where a< r < b . Using Gauss’s law, we have: (a) A cylindrical capacitor.
(a) spherical capacitor with two concentric
spherical shells of radii a and b
An “isolated” conductor
(with the second conductor placed at infinity) also has a capacitance. In the limit where b  ∞)
(b) End view of the capacitor.
The electric field is non-vanishing only in the region a < r < b.
(b) Gaussian surface for calculating the electric
a single isolated spherical conductor of radius R, the capacitance is field. One Love. One Future. 4/1/2020
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In a dielectric, 0 replaced by  = K0 K: the dielectric constant of the material
Capacitors in Electric Circuits
24.51(51). A parallel-plate air capacitor is made by using two plates 16 cm square, spaced 9.4 mm apart. It is connected to Parallel
a l2-V battery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between C  C  C  ...  C
the plates? (d) What is the energy stored in the capacitor? eq 1 2 n U  U  ...  U 1 2 n A = 16 cm x 16cm = (0.16)2 m2 Series d = 9.4 mm=9.4 x 10-3 m 1 1 1 1 V = 12 (V)    ...  C C C C 0 = 8.86 x 10-12 C2/N.m2 eq 1 2 n 1  2 2 Q  Q  ...  Q  A 8.86 10  (0.16) 11 0  1 2 n
(a) What is the capacitance? C    2.4 10 F 3 d 9.410 Storing Energy in a Capacitor
(b) What is the charge on each plate?  11  10 Q CV 2.4 10 12 2.9 10       C Electrical potential energy V 12
(c) What is the electric field between the plates? 3 E    1.310 V / m
Energy Density of the Electric Field 3 d 9.4 10 2 1 1 1 Q
(d) What is the energy stored in the capacitor? 2 9 U  CV  QV   1.7 10 J E   A A 1 2 A 2 2 2 C A1 A2 A 1 1 1 1 C  C  C    1 2 C C C C 1 2 3 One Love. One Future. 4/1/2020
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24.57. A 4.00 F and a 6.00  F capacitors are connected in parallel across a 660-Vsupply line. (a) Find the charge on each
24.59(59). In Fig. below C1 = C5 =8.4 pF and C2 = C3 = C4 = 4.2 pF. The applied potential is
capacitor and the voltage across each. (b) The charged capacitors are disconnected from the line and from each other, and
Vab = 220 V. (a) What is the equivalent capacitance of the network between points a and b?
then reconnected with terminals of unlike sign together. Find the final charge on each and the voltage across each.
(b) Calculate the charge on each capacitor and the potential difference across each capacitor + C C 1 C2 1 + + C2 C1 + - C2 C1 + + C2 - - - + - - -
Q  Q  Q  Q  Q 1 2 2 1 V  V  V  660 V 3 3  1 2  
 3.96  2.64 10 C  1.3210 C Q C V  6 4 10  3 660 2.64 10      C 1 1 1
C V  C V  Q  V  132 V 1 2   Q C V  6 6 10  3 660 3.96 10      C 2 2 2 4 Q  C V  5.38 10 C 1 1   4 Q  C V  7.92 10 C 1 2   One Love. One Future. 4/1/2020
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24.61(61). Three capacitors having capacitances of 8.4, 8.4, and 4.2 F are connected in series across a 36-V potential
24.63(63). In Fig. below each capacitance C is 6.9 F, and each capacitance C is 4.6 F. 1 2
difference. (a) What is the charge on the 4.2- F capacitor? (b) What is the total energy stored in all three capacitors?(c)
(a) Compute the equivalent capacitance of the network between points a and b.
The capacitors are disconnected from the potential difference without allowing them to discharge. They are then
(b) Compute the charge on each of the three capacitors nearest a and b when V = 420V. ab
reconnected in parallel with each other. with the positively charged plates connected together. What is the voltage across
(c) With 420V across a and b, compute Vcd.
each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?
(a) What is the charge on the 4.2- F capacitor b) What is the total energy
c) The capacitors are disconnected and then reconnected in parallel. Q 228 C  1 V    10.8V  11V 3 U  QV  1.310 J C 8.4  8.4  4.2 F  2 One Love. One Future. 4/1/2020
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24.65(65). A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is
24.67(67). Capacitance of the Earth. (a) Discuss how the concept of capacitance can also be applied to a single conductor.
then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads 45.0 V when placed
(Hint: In the relationship C = Q/Vab, think of the second conductor as being located at infinity.) (b) Use Eq. (24.1) to show that
across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 V.
C = 4π0R for a solid conducting sphere of radius R. (c) Use your result in part (b) to calculate the capacitance of the earth,
What is the dielectric constant of this material? (b) What will the voltmeter read if the dielectric is now pulled partway out so
which is a good conductor of radius 6380 km. Compare to typical capacitors used in electronic circuits that have capacitances
it fills only one-third of the space between the plates? ranging from 10 pF to 100 μF. (a) Q is constant
Let’s consider a spherical capacitor which consists of two concentric spherical shells of radii a and b. An “isolated” conductor C C
(with the second conductor placed at infinity) also has a capacitance. One can think of “infinity” as a giant conductor with 1 C2 C0 + + C1=KC0
V= 0. In the limit where b∞ - - Q Q 2A A V   45.0V V   11.5V  K 0 2 0 1 0 1 C C 3 C   C 3 C   KC 0 1 1 0 d 3 2 0 d 3 2 K 2  K C V 45.0 C  C  C  C  C  C 0 0 V C  C V  K     3.91 1 2 0 0 0 1 1 0 0 3 3 3 C V 11.5 1 1 V C 3 0 Q  V C  CV   
Solid conducting sphere of radius R: C = 4π 0 0 V 2  K 2  K 0R 0 C0 3 The Earth, R = 6380 km 12  3
 C  4 R  4 3.14  8.86 10  6380 10  710 F  3 3 0 V  V  45.0  22.8 V 0   2  K 3.91 2 One Love. One Future. 4/1/2020
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24.69. A uniformly charged sphere of radius R has total charge Q. Calculate the electric-field energy density at a point a
24.71(71). A parallel-plate capacitor has the space between the plates filled with two slabs of
distance r from the center of the sphere for (a) r < R (b) r > R (c) Calculate the total electric-field energy.
dielectric, one with constant K1, and one with constant K2 (Fig.). Each slab has thickness d/2,
where d is the plate separation. Show that the capacitance is (a) r < R   2    E.dA  E.4 r   R    r ⃗ 3   q  Qr Qr 2 2
Let the potential of the positive plate be V O 1 Q r
a, the potential of the negative plate be Vc, and the potential midway between 2 enclose 1 Q   4 E.4 r    E  2  u   E  inside 3 3 3  r 
the plates where the dielectrics meet be V    R 4 R E 0 2 6 2 32  R b 0 0     4   0 3  3  0 0 R        3   C1 C2 (b) r > R  1 1 1 C C   1 2    C  C C C C  C  1 2 1 2 2
   E.dA  E.4 r    Q  Q  Q R    1 2   Q Q 2 1 Q 2 2   q E.4 r    E   u   E  O enclose  2 Q E 0 2 4 2 32  r inside  4 r   0 0    0    Other method:  0 0  
(c) Calculate the total electric-field energy 2 2 2 2 4 2 2 R  R  R Q r Q Q  r dr         dr   Q  1 1  3Q
U   u dV   u dV     2 4 r  dr     2 4 r  dr           E1 E 2      2 6 2 4  6 2 0 R 0 R 0 R 32  R 32  r 8       R   r   8  5R R  20 R 0 0 0 0 0 One Love. One Future. 4/1/2020
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