Chapter 23 Electrict field SV | Bài giảng vật lí đại cương | Đại học Bách Khoa Hà Nội

3/29/2020
Electric potential energy:
The electric force is conservative; therefore, there must be a potential energy
Chapter 23: ELECTRIC POTENTIAL associated with it.
For a conservative force, the work is path-independent
Exercises: 1, 3, 5, 9, 11, 13, 17, 21, 29, 31, 33, 35, 37, 39, 41, 43
The work W done by the electric force on a charged particle moving in an
electric field can be represented by the change in a potential-energy function
Problems: 49(53), 51(55), 55(59), 57(61), 59(63), 61(65), 63(66), 79(79), 81(81), 83(83) U.
U : is the initial electric potential energy Dang Duc Vuong ΔU = U − U = −W
U : is the final electric potential energy
W : work done by electric field
Email: vuong.dangduc@hust.edu.vn 1 Two point charges q and qo = 4 1 q A charge q U = q
o in the presence of a collection of charges q1, q2, q3, 4πε r One Love. One Future. 1 One Love. One Future. 2 1 2 U
23.49(53). A particle with charge +7.60 nC is in a uniform electric field directed to the left. Another force, in Electric potential M V  q r1
addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After M q r M 1 M 0 q r r 3 1 dq
it has moved 8.0 cm, the additional force has done 6.50 x 10-5 J of work and the particle has 4.35 x 10-5 J of a point charge V = q 2 dV  2 q3 ri 4 r
kinetic energy. (a) What work was done by the electric force? (b) What is the potential of the starting point with 0 q r M
respect to the end point? (c) What is the magnitude of the electric field? i dq U 1 q
IDENTIFY: Apply the work-energy theorem V = = K a collection of point charges
1: Kinetic energy at initial point q 4πε r or energy consevation
U1: Potential energy at initial point 2
K U W  K U W 1 dq 1 1 other 2 2 other: Other work a charge distribution V = dV = 4πε r V V V 1
K2: Kinetic energy at final point E = − ı⃗ + ȷ⃗ + k E x y z
U2: Potential energy at final point E K Conservative force Energy Conservation ℓ 1 2 1= 0 V − V = Edℓ 8.0 cm U1 = V1.q q = +7.60 nC 
K U W  K  U W  other= 6.5 x 10-5J 1 1 other 2 2
* The surface of a conductor is always an equipotential surface and all FE F K2 = 4.35 x 10-5 J
points in the interior of a conductor are at the same potential. U2=V2.q One Love. One Future. 3 One Love. One Future. 3 4 1 3/29/2020 K1= 0
23.51(55). A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. U
Because of the accumulation of charge near the cathode, the electric potential between the electrodes is not a linear function 1 = V1.q
K U  W  K  U W 5  5 
of the position, even with planar geometry, but is given by V(x) = Cx4/3, where x is the distance from the cathode and C is a other= 6.5 x 10-5J 1 1 other 2 2 0  qV  6.510  4.35 10  qV 1 2
constant, characteristic of a particular diode and operating conditions. Assume that the distance between the cathode and K2 = 4.35 x 10-5 J
anode is 13.0 mm and the potential difference between electrodes is 240 V. U2=V2.q (a) Determine the value of C. 5 2.15 10 5 3
q(V V )  2.1510  V V   2.8310 (V )
(b) Obtain a formula for the electric field between the electrodes as a function of x. 1 2 1 2 9 7.6 10
(c) Determine the force on an electron when the electron is halfway between the electrodes.
The starting point (point 1) is at 2.83× 103 V lower potential than the ending point (point 2). We know that V2 >V1
because the electric field always points from high potential toward low potential. (a) Determine the value of C V(x) = Cx4/3 4 4/3
Work done by the electric force: change potential energy (1)  (2): W C  7.85 10 V / m 
Fe = U1-U2 = q (V1-V2) = -2.15 x 10-5 (J) + Anode V = 240 V when x = 13.0 mm
(similar work done by gravity and a change in gravitational potential energy) V  4  V  1/3 5 1/3 (b) E(x) = ? E(x)  
  Cx  1.05 10 .x   x  3  m 
(c) What is the magnitude of the electric field? -Cathode  
(c) Determine the force when x=13.0/2 = 6.5 mm 3 2 2 V  V 2.  8310  v V − V = Edℓ  V  V  4 1 2 x
 Ed  E d  E  0.08 E    3.54 10     1 2 1 1  4  1 1 8 0.08  m  F  qE  q.  Cx     19 1.602  10  5 3  15 3 3
 (1.05 10 ).(6.510 )  3.1310 (N)  3  One Love. One Future. 5 One Love. One Future. 6 5 6
23.55(59). The H2+ Ion. The H2+ ion is composed of two protons, each of charge +e = 1.60 x 10-19 C, and an electron of
(b) Suppose the electron in part (a) has a velocity of magnitude 1.50 x 106 m/s in a direction along the perpendicular bisector
charge -e and mass 9.11 x 10-31 kg. The separation between the protons is 1.07 x 10-10 m. The protons and the electron may
of the line connecting the two protons. How far from the point midway between the two protons can the electron move?
be treated as point charges. (a) Suppose the electron is located at the point midway between the two protons. What is the
potential energy of the interaction between the electron and the two protons? (Do not include the potential energy due to
Electron have va = 1.50 x 106 m/s; move from point a to point b
the interaction between the two protons.) (b) Suppose the electron in part (a) has a velocity of magnitude 1.50 x 106 m/s in 2 2 r  r  d b a
a direction along the perpendicular bisector of the line connecting the two protons. How far from the point midway 10 1.07 10 10 r   0.535 10 m
between the two protons can the electron move? Because the masses of the protons are much greater than the electron a 2 1 2 18 
mass, the motions of the protons are very slow and can be ignored. (Note: A realistic description of the electron motion K  mv 1.02510 (J) a a Proton 1 Proton 2 2 2 2 e e
requires the use of quantum mechanics, not Newtonian mechanics.) Energy conservation: 18 U    8.06 10 (J) a 2 r 2 r 0 0 a
K U W  K  U W a a other b b other= 0
The positions of the particles
(a) The potential energy of interaction of the electron with protons Kb = 0 2 2 e e 1  8 18  18  2 2 (e).e (e).e 2e e
U  K  U  K 1.025 10  8.6010  7.  575 10 (J) U   b a a b b 1  8 U  U  U      8.60  10 (J) 2 r 2 r 2 1 2 0 0 b 4 r 4 r 4 r 2 r e 11  0 0 0 0 U   r  6.07510 (m) b b 2 r 0 b 2 2 2 2 10
r  r  d  d  r  r  2.88 10 (m) b a b a One Love. One Future. 7 One Love. One Future. 8 7 8 2 3/29/2020
23.57(61). Coaxial Cylinders. A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal
(a) Calculate the potential V(r) for (i) r < a; (ii) a < r < b; (iii) r > b
tube with radius b. The positive charge per unit length on the inner cylinder is  and there is an equal negative charge per unit length on (i) r < a E  0  V  V r a the outer cylinder.    b    b 
(ii) a < r < b E   V  V  V  ln  V  ln    
(because V = 0 at r = b) 2 r rb r b r 2  r  2  r
(a) Calculate the potential V(r) for (i) r < a; (ii) a < r < b; (iii) r > b. (Hint: The net potential is the sum of the potentials due to the  0 0 0 (iii) r > b E  0  V  V  0
individual conductors). Take V = 0 at r = b. r b   b    b 
(b) Show that the potential of the inner cylinder with respect to the outer is V  V  V  ln
(b) Show that the potential of the inner cylinder with respect to the outer is V  V  V  ln   ab a b   ab a b 2  a  2  a  0 0   b b b      b E  
V  V  V   Edr   dr   ln(r)  ln V 1 2 r   ab a b 2 r 2 2  a
(c) Show that the electric field at any point between the cylinders has magnitude ab E  a a a 0  0 0 0 l n b / a  r V 1 ab
(d) What is the potential difference between the two cylinders if the outer cylinder has no net charge?
(c) Show that the electric field at any point between the cylinders has magnitude E  lnb / a r   b  Gauss’ law E  0 V  V  V  ln When r < a (inside) ab a b   2  a  V 1 0
a long metal cylinder with radius a  ab E   E  ln b / a  r E  When r > a (outside) 2 r 2 r 0 0
(d) What is the potential difference between the two cylinders if the outer cylinder has no net charge?
When r > a (outside a long metal cylinder) -   b b b      b   b  E    b V  V  V    Edr   dr   ln(r)  ln   a < r < b E  V  V  V  ln 2 r ab a b ab a b   a a a 2 r 2 2  a  2 r 2  a  0 0 0 0 Vb=0 0 0 One Love. One Future. 9 One Love. One Future. 10 9 10
21.59(63).Deflection in a CRT. Cathode-ray tubes (CRTs) are often found in oscilloscopes and computer monitors. In Fig. an
(a) What is the force (magnitude and direction) on the electron -
electron with an initial speed of 6.50 x 106 m/s is projected along the axis midway between the deflection plates of a cathode- when it is between the plates?   
ray tube. The uniform electric field between the plates has a magnitude of 1.10 x 103 V/m and is upward. (a) What is the force   19 q 1.6 10   
C  F & E : opposite direction  E
(magnitude and direction) on the electron when it is between the plates? (b) What is the acceleration of the electron (magnitude F  qE v0 19 3 16 F  q E  1.610
1.10 10  1.76 10 N
and direction) when acted on by the force in part (a)? (c) How far below the axis has the electron moved when it reaches the 0  y x  1
end of the plates? (d) At what angle with the axis is it moving as it leaves the plates? (e) How far below the axis will it strike
b) What is the acceleration of the electron? F v v the fluorescent screen S?  1x y2 y y + v v - 1 1y 
a >> g = 9.80 m/s2, so the gravity force on the electron can be  E   neglected. F is downward, so a is downward Orbit: Parabol Orbit: Straight line v0 0
(c) How far below the axis has the electron moved when it reaches the end of the plates?  x  1 2 x(t)  x  v t  a t 0 ox x F v v 2 1x 1 x(t)  v t + v v 2 y(t)  y  v t  a t 0 1 0 oy y 2 1y 2 at y(t) 
x  0; y  0; v  v ; v  0 2 0 0 0 x 0 oy Orbit: Parabol Orbit: Straight line a  0;a  a x y One Love. One Future. 11 One Love. One Future. 12 11 12 3 3/29/2020
Reach the end of the plates
23.61(65). Electrostatic precipitators use electric forces to remove pollutant particles from
smoke, in particular in the smokestacks of coal-burning power plants. 0.060 - x  v t  0.060m 9  t   9.32110 (s) 
One form of precipitator consists of a vertical, hollow, metal cylinder with a thin wire, insulated 0 1 1 6 6.50 10  E
from the cylinder, running along its axis (Fig.). A large potential difference is established  1 1 t 9.321 10 (s) y at 1.93 10 .9.321 10        2 9 2 14 9  0.822 cm
between the wire and the outer cylinder, with the wire at lower potential. This sets up a strong 1 1 1 v x 2 2 0 0
radial electric field directed inward. The field produces a region of ionized air near the wire.
d) At what angle with the axis is it moving as it leaves the plates?   y1
Smoke enters the precipitator at the bottom, ash and dust in it pick up electrons, and the charged v  v  a t  v  v F v v
pollutants are accelerated toward the outer cylinder wall by the electric field. Suppose the radius 1x 0 x x 1 0 x 0  1x y
of the central wire is 90.0 m, the radius of the cylinder is 14.0 cm, and a potential difference of 14 9 6
v  v  a t  (1.9310 ).(9.32110 )  1.782 10 (m / s) 2 y + v v 1 y 0 y y 1
50.0 kV is established between the wire and the cylinder. 1 1y 6 v 1.782 10
Also assume that the wire and cylinder are both very long in comparison to the cylinder radius. 1 y o tan     0.2742    15.3 6 v 6.50 10
(a) What is the magnitude of the electric field midway between the wire and the cylinder wall? 1x Orbit: Parabol Orbit: Straight line
(b) What magnitude of charge must a 30.0 g ash particle have if the electric field computed in
(e) How far below the axis will it strike the fluorescent screen S?
part (a) is to exert a force ten times the weight of the particle?
Gravity can still be neglected, outside the plates there is no electric field, so a= 0 motion straight line 6
v (t)  v  6.50 10 (m / s) x 1 x 0.120 Time travel outside plates 8 t 1.846 10    (s) 6
v (t)  v  1.782 10 (m / s) 2 6 6.50 10 y 1 y y  v .t   6 1.78210  8 (1.846 10 )  3.29cm
y  y  y  0.822  3.29  4.11 cm 2 1y 2 1 2 One Love. One Future. 13 One Love. One Future. 14 13 14
Wire: radius a = 90.0 m; cylinder: radius b = 14.0 cm
23.63(66). A disk with radius R has uniform surface charge density . (a) By regarding the disk as a series of thin concentric
(a) What is the magnitude of the electric field midway between the wire and the cylinder wall?
rings, calculate the electric potential V at a point on the disk’s axis a distance x from the center of the disk. Assume that the
At point M, a < r < b (between the wire and the cylinder wall)
potential is zero at infinity. (Hint: Use the result of Example 23.11 in Section 23.3.) (b) Calculate -dV/dx. Show that the   b b b      b
result agrees with the expression for E E  
x calculated in Example 21.12 (Section 21.5).
V  V  V   Edr   dr    ln(r)  ln 2 r ab a b   (a) V a a a 2 r 2 2  a ring = ? 0  0 0 0
Consider a thin ring of radius r and width dr. The ring has area dS =2πrdr M   b  1  V  V
so the charge on the ring is dq = dS = 2πrdr a b  V  V  V  ln  E(r)  M ab a b   2  a  r  b 
Assume that the potential is zero at infinity. 0 ln   d  a  dq
The potential due to this thin ring at the point M x 3 a  b 50.0 10  V  4 r   0.07004m  E   9.7110 dq 2 r  dr M M   2  0.140   m dV   0.07004 ln   2 2 r   4 d  4 x  r 6  90.0 10  0 0 dr R R R 2 rdr  
(b) What magnitude of charge must a 30.0 g ash particle have if the electric field computed V   dV     2 2 x  R  x 2 2  0 0 4 x  r 2
in part (a) is to exert a force ten times the weight of the particle? 0 0 (b) Calculate -dV/dx. V            E       x x 1 1 2 2 x  R  x   1   x      2 2 2 2 x  x  2 2    x  R  2  x x  R  0 0 0 One Love. One Future. 15 One Love. One Future. 16 15 16 4 3/29/2020
23.79(79). Electric charge is distributed uniformly along a thin rod of length a, with total
(b) point R, a distance y above the right-hand end of C the rod. Q
charge Q. Take the potential to be zero at infinity. Find the potential at the following 2 2 R  (a  r)  y R
points (a) point P, a distance x to the right of the rod, and (b) point R, a distance v 1 dq Qdr Q y
above the right-hand end of C the rod. (c) In parts (a) and (b). what does your result dV   2 2 dq  dr 4 R 4 a (a  r)  y 0 0 a
reduce to as x or y becomes much larger than a? 2 2 a a a Qdr Q dr Q  a  a  y 
(a) point P, a distance x to the right of the rod Q V   dV      ln   dq  dr 2 2 2 2 0 0 0 4 a (a  r)  y 4 a (a  r)  y 4 a y r dr a - r a P 0 0 0  
An infinitesimal slice of the rod and its distance from point P 1 dq Qdr
y becomes much larger than a (y>>a) dV   4 R 4 a(a  x  r) r dr R=a+x-r 0 0 a a a Qdr Q d(a  x  r) Q a Q  a  x  V   dV        ln(a  x  r)  ln   0 0 0 0 4 a(a  x  r) 4 a (a  x  r) 4 a 4 a  x  0 0 0 0
x becomes much larger than a (x>>a) Binomial expansion a a Q  a  x  Q  a  V  ln  ln 1 and ln 1 + ≃     4 a y y  x  4 a  x  2 Q  a  Q  a a  Q a Q 0 0 V  ln 1    .   Point charge 2 2 2 u         2 4 a Q a a y Q  a  Q a Q  x  4 a  x 2x  4 a x 4 x   ln(1 u)  u  when u  1 V  ln ln 1 .  0 0 0 0      Point charge 2 4 a y 4 a y 4 a y 4 y     0 0 0 0 One Love. One Future. 17 One Love. One Future. 18 17 18
23.81(81). Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each.
23.83(83). A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the
Sphere A has a radius three times that of sphere B. Let Q
sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a
A and QB be the charges on the two spheres and let EA and EB be
the electric-field magnitudes at the surfaces of the two spheres. What are (a) the ratio Q
long, thin conducting wire to another sphere of radius R
B /QA and (b) the ratio EB/EA ?
2 that is several meters from the first sphere. Before the connection
is made. this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on RA= 3RB
each sphere; (c) the electric potential at the surface of each sphere (d) the electric field at the surface of each sphere?
At the surface of a charged conducting sphere
Assume that the amount of charge on the wire is much less than the charge on each sphere. Q RA Electric field E  2 R 4 r B
(a) Electric field and electric potential at the surface of the isolated sphere? 0  Q
The electric potential V   Q Q  Edr  1 1 E  V  R 4 r 1 2 1 0 Q 4 R 4 R  0 1 0 1 A V  A 4 R  
Connect two spheres with a long, thin conducting wire V  V  Q V R 1 1 1 2 R1 R 0 A B B B 2     1.  Q Q V R 3 3 q  1 B A A A V   V  1  q q q  q Q B 4 R  4 R  q q 1 2 1 2 1 0 1    1 2 0 B     R R R  R R  R Q 1 2 1 2 1 2  q R R 2 1 2  A E  V  Q R Q R 2 1 1 1 2 A 2  4 R  q  ; q  2 4 R  E Q R 1 0 2  1 2 0 A 2 B B A R  R R  R     .3  3 1 2 1 2 2 Q E Q R 3 B A A B E   q  q  Q  0  Q 1 Q 1 Q 1 Q 1 1 1 V  V  ; E  ; E  B 2 4 R  1 2 1 1 1 2 4 R  R  1 4  R  R  2 R 4 R  R R 0 1 2 0 1 2 1 0  1 2  0 B  2 One Love. One Future. 19 One Love. One Future. 20 19 20 5