Chapter 21 Electric Charge | Bài giảng vật lí đại cương | Đại học Bách Khoa Hà Nội

10/5/2021
ELECTRIC CHARGE AND ELECTRIC FIELD Electric Charges
Electric charge is a basic property of matter
Two basic charges: Positive and Negative
Each having an absolute value of 1.6 x 10-19 Coulombs
Chapter 21: ELECTRIC CHARGE AND ELECTRIC FIELD Experiments have shown that
Like signed charges repel each other
Exercises 1, 5, 7, 9, 17, 19, 23, 25, 27, 29, 31, 35, 39, 43, 45, 47, 57, 59, 61, 67
Unlike signed charges attract each other
Problems: 69(73), 71(75), 73(77), 75(79), 77(80), 81(85), 83(87), 85, 87(90), 95(97), 101(103)
For an isolated system, the net charge of the system remains constant: Charge Conservation Dang Duc Vuong
Conductors: Materials, such as metals, that allow the free movement of charges
Email: vuong.dangduc@hust.edu.vn
Insulators: Materials, such as rubber and glass, that don’t allow the free movement of charges
Point charge: a charge can be considered a point charge if it’s size is very very small as compare to the distance from
which it is viewed (it should look like a point) One Love. One Future. 10/5/2021
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Dang Duc Vuong - SEP - HUST 2 1 2 Coulomb’s Law Electric Field
Coulomb found that the electric force between two charged objects is
Definition of the electric field:
Proportional to the product of the charges on the objects, and
Inversely proportional to the separation of the objects squared
Here, q0 is a “test charge” – it serves to allow the electric force to be measured, but is not large enough to create a q q
significant force on any other charges. 1 2 F  k
A negative charge sets up an electric field pointing 2
A positive charge sets up an electric field r
pointing away from the charge towards the charge
k being a proportionality constant, having a value of 8.988 x 109 Nm2/c2
If there are multiple point charges, the forces add by superposition. One Love. One Future. 10/5/2021
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Dang Duc Vuong - SEP - HUST 4 3 4 10/5/2021 Electric Dipole
Superposition of electric fields
An electric dipole is a pair of point charges having equal magnitude
If we have a continuous charge distribution the
but opposite sign that are separated by a distance d.
summation becomes an integral  dq The electric dipole moment: E  k  rˆ r2 Torque on a Dipole   
  qd  E Ring of Charge Line of Charge
d is a vector pointing from the
negative charge to the positive charge  
Potential Energy of a Dipole
U  qd  E One Love. One Future. 10/5/2021
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21.69(73). Two positive point charges Q are held fixed on the x-axis at x = a and x = -a. A third positive point charge q, with
(b) Suppose instead that the charge q were placed on the y-axis at a coordinate y such that |y | << a, and then released. If this
mass m, is placed on the x-axis away from the origin at a coordinate x such that |x| << a. The charge q, which is free to move
charge is free to move anywhere in the xy-plane, what will happen to it? Explain your answer.
along the x-axis, is then released. (a) Find the frequency of oscillation of the charge q. (Hint: Review the definition of simple
harmonic motion in Section 13.2. Use the binomial expansion (1 + x)n = 1 + nx + n(n - 1)x2/2 + ..., valid for the case |x| < 1.)
(b) Suppose instead that the charge q were placed on the y-axis at a coordinate y such that |y | << a, and then released. If this
charge is free to move anywhere in the xy-plane, what will happen to it? Explain your answer.
(a) Find the frequency of oscillation of the charge q y    Find the net force on q.
F  F  F  F  F  F 1 2 x 1 2 Q (1) Q (2) q + + 0 x x -a a   F F 2 1
The x-components of the forces exerted by the two charges cancel, the y-components add, and the net force is in
the +y-direction when y > 0 and in the -direction y − when y < 0. The charge moves away from the origin on the
Since x <y-axis and never returns. 2 2 d x qQ d x qQ  m  x  0   x  0 2 3 2 3 dt  a dt m a 0 0 2 d x qQ 2 2 F  ma  m. 2
 x   x  0     f  3 dt m   a 0 One Love. One Future. 10/5/2021
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21.71(75). Two small spheres with mass charge q, m = 15.0 g are hung by silk threads of length L =1.20 m from
21.73(77). Sodium chloride (NaCl, ordinary table salt) is made up of positive sodium ions (Na+) and negative chloride ions (Cl-).
a common point (Fig.). When the spheres are given equal quantities of negative charge, so that q
(a) H a point charge with the same charge and mass as all the Na+ ions in 0.100 mol of NaCl is 2.00 cm from a point charge 1 = q2 = q, each
thread hangs at  = 25.0° from the vertical. (a) Draw a diagram showing the forces on each sphere. Treat the
with the same charge and mass as all the Cl- ions, what is the magnitude of the attractive force between these two point
spheres as point charges. (b) Find the magnitude of q. (c) Both threads are now shortened to length L = 0.600 m,
charges? (b) If the positive point charge in part (a) is held in place and the negative point charge is released from rest, what is while the charges q
its initial acceleration? (c) Does it seem reasonable that the ions in NaCl could be separated in this way? Why or why not? (In
1 and q2 remain unchanged. What new angle will each thread make with the vertical?
(Hint: This part of the problem can be solved numerically by using trial values for  and adjusting the values of
fact, when sodium chloride dissolves in water, it breaks up into Na+ and Cl- ions. However, in this situation there are additional
 until a self-consistent answer is obtained.) (g=9.8 m/s2)
electric forces exerted by the water molecules on the ions.) For Cl, atom mass M = 35.45x10-3 kg/mol.
• Find the number of Na+ and Cl− ions and the total positive and negative charge;
Use Coulomb's law for the force that one sphere exerts on the other and
• Avogadro's number NA= 6.022 × 1023 ions/ mol
apply the 1st condition of equilibrium to one of the spheres. 0.100 mol of NaCl
→ The number of Na+ ions is N=n.NA → q1 = N.e = 9.647× 103 C    F  T sin 
→ The number of Cl- ions is N=n.NA → q2 = -N.e = -9.647× 103 C T  F  W  0  
 F  w tan   mg tan  2 T 1 q q T W  T cos  q  1 2 21 F    2.09 10 N 2 2 2 1 q q q 4 r 4 r 1 2 2 6 0 0 F F    q  4 (2L sin )
 mg tan   2.80 10 C 2 2 0
(b) If the positive point charge in part (a) is held in place and the negative point charge is released from rest, what is its initial F 4 r 4 r 0 0 acceleration? F  ma;  W W
(c) Both threads are now shortened to length L = 0.600 m 23 2
  a  5.90 10 m / s . 3 3
m  n.M  0.100  35.45 10  35.45 10 kg
F is the repulsive Coulomb force 2 6 q  4 (2L sin )
 mg tan   2.80 10 C =39.50 0
(c) Does it seem reasonable that the ions in NaCl could be separated in this way?: Is is not reasonable to have such a huge force
exerted by one sphere on the other.
Reason: A small amount of material contains huge amounts of positive and negative charges. One Love. One Future. 10/5/2021
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21.75(79). Three identical point charges q are placed at each of three corners of a square of side L. Find the magnitude and
21.77(80). Three point charges are placed on the y-axis: a charge q at y = a, a charge -2q at the origin, and a charge q at y = -a.
direction of the net force on a point charge -3q placed (a) at the center of the square and (b) at the vacant corner of the square.
Such an arrangement is called an electric quadrupole. (a) Find the magnitude and direction of the electric field at points on the
In each case, draw a free-body diagram showing the forces exerted on the -3q charge by each of the other three charges.
positive x-axis. (b) Use the binomial expansion to find an approximate expression for the electric field valid for x >> a.
Contrast this behavior to that of the electric field of a point charge and that of the electric field of a dipole. One Love. One Future. 10/5/2021
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21.81(85). Two small, copper spheres each have radius 1.00 mm. (a) How many atoms does each sphere contain? (b) Assume
21.83(87). A proton is projected into a uniform electric field that points vertically upward and has magnitude E. The initial
that each copper atom contains 29 protons and 29 electrons. We know that electrons and protons have charges of exactly the
velocity of the proton has a magnitude v0 and is directed at an angle α below the horizontal. (a) Find the maximum distance
same magnitude, but let’s explore the effect of small differences. If the charge of a proton is +e and the magnitude of the
hmax that the proton descends vertically below its initial elevation. You can ignore gravitational forces. (b) After what
charge of an electron is 0.100% smaller, what is the net charge of each sphere and what force would one sphere exert on the
horizontal distance d does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the
other if they were separated by 1.00 m?
numerical values of hmax and d if E = 500 N/C, v0 = 4.00 x 105 m/s, and α= 30.0°.
Use the density of copper to calculate the number of moles and then the number of atoms.
The electric field is upward so the electric force on the positively charged proton is upward and
Calculate the net charge and then use Coulomb's law to calculate the force. has magnitude F = eE. One Love. One Future. 10/5/2021
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21.87(90). Positive charge Q is distributed uniformly along the positive y-axis between y = 0 and y = a. A negative point a
charge -q lies on the positive x- axis, a distance x from the origin (Fig.). (a) Calculate the x- and y-components of the electric
field produced by the charge distribution Q at points on the positive x- axis. (b) Calculate the x- and y-components of the force
that the charge distribution Q exerts on q. (c) Show that if x >> a, Fx  -qQ/4π0x2 and Fy  +qQa/8π0x3. Explain why this result is obtained.
Slice the charge distribution up into small pieces of length dy a y 0 x
For x >> a the charge distribution Q acts like a point charge. One Love. One Future. 10/5/2021
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21.95(97). Negative charge -Q is distributed uniformly around a quarter-circle of radius a that lies in the first quadrant, with the  dE
center of curvature at the origin. Find the x- and y-components of the net electric field at the origin. dE 1y 1  E dE M=? 2  1 dq
Divide the charge distribution into small segments, use the point charge formula for the electric field due to each small M dE  1 2 4 r
segment and integrate over the charge distribution to find the x and y components of the total field. 0  r h Q A 0 R Q Qd  1 dq dq  (Rd)   dE   dE  dE cos 1 2 2 1y 1 2 R  2 4 (h  R ) dE 0 dEy h dα cos  dE 2 2 x α h  R 2  2 1 Qd h 1 Qh    E   d  2 2 2 2 2 2 3/2  2 2 8  (h  R ) 8  (h  R ) 0 0 h  R 0 0 One Love. One Future. 10/5/2021
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Dang Duc Vuong - SEP - HUST 18 17 18 www.hust.edu.vn L  Q x M a dE dx EM=? Q 1 dq Q dx dq  dx dE   L 2 2 4 (x  a) 4 L (x  a) 0 0 L L Q dx E  dE   .... M L/2   2 4 L (x  a) 0 0 0 dx Q 1 dq Q dx L,Q dq  dx dE   M 2 2 2 2 x L 4 (x  h ) 4 L (x  h ) α 0 0 dE 0 h h Qh dx dE dE  dE.cos   dE.  y 2 2 3/2 2 2 4 L (x  h ) h  x 0 L/2 L/2 L/2 Qh dx -L/2 E  2 dE  dE  M y y    2 2 3/2 4 L (x  h ) 0 L/2 0  L/2
Thank you for your attentions! One Love. One Future. 10/5/2021
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