Chapter 21: Electric Charges and Forces

Transcription

1 Chapter 21: Electric Charges and Forces

2 Electric Force The electric force is one of the fundamental forces of nature. Examples: Running a comb through hair Rubbing rubber/plastic/glass rods with fur and silk Walking on carpet with slippers and touching door handle Unrolling plastic rap Thunderstorms

3 Experimenting with Charges: Rub both plastic rods with wool. Now the two rods repel each other. Rubbing two glass rods with silk produces the same result

4 Experimenting with Charges: Bring a glass rod that has been rubbed with silk close to a hanging plastic rod that has been rubbed with wool. These two rods attract each other. The strength of the force is greater for rods that have been rubbed more vigorously.

5 Experimenting with Charges: Interpretation: Rubbing a rod somehow changes its properties so that forces now act between two such rods. We say that the rubbed rod is charged. Apparently there are two different kinds of charge that a material can acquire: Positive and Negative charge (arbitrary names). Like charges exert repulsive forces on each other, while opposite charges exert attractive forces on each other.

6 Experimenting with Charges: If the two rods are held farther from each other, the force between them decreases. Interpretation: Like the gravitational force, the electric force decreases with the distance between the charged objects. The greater the charge on the two objects, the greater the force between them.

7 Clicker Question From the picture, what can you conclude about the three charges? 1) Purple and yellow have opposite charges 2) Purple and yellow are both positive. 3) Purple, yellow, and green are all positive. 4) Purple, yellow, and green are either all positive or all negative, but there is no way to determine which.

8 Insulators and Conductors Conductors are those materials through or along which charge easily moves. Insulators are materials in which charge is immobile. Glass and plastics are insulators, metals are conductors.

9 Charge and Atoms Atoms are made of negatively-charged electrons in orbit about positively-charged nuclei (comprised of protons and neutrons). Removing one electron makes a singly charged ion, q ion = +e.

10 Atoms A nucleus contains Z protons plus N neutrons. A neutral atom has Z electrons. Outer-most electrons are called the valence electrons

11 Insulators and Conductors Insulators are materials where charge isn t free to move. Examples include wood, plastic, & glass. The valence electrons in an insulator are tightly bound to the positive nuclei and are not free to move around. Charging an insulator may leave a patch of molecular ions on the surface, but the patches are immobile.

12 Insulators and Conductors In metals, the outer valence electrons are weakly bound to the nuclei. They are easily detached from their parent nuclei and are free to wander through the entire solid, creating a sea of electrons permeating an array of positively charged ion cores. All metals are conductors.

13 Clicker Question An electroscope consists of a neutral conducting sphere connected to metal leaves via a metal rod. Why do the leaves of the electroscope repel each other when the positively-charged rod is brought near the sphere? 1. The leaves become negatively charged 2. The leaves become positively charged 3. The leaves are neutral, but are attracted to the charged rod

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15 Clicker question A negatively charged glass rod is brought near a spherical conductor, which is initially neutral. The spherical conductor is initially touching another spherical conductor which is also initially neutral. While the glass rod is near the conductor, the two conductors are separated (and kept separated). The glass rod is then removed. What are the charged states of the two conductors? 1. Left is positively charged, right is negatively charged. 2. Left is negatively charged, right is positively charged. 3. Left is positively charged, right is neutral.

16 Clicker Question

17 Coulomb s Law The magnitude of the electric force between two point-like particles depends on: The product of the two charges The inverse square of the distance between them Coulomb s law: ~F 12 = kq 1q 2 r 2 ˆr r Here F~F is the force q exerts 1 on q and rˆ is a unit vector 2 pointing from q toward q. 1 2 k is the Coulomb constant, 9 approximately N m / C 2 2.

18 Clicker question

19 Example: 19 Determine the charge of the spheres in terms of mass m, distance d, g, and θ.

20 The superposition principle The electric force obeys the superposition principle. That means the force two charges exert on a third force is just the vector sum of the forces from the two charges, each treated without regard to the other charge. The superposition principle makes it mathematically straightforward to calculate the electric forces exerted by distributions of electric charge. The net electric force is the sum of the individual forces.

21 Clicker Question

22 CT Consider the charge configuration shown below. What is the direction of the net force on the +q charge? B +q h s/2 s/2 +Q -Q A E D C University of Colorado, Boulder (2008)

23 Example: Two positive charges of charge q are a distance 2a apart. A third charge of charge Q is a distance y above the midpoint of the other two charges. What is the net force on Q? y +Q q x= a y q x=a x

24 F y =2kqQ/r 2 (y/r) 2kqQy F = (a 2 + y 2 ) 3/2 F y/a

25 The Electric Field How exactly does a charge exert a force on another charge? Long range force Action at a distance q q 0 Michael Faraday s Idea: F Source charges alter the space around them by creating an electric field (at all points in space) The field is the agent that exerts an electric force on another charge q 0 ~F = q 0 ~ E

26 The Electric Field, E The electric field due a charge distribution is a vector field There is a vector at every location in space. We can use use the force exerted on a test charge q 0 to determine the electric field phet simulation ~E(x, y, z) = ~ F q 0 ~E(x, y, z) q 0

27 Fields of point charges and charge distributions Consider a positive single point-charge q: ~F = kq r 2 ˆr q 0 q q 0 F ~E = kq r 2 ˆr

28 Superposition The superposition principle states that, for a collection of charges, the total electric field at a point is simply the summation of the individual fields due to each charge ~E = X i ~E i = X i kq i r 2 i ˆr i

29 Clicker Question

30 CT 26.7 What is the E-field at point P? A) E = 2kQ / s 2 B) E = sqrt(2) kq / s 2 C) E = kq / ( sqrt(2) s) 2 D) zero +2Q s -Q s P s E) none of the above. -Q s +Q University of Colorado, Boulder (2008)

31 The Dipole Electric Field at Two Points Slide 26-32

32 QuickCheck Which electric field is responsible for the proton s trajectory? A. B. C. D. E. Slide 26-84

33 JITT Question An electron and a proton have equal but opposite electrical charges, but the proton is about 2000 times more massive than the electron. If the two were separated by about 1 mm out in space and released, 1. the magnitude of the electric force on the electron would be less than that of the proton, but the electron would accelerate at a higher rate than the proton. 2. the magnitude of the electric force on the electron would be greater than that of the proton, and the electron would accelerate at a higher rate than the proton. 3. the magnitude of the electric force on the electron would be equal to that of the proton, but the electron would accelerate at a higher rate than the proton. 4. the electron and the proton would accelerate towards each other with equal rates.

34 Continuous Charge Distribution Macroscopic objects typically contain so many excess protons or electrons (say, over ), it is a very good approximation to describe the distribution to be continuous. Instead of summing over every single point charge, we integrate over the infinitesimal spatial chunks of charge. E field at P due to Δq ~E = k q r 2 ˆr ~E = X ~ E In limit that volume becomes infinitesimally small: ~E = Z ~ de = Z k ˆr r 2 dq

35 ~E = Z ~ de = Z k ˆr r 2 dq What is dq in terms of charge density? 1D 2D 3D dq=λdx dq=σda dx dq = dl dq = da dq = dv ~E = Z k dl ˆr r 2 ~ E = Z k da ˆr r 2 ~ E = Z k dv ˆr r 2

36 Example: Line Charge Determine the electric field a distance a to the right of the right end of a uniform line charge of length L and charge Q. Q L Since charge distribution is uniform: a ~E = = Q L Z k ˆr dq r 2 = Z k dl ˆr r 2 Pick a coordinate system: say the x-axis along the rod with origin at left end of rod. E x = E x = Z L 0 k (a + L x) 2 dx = k a + L k L a(l + a) = kq a(l + a) x L 0 = k a k L + a

37 a + L/2 Clicker Question If the line charge collapsed to a point charge, situated where the middle of the line charge was, the electric field (at the dot) would 1. increase 2. decrease 3. remain the same Q Q L a E =? E =?

38 QuickCheck Clicker Question 26.8 At the dot (a distance y from origin), the y-component of the electric field due to the shaded region (of length dx) is de y =? A. B. C. D. E. k(q/l) dx x 2 + y 2 k(q/l) dx x p x2 + y 2 y k(q/l) dx x 2 + y 2 k(q/l) dx x 2 + y 2 k(q/l) dx p x2 + y 2 x y x p x2 + y 2 y p x2 + y 2 y p x2 + y 2 Slide 26-52

39 QuickCheck Clicker Question 26.8 At the dot (a distance y from origin), the y-component of the electric field due to the shaded region (of length dx) is de y =? A. B. C. D. E. k(q/l) dx x 2 + y 2 k(q/l) dx x p x2 + y 2 y k(q/l) dx x 2 + y 2 k(q/l) dx x 2 + y 2 k(q/l) dx p x2 + y 2 x y x p x2 + y 2 y p x2 + y 2 y p x2 + y 2 Slide 26-52

40 E y = Z L/2 L/2 k(q/l)y dx (Q/L)yx =2k (x 2 + y 2 ) 3/2 y 2p x 2 + y 2 L/2 0 E y =2k 1 y p 1+4y 2 /L 2 For an infinitely-long rod: E y = 2k y

41 Figure The electric field at point P is 1. kq/a 2 2. less than kq/a 2 but greater than zero 3. zero 4. Depends on the value of the charge at point P

42 The electric field on the axis of a charged ring: ~E = Z ~ de = Z k ˆr r 2 dq E x = Z kdq r 2 cos = Z 2 0 k ad x 2 + a 2 cos cos = x/r = x/ p x 2 + a 2 E x = kqx (x 2 + a 2 ) 3/2

43 A Ring of Charge The electric field on the x-axis points away from the center of the ring, increasing in strength until reaching a maximum when z R, then decreasing: x E x = kq (x 2 + R 2 ) 3/2 Slide 26-55

44 Electric field due to disk of radius R and charge Q (a distance x from center): E = E = Z de Ring = Z R 0 E =2 kx E =2 k 1 Z kdqring x (x 2 + r 2 ) 3/2 k 2 r dr x (x 2 + r 2 ) 3/2 1 (x 2 + r 2 ) 1/2 R 0 x p x2 + R 2

45 Field lines Field lines show the direction of the electric field. The density of field lines is proportional the the strength of the field. Examples: This figure represents the electric field of a dipole as a field-vector diagram. This figure represents the electric field of a dipole using electric field lines.

46 QuickCheck A set of electric field lines is directed as shown. At which of the noted points is the magnitude of the field the greatest?

47 QuickCheck Which of the following is the correct representation of the electric field created by two positive charges? A. B. C. D.

48 phet: charges

49 The dipole: an important charge distribution An electric dipole consists of two point charges of equal magnitude but opposite signs, held a short distance apart. The dipole is electrically neutral, but the separation of its charges results in an electric field. Many charge distributions, especially molecules, behave like electric dipoles. The product of the charge and separation is the dipole moment: p = qs. Far from the dipole, its electric field falls off as the inverse cube of the distance. (E / 1/r 3 )

50 Dipoles Dipoles in in a Uniform Uniform Electric Electric Field Field The figure shows an electric dipole placed in a uniform external electric field. The net force on the dipole is zero. The electric field exerts a torque on the dipole which causes it to rotate. Slide 26-86

51 The Torque on a Dipole ~ = ~r ~ F The torque on a dipole placed in a uniform external electric field is =2 1 2 s sin (qe) =pe sin Slide 26-93

52 Clicker Question 1. (a) only 2. (c) only 3. (a) and (b) 4. (c) and (d) 5. (a), (b), (c) and (d)

53 How can a charged object attract a neutral, macroscopic object?

54 Dipoles in a Non-uniform Electric Field Field Suppose that a dipole is placed in a non-uniform electric field, such as the field of a positive point charge. The first response of the dipole is to rotate until it is aligned with the field. Once the dipole is aligned, the leftward attractive force on its negative end is slightly stronger than the rightward repulsive force on its positive end. This causes a net attractive force. Slide 26-99

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