PH 222-3A Spring 2007

Transcription

1 PH -3A Spring 7 ELECTRIC FIELDS Lectures,3 Chapter (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition) 1

2 Chapter Electric Fields In this chapter we will introduce the concept of an electric field. As long as charges are stationary, Coulomb s law describes adequately the forces among charges. If the charges are not stationary we must use an alternative approach by introducing the electric field (symbol E ). In connection with the electric field, the following topics will be covered: -Calculating the electric field generated by a point charge. -Using the principle of superposition to determine the electric field created by a collection of point charges as well as continuous charge distributions. -Once the electric field at a point P is known, calculating the electric force on any charge placed at P. -Defining the notion of an electric dipole. Determining the net force, the net torque, exerted on an electric dipole by a uniform electric field, as well as the dipole potential energy.

3 In Chapter 1 we discussed Coulomb s law, which gives the force between two point charges. The law is written in such a way as to imply that q acts on q 1 at a distance r instantaneously ( action at a distance ): 1 q 1 q q11 F = 4πε r Electric interactions propagate in empty space with a large but finite speed (c = m/s). In order to take into account correctly the finite speed at which these interactions propagate, we have to abandon the action at a distance point of view and still be able to explain how q 1 knows about the presence of q. The solution is to introduce the new concept of an electric field vector as follows: Point charge q 1 does not exert a force directly on q. Instead, q 1 creates in its vicinity an electric field that exerts a force on q. charge generates electric field E E exerts a force F on q q 1 3

4 E = F q Definition of the Electric Field Vector Consider the positively charged rod shown in the figure. For every point P in the vicinity of the rod we define the electric field vector E as follows: 1. We place a positive test charge q at point P.. We measure the electrostatic force F exerted on q by the charged rod. 3. We define the electric field vector E at point P as: F E =. SI Units : N/C q From the definition it follows that E is parallel to F. Note : We assume that the test charge q is small enough so that its presence at point P does not affect the charge distribution on the rod and thus alter the electric field vector E we are trying to determine. 4

5 Electric Field Generated by a Point Charge Consider the positive charge q shown in the figure. At point P a distance r from q q P o r E q we place the test charge q. The force exerted on q by q is equal to: F = 1 4πε qq r E F 1 qq 1 q = = = q 4 πε q r 4 πε r The magnitude of E is a positive number. In terms of direction, E points radially E = 1 4πε q r outward as shown in the figure. If q were a negative charge the magnitude of E would remain the same. The direction of E would point radially inward instead. 5

6 O O Electric Field Generated by a Group of Point Charges. Superposition The net electric electric field E generated by a group of point charges is equal to the vector sum of the electric field vectors generated dby each charge. In the example shown in the figure, E = E1+ E + E3. Here E, E, and E are the electric field vectors generated by q, q, and q, respectively. No te: E, E, and E m ust be added as vectors: 1 3 Ex = E1x + Ex + E3x, Ey = E1y + Ey + E3y, Ez = E1z + Ez + E3z 6

7 Problem (Electric field). To measure the magnitude of the horizontal electric field, an experimenter attaches a small charged cork ball to a string and suspends this device in the electric field. The electric force pushes the cork ball to one side, and the ball attains equilibrium when the string makes an angle of 35 with the vertical. The mass of the ball is 3x1-5 kg, and the charge on the ball is 4x1-7 C. What is the magnitude of the electric field? X 5.1x1 N/C 7

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9 v = v o + ax (if x o = at t o =) m/s 9

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11 Electric Dipole A system of two equal charges of opposite sign ( ±q ) placed at a distance d apart is known as an"electric + q dipole." For every electric dipole we associate d a vector known as "the electric dipole moment" (symbol p ) defined as follows: - q The magnitude p= qd +q/ +q/ The direction i of p is along the line that connects the two charges and points from - q to + q. Many molecules have a built-in electric dipole moment. An example is the water molecule (H O). -q The bonding between the O atom and the two H atoms involves the sharing of 1 valence electrons (8 from O and 1 from each H atom). The 1 valence electrons have the tendency to remain closer to the O atom. Thus the O side is more negative than the H side of the H O molecule. 11

12 ( x ) 1+ 1 x Electric Field Generated by an Electric Dipole We will determine the electric field E generated by the electric dipole shown in the figure using the principle of superposition. The positive charge generates at P an electric field whose magnitude E 1 4πε ( + ) r+. The negative charge 1 q creates an electric field with magnitude E( ) =. 4πε r The net electric field at P is E = E E. = q ( + ) ( ) 1 q q E = 4πε r 1 q q = + r 4 πε ( z d /) ( z+ d /) q d d d E = 1 1 We assume: 1 + 4πε z z z z q 1 d 1 d 1 = qd p E = 3 3 4πε z + z z = πε z πε z 1

13 dv P r ˆr dq Electric Field Generated by a Continuous Charge Distribution de Consider the continuous charge distribution shown in the figure. We assume that we know the volume density ρ of dq 3 the electric charge. This is defined as ρ = ( Units: C/m ). dv Our goal is to determine the electric field de generated by the distribution at a given point P. This type of problem can be solved using the principle of superposition as described below. 1. Divide the charge distribution into "elements" of volume dv. Each element has charge dq = ρdv. We assume that point P is at a distance r from dq.. Determine the electric field de generatedb y dq at point P. dq The magnitude de of de is given by the equation de =. 4πε r 1 ρdvrˆ 3. Sum all the contributions: E =. 4πε r 13

14 C A Example : Determine the electric field E generated at point P by a uniformly charged ring of radius R and total charge q. Point P lies on the normal to the ring plane that passes through the ring center C, at a distance z. Consider the charge element of length ds and charge dq shown in the figure. The distance P r = z + R between the element and point is. The charge dq generates at P an electric field of magnitude de that points outward along the line AP: dq de =. The z-component of de is given by 4πε r de = de cos θ. From triangle PAC we have: cos θ = z / r z zdq zdq de =. E de 3 4πε r = = 4πε z + R z z z dq ( ) 3/ E z z zq = dq = 4πε z + R 4πε z + R ( ) ( ) 3/ 3/ 14

15 Example : Determine the electric field E generated at point P by a uniformly positively charged disk of radius R and charge density σ. Point P lies on the normal to the ring plane that passes through the disk center C, at a distance z. Our plan is to divide the disk into concentric flat rings and then to calculate the electric field at point P by integrating the contributions of all the rings. For a ring with radius r and radial width dr, dq = σda = σ( πrdr). We have already solved the problem of the electric field due to a ring of charge: dqz zσ πrdr σ rdr de = = = πε ( z + r ) 4 πε ( z + r ) εo ( z + r ) o o We can now find E by integrating gthis eq. over the surface of the disk, that is, by integrating with respect to r, from r = to r = R, while z = const. 3 15

16 σ z R 3 E = de = ( z + r ) ( r) dr 4ε o m To solve this integral, we cast titi in the form X dx by setting 3 X = ( z + r ), m =, and dx = ( r) dr m 1 1 z m z X z ( z r ) E σ X dx σ + σ + = 4ε = = ε m + ε o o o σ z E = 1 ε o z + R If we let R (infinite sheet) while keeping z finite, the seconf term in the parentheses approaches zero and E = σ ε o (infinite sheet) This is the electic field produced by an infinite sheet of uniform charge located on one side of a nonconductor such as plastic. R 16

17 Electric Field Lines. In the 19th century Michael Faraday introduced the concept of electric field lines, which help visualize the electric field vector E without using mathematics. For the relation between the electric field lines and E : 1. At any point Pthe electric field vector E is tangent to the electric field lines. P E P electric field line. The magnitude of the electric field vector E is proportional to the density of the electric field lines. E P > E Q P E P Q E Q electric field lines 17

18 Two graphical methods of graphical representation of electric field: a) electric field vectors; b) electric field lines (directed radially outward from the positive charge +q) 18

19 Electric field lines extend away from positive charges (where they originate) and toward negative charges (where they terminate). Example 1: Electric field lines of a negative point charge - q : E = 1 q 4πε r q -The electric field lines point toward the point charge. -The direction of the lines gives the direction of E. -The density of the lines/unit area increases as the distance from q decreases. Note : In the case of a positive point charge the electric field lines have the same form but they point outward. q 19

20 The electric field can be represented graphically by field lines

21 Major properties of the field lines 1

22 Example : Electric field lines of an electric field generated by an infinitely large plane uniformly charged. In the next chapter we will see that the electric field generated by such a plane has the form shown in fig. b. 1. The electric field on either side of the plane has a constant magnitude.. The electric field vector is perpendicular p to the charge plane. 3. The electric field vector E points away from the plane. The corresponding electric field lines are given in fig. c. Note : For a negatively charged plane the electric field lines point inward.

23 The concept of the field lines for the computation of the electric field of a large, flat, charged sheet 3

24 The Electric Field of the Two Charged Sheets 4

25 E D =(1.1x1 5 N/C)[-1-1+1] = -1.1x1 5 N/C 5

26 Example 3 : Electric field lines generated by an electric dipole (a positive and a negative point charge of the same size but of opposite sign) Example 4 : Electric field lines generated by two equal positive point charges 6

27 F - x-axisi F + Forces and Torques Exerted on Electric Dipoles by a Uniform Electric Field Consider the electric dipole shown in the figure in the presence of a uniform (constant magnitude and direction) electric field E along the x-axis. The electric field exerts a force F = qe on the positive charge and a force F = qe on the negative charge. The net force on the dipole is F = qe qe = net. + The net torque generated by F and F about the dipole center is + d d τ = τ + + τ = F + sinθ F sinθ = qed sinθ = pe sinθ In vector form: τ = p E The electric dipole in a uniform electric field does not move but can rotate about its center. F = τ = p E net 7

28 U U = pecosθ U = p E A B 18 Potential Energy of an Electric Dipole in a Uniform Electric Field The dipole has its least potential energy, when it is in equilibrium orientation ( p is lined up with the E; τ = p E = ). To rotate the dipole to any other orientation requires work by some external agent. We are free to define the zero-poyential-energy configuration in an arbitrary way. The expression for the potential energy of the electric dipole in an external electric field is simplest if we choose it to be zero when angle θ=9. We can find U at any other value of θ by calculating the work W done by the field on the dipole when the dipole is rotated to that value θ from 9. θ θ U = τdθ = pe sinθdθ θ 9 9 θ U = pe sinθdθ = pecosθ = p E At point A ( θ = ), U has a minimum At point B ( θ = 18 ), U has a maximum value Umin = pe. value Umax =+ pe. It is a position of stable equilibrium. It is a position of unstable equilibrium. 9 p E p E 8

29 θ f θ θ i Fig. a p E Work Done by an External Agent to Rotate an Electric Dipole in a Uniform Electric Field Consider the electric dipole in fig. a. It has an electric dipole moment p and is positioned so that p is at an angle θ E p i with respect to a uniform electric field. E An external agent rotates the electric dipole and brings it to its final position shown in fig. b. In this position p is at an angle θ with respect to E. f Fig. b The work W done by the external agent on the dipole is equal to the difference between the initial and final potential energy of the dipole: W = U U = pecosθ pe W ( cosθi cosθ f ) ( cosθ ) f i f i = pe 9