Physics 1302W Introductory Physics for Science and Engineering II Clem Pryke, Associate Professor. Lecture 4 Electric Fields & Electric Dipoles

Transcription

1 Physics 1302W Introductory Physics for Science and Engineering II Clem Pryke, Associate Professor Lecture 4 Electric Fields & Electric Dipoles

2 What to do this week Go to your lab session as scheduled first proper lab this week Ø Do the Warm Up and Prediction questions ahead of time, write the answers in your lab notebook and send a copy to your TA 24 hours ahead of the session. Ø Really do this it is a requirement for being allowed to participate in the lab session! Homework assignment #1 is live in the MasteringPhysics system due on Friday This week s homework will be set on Wednesday due next Wednesday. Go to your discussion session on Thursday.

3 Section 23.5: Electric field of a charged particle Let s derive the equation for the electric field produced by a single charged object from Coulomb s law. The figure shows the geometry of the situation. The electric field at a certain point P in space is the electric force experienced at P by a test particle carrying a charge q t divided by the charge of the test particle: E F t E q t Slide 23-3

4 Section 23.5: Electric field of a charged particle If we place a test particle carrying a charge q t at P, Coulomb s law tells us that the force exerted on the test particle is F E st = k q q s t If we divide the electric force exerted by the source particle on the test particle by the charge q t of the test particle, we obtain an expression for the electric field created by the source charge at P: For more than one source charge we use the superposition principle to obtain the vector sum of the individual electric fields: E = E 1 + E 2 + = k q ˆr i ip 2 r ip! E s =! F st E q t r st 2 ˆr st = k q s r st 2 ˆr st Slide 23-4

5 Section 23.5: Electric field of a charged particle Example 23.3 Electric field due to two charged particles A point P is located at x P = 2.0 m, y P = 3.0 m. What are the magnitude and direction of the electric field at P due to a particle 1 carrying charge q 1 = +10 µc and located at x 1 = 1.0 m, y 1 = 0 and a particle 2 carrying charge q 2 = +20 µc and located at x 2 = 1.0 m, y 2 = 0? Slide 23-5

6 Section 23.5: Electric field of a charged particle Example 23.3 Electric field due to two charged particles (cont.) ❶ GETTING STARTED I begin by making a sketch of the situation (Figure 23.23). Each particle carries a positive charge, so the electric field due to each particle points away from the particle. Slide 23-6

7 Section 23.5: Electric field of a charged particle Example 23.3 Electric field due to two charged particles (cont.) ❷ DEVISE PLAN To determine the electric field E P at P, I must take the vector sum of E 1 and E 2 at P. I can use Eq to calculate the magnitudes E 1 and E 2. To obtain the vector sum of the two fields, I add their x and y components. Slide 23-7

8 Section 23.5: Electric field of a charged particle Example 23.3 Electric field due to two charged particles (cont.) ❸ EXECUTE PLAN The distances from the particles to P are r 1P = (x P x 1 ) 2 + y 2 P = 10 m 2 = 3.2 m and r 2P = (x P x 2 ) 2 + y 2 P = 18 m 2 = 4.2 m. The magnitudes of the electric fields created by the particles at P are thus E 1 = k q 1 r = ( N m 2 /C 2 ) ( C) 1P 10 m 2 = N/C) E 2 = k q 2 = ( N m 2 /C 2 ) ( C) 2 r 2P 18 m 2 = N/C). Slide 23-8

9 Section 23.5: Electric field of a charged particle Example 23.3 Electric field due to two charged particles (cont.) ❸ EXECUTE PLAN To calculate E P, I take the vector sum of and at P. In component form, I have E 1 E 2 E Px = E 1x + E 2x = E 1 cosθ 1 + E 2 cosθ 2 (x = E P x 1 ) (x 1 + E P x 2 ) r 2 1P r 2P E Py = E 1y + E 2 y = E 1 sinθ 1 + E 2 sinθ 2 = E 1 y P r 1P + E 2 y P r 2P. Slide 23-9

10 Section 23.5: Electric field of a charged particle Example 23.3 Electric field due to two charged particles (cont.) ❸ EXECUTE PLAN Substituting the values given, I have E Px = ( N/C) 1.0 m 3.2 m + ( N/C) 3.0 m 4.2 m = N/C E Py = ( N/C) 3.0 m 3.2 m + ( N/C) 3.0 m 4.2 m = N/C. Finally, I write this in vector form as E P = ( N/C)î + ( N/C) ĵ. Slide 23-10

11 Section 23.5: Electric field of a charged particle Example 23.3 Electric field due to two charged particles (cont.) ❹ EVALUATE RESULT Both E Px and E Py are positive, as I expect based on my sketch. The magnitudes of E 1 and E 2 are comparable, which is what I would expect: Particle carries twice the charge of particle 1, but the square of its distance to P is greater by a factor of 1.8. Slide 23-11

12 Section 23.5 Clicker Question 1 A delicate instrument is two paces away from a highly charged metallic sphere. If you want to reduce the magnitude of the electric field at the instrument to 1% of its present value, how many paces away from the charge must you move the instrument? A. 100 paces B. 200 paces C. 10 paces D. 20 paces E. Some other number Slide 23-12

13 Section 23.5 Clicker Question 1 A delicate instrument is two paces away from a highly charged metallic sphere. If you want to reduce the magnitude of the electric field at the instrument to 1% of its present value, how many paces away from the charge must you move the instrument? A. 100 paces B. 200 paces C. 10 paces D. 20 paces E. Some other number Slide 23-13

14 Section 23.6: Dipole field As previously mentioned electric dipoles are important (molecules). The electric field due to a permanent dipole can be easily calculated from the equations derived previously. The figure shows the necessary geometry. Slide 23-14

15 Section 23.6: Dipole field We will do the calc along the axis and across the bisector First the bisector: symmetry demands that the magnitude of the electric fields produced by each charge be the same. Also, the symmetry requires that the horizontal components of the individual electric fields sum to zero. The resultant then points in the negative y direction. Slide 23-15

16 Sec-on 23.6: Dipole field along bisector Summing the ver-cal components E y = E + y + E y = (E + + E )cosθ = 2k = k q p x 2 + (d 2) 2 q p d [x 2 + (d 2) 2 ] 3 2 d 2 [x 2 + (d 2) 2 ] 1 2 Hypotenuse by Pythag Define the magnitude of the dipole moment as p q p d Take the limit x >> d/2, i.e observa-on point is far away compared with the separa-on of the charges, this reduces to E y k p x 3

17 Section 23.6: Dipole field along axis For the case with the observation point on the axis of the dipole the problem is only one-dimensional. For the y-component we have for y > +d /2 E y = k q p y d 2 q 2 p y + d 2 2 = k q p 1 d 2 1+ d 2 y 2 2y 2y Taking the limit as y >> d/2, which implies that the observation point is large compared with the separation of the charges, gives E y k q p y 2 = k q p y d 1 2 d 2y 2y 2d y = 2k q d p y = 2k p 3 y 3 Binomial expansion: for x <<1 ( 1+ x) n 1+ nx Slide 23-17

18 Erratum: Dipole field along axis For the case with the observation point on the axis of the dipole the problem is only one-dimensional. For the y-component we have for y > +d /2 E y = k q p y d 2 q 2 p y + d 2 2 = k q p 1 d 2 1+ d 2 y 2 2y 2y Taking the limit as y >> d/2, which implies that the observation point is large compared with the separation of the charges, gives E y k q p y 2 = k q p y d 1 2 d 2y 2y 2d y = 2k q d p y = 2k p 3 y 3 There were a couple of errors when shown in class fix your notes! Binomial expansion: for x <<1 ( 1+ x) n 1+ nx Slide 23-18

19 Section 23.6: Dipole field - summary Far from the dipole: As we move along the bisector the field goes as: E y k p x 3 As we move along the axis the field goes as: E y 2k p y 3 Slide 23-19

20 Section 23.6 Clicker Question 2 An electric dipole is fixed at the origin of a coordinate system, and an electric field detector can be moved anywhere along the surface of an imaginary sphere of radius R that is centered on the origin. The radius of this sphere greatly exceeds the dipole separation, R >> d. As the detector is moved over the spherical surface, what is the ratio of the largest and smallest magnitudes of the electric field it detects, E maximum /E minimum? A. 1 B. 0 C. 2 D. 1/2 E. Some other number Slide 23-20

21 Section 23.6 Clicker Question 2 An electric dipole is fixed at the origin of a coordinate system, and an electric field detector can be moved anywhere along the surface of an imaginary sphere of radius R that is centered on the origin. The radius of this sphere greatly exceeds the dipole separation, R >> d. As the detector is moved over the spherical surface, what is the ratio of the largest and smallest magnitudes of the electric field it detects, E maximum /E minimum? A. 1 B. 0 C. 2 D. 1/2 E. Some other number Slide 23-21

22 Section 23.8: Dipoles in electric fields The figure shows a dipole consisting of two particles that carry charges of equal magnitude but opposite sign connected by a rod of length d. The dipole makes an angle θ with a uniform electric field E created by some unseen distant source. Slide 23-22

23 Section 23.8: Dipoles in electric fields The forces exerted by the electric field on the charged ends of the dipole are equal in magnitude but opposite in direction, and so the vector sum of the forces exerted on the dipole is zero. Consequently the acceleration of the center of mass of the dipole is zero. However, because the forces are exerted on opposite ends of the dipole they create torques that cause the dipole to rotate counterclockwise about its center of mass. Slide 23-23

24 Section 23.8: Dipoles in electric fields This torque can be expressed by Defining a vector called the dipole moment We can write the torque compactly in vector form: The torque on the dipole is maximum when the dipole moment is perpendicular to the electric field and zero when it is parallel or antiparallel to the electric field. τ ϑ = 2( 1 d sinθ )( q 2 p E) = ( q p d) E sinθ pe sinθ τ = p E! p q p ˆr p d Vector cross product Slide 23-24

25 Demo: Dipole in Electric Field A rod consists of half acrylic, half PVC. When each half is charged, a dipole is formed. The dipole is placed between two conducting plates, which are charged with a Wimshurst machine. The rod will align itself with the electric field between the plates. Slide 23-25