23.5 Electric field of a charged particle

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1 Electric field II The heart is a large electric dipole that changes its orientation and strength during each heart beat. An electrocardiogram measures this dipole electric field of the heart. Reading: Mazur Electric field of a charged particle Dipole field Electric fields of continuous charge distributions Dipoles in electric fields LECTURE 4 1

2 23.5 Electric field of a charged particle The electric field created by the source charge is given by: E " = F "& ' q ) = k q + r "& - r "& For more than one source charge we use the superposition principle to obtain the vector sum of the individual electric fields: E = E / + E - + = 2 k q 3 r 34 - r 34 3 LECTURE 4 2

3 Reading quiz (Problem 23.22) Two uniformly charged pellets A and B are held some distance from each other, and then the charge on A is doubled. Which of the following statements is most correct? A. The magnitude of the electric force exerted on A is doubled because the electric field at the position of A is doubled. B. The magnitude of the electric force exerted on B is doubled because the electric field at the position of B is doubled. C. The magnitude of the electric force exerted on A is doubled because the electric field at the position of B is doubled. D. The magnitude of the electric force exerted on B is doubled because the electric field at the position of A is doubled. Charge A is doubled, so we know the electric field at position B is doubled. Therefore the force that B feels is also doubled. E " = = k : < : ; = > r "& 78 LECTURE 4 3

4 Question: (Field from two charges) Particles with positive and negative charges are arranged as shown in Case 1. The arrangement in Case 2 is identical except the particle with negative charge is replaced by an equal magnitude positive charge. In which case is the magnitude of the electric field at the point labeled A the largest? A. Case 1 B. Case 2 C. The magnitudes are equal and non- zero. D. The magnitudes are both zero. LECTURE 4 4

5 Question: (Field from two charges) answer A positive and negative charge are arranged as shown in Case 1. The arrangement in Case 2 is identical except the negative charge is replaced by an equal magnitude positive charge. In which case is the magnitude of the electric field at the point labeled A the largest? A. Case 1 B. Case 2 C. The magnitudes are equal and non- zero. D. The magnitudes are both zero. The direction of the component of the electric field from bottom charge changed direction but not magnitude, since only the sign of the bottom charge changed. LECTURE 4 5

6 Question: (Finding zero electric field) Particles with charges q and +4q are placed as shown. Mark an approximate position along the x- axis where the electric field is zero. q Q / +4q Q - x LECTURE 4 6

7 Question: (Finding zero electric field) answer Particles with charges q and +4q are placed as shown. Mark an approximate position along the x- axis where the electric field is zero. q Q / +4q Q - x E " = k : < = > r "& 78 First consider just q; it has negative charge so the electric field points towards it. Now consider just +4q; it has positive charge so the electric field points away from it. For the total electric field to be zero the electric field from each charge must point in opposite directions. So the region between charges is ruled out. +4q is four times q in magnitude so to have the same magnitude electric field the position must be twice as close to q as +4q. LECTURE 4 7

8 23.6 Dipole field The dipole moment vector is given by p q G r G The electric field far from the dipole is given by E H k G J K (along the x- axis) E H 2k G HK (along the y- axis) LECTURE 4 8

9 Reading quiz: (Problem 23.44) Two plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.60 nc. Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 800- mm stick stuck through the holes so that it runs from the center of one ball to the center of the other. What is the magnitude of the dipole moment of the arrangement? p = q G r G = RS C RV m = R/X C Y m LECTURE 4 9

23.7 Electric field of continuous charge distributions We can calculate the electric field for a continuous charge distribution by integrating the electric field from infinitesimal charges E = k Z dq

10 23.7 Electric field of continuous charge distributions We can calculate the electric field for a continuous charge distribution by integrating the electric field from infinitesimal charges E = k Z dq " r "4 - r "4 Math 125 (or equivalent) is a co- request for this class. So hopefully you know that the integral represents the sum over all the infinitesimal components of the object. But maybe you have not seen an integral that involves a vector. However, you could often use symmetry to simplify. LECTURE 4 10

11 Question: Using symmetry, determine the direction of the resulting electric field due to a thin straight line of charge of length l at a distance x away from the center. A. x direction only B. y direction only C. z direction only D. Combination of x and y directions E. Combination of x and z directions F. Combination of y and z directions G. Combination of all directions l x P z Coordinates y (into the page) x LECTURE 4 11

12 Question: answer x direction only LECTURE 4 12

13 Question: Using symmetry, determine the direction of the resulting electric field due to a thin ring of charge of radius R at a distance z away from the center. A. x direction only B. y direction only C. z direction only D. Combination of x and y directions E. Combination of x and z directions F. Combination of y and z directions G. Combination of all directions z P R z Coordinates y (into the page) x LECTURE 4 13

14 Question: answer z direction only LECTURE 4 14

15 Question: (Charge on segment) Suppose we take a 2- m long rod with a charge of 4 10 Ra C distributed uniformly along the rod. What is the charge contained on a segment, q ", that is 8 10 RV m long? Express answer in nc (leave unit off answer). LECTURE 4 15

16 Question: (Charge on segment) answer Suppose we take a 2- m long rod with a charge of 4 10 Ra C distributed uniformly along the rod. What is the charge contained on a segment, q ", that is 8 10 RV m long? The segment fraction of the total length is 8 10 RV m 2 m = 4 10 RV This is also the segment fraction of the total charge, so q " 4 10 Ra C = 4 10RV And therefore q " = 16 nc LECTURE 4 16

17 Charge density In order to do this integration (E = k d: 7 = > r "4) it is necessary to express dq ", r - "4, and r "4 in the same coordinates. 7e We do this by expressing dq " in terms of the charge density. For uniformly charged objects 1- D: the linear charge density for an object of charge q and length l is λ : g 2- D: the surface charge density for an object of charge q and area A is σ : j 3- D: the volume charge density for an object of charge q and volume V is ρ : m In non- uniform charge distributions the following equations are still true 1- D: dq = λdl 2- D: dq = σda 3- D: dq = ρdv we can determine the total charge for non- uniform charge distributions by integrating dq over the line, area, or volume. LECTURE 4 17

18 Magnitude of electric field near common objects A distance r away from a very long line of uniform charge density λ: E = 2kλ r A distance z along the axis away from the center of a uniform ring of charge q and radius R: qz E = k z - + R - V - A distance z along the axis away from the center of a uniform disk of charge q and radius R: E = k 2q R R- R/ - z - An infinite sheet of charge density σ: E = 2kπσ LECTURE 4 18

19 23.8 Dipoles in electric fields The torque on a dipole is given by τ = p E The magnitude of the torque is given by τ = pe sin θ LECTURE 4 19

20 Question: (Magnitude of torques) A dipole with positive charge q is placed in a uniform electric field that points to the right. Rank the four different orientations (labeled A through D) according to the magnitude of the torque (from least to greatest). A) +q q B) +q q C) q +q D) q +q LECTURE 4 20

21 Question: (Magnitude of torques) answer A dipole with positive charge q is placed in a uniform electric field that points to the right. Rank the four different orientations (labeled A through D) according to the magnitude of the torque (from least to greatest). A) +q q 180 θ Magnitude of torque us given by τ = pe sin θ p and E is identical in each case B) +q q 180 θ Since sin 180 θ = sin θ the magnitude of the torques are the same. C) q θ +q D) q θ +q LECTURE 4 21

22 23.8 Dipoles in electric fields induced dipole The presence of an external electric field induces a dipole moment (induced dipole). p vwx = αe where α is the polarizability of the atom. LECTURE 4 22

PH 222-3A Spring 2007

PH 222-3A Spring 2007 PH -3A Spring 7 ELECTRIC FIELDS Lectures,3 Chapter (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition) 1 Chapter Electric Fields In this chapter we will introduce the concept of an electric