PH 222-3A Spring 2007 ELECTRIC FIELDS Lectures 2,3 L 23 Chapter 22 (Halliday/Resnick/Walker, Fundamentals of Physics 8th edition) 1 Chapter 22 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 f h forces among charges. If the charges are not stationary we must use h fh h i an alternative approach by introducing the electric field (symbol E ). In connection with the electric field, the following topics will be , gp 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 <a href="/keyword/electric-dipole/" >electric dipole</a> . Determining the net force, Defining dipole force the net torque, exerted on an <a href="/keyword/electric-dipole/" >electric dipole</a> by a <a href="/keyword/uniform-electric-field/" >uniform electric field</a> , 2 as well as the dipole potential energy. In Chapter 21 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 q2 acts on q1 at a di h distance r i instantaneously ( i at a l ( action distance ): 1 q1 q2 q q1 F= 4 0 r 2 1 Electric interactions propagate in empty space with a large but finite speed (c = 3 108 m/s). In order to take into account correctly the finite speed at w c these interactions p op g e, we have to abandon the which ese e c o s propagate, ve o b do e action at a distance point of view and still be able to explain how q1 knows about the presence of q2 . The solution is to introduce the new concept of an <a href="/keyword/electric-field-vector/" >electric field vector</a> as follows: Point charge q1 does not exert a force directly on q2. Instead, q1 creates in its vicinity an electric field that exerts a force on q2 . charge q1 generates electric field E E exerts a force F on q2 3 F E= q0 Definition of the <a href="/keyword/electric-field-vector/" >electric field vector</a> Consider the positively charged rod shown in the figure. For every point P in the vicinity of the rod we define the <a href="/keyword/electric-field-vector/" >electric field vector</a> E as follows: 1. We place a positive test charge q0 at point P. p p g p 2. We measure the electrostatic force F exerted on q0 by the charged rod. 3. We define the <a href="/keyword/electric-field-vector/" >electric field vector</a> E at point P as: E= F . q0 SI Units: N/C From the definition it follows that E is parallel to F . Note : We assume that the test charge q0 is small enough g g so that its presence at point P does not affect the charge distribution on the rod and thus alter the <a href="/keyword/electric-field-vector/" >electric field vector</a> E we are trying to determine. 4 Electric Field Generated by a Point Charge Consider the positive charge q shown in the figure. At point P a distance r from E qo P q r q we place the test charge q0 . The force exerted on q0 by q0 is equal to: F= E= 1 4 0 q q0 r2 = 1 q 4 0 r 2 F 1 q q0 = q0 4 0 q0 r 2 The magnitude of E is a positive number. In terms of direction, E points radially outward as shown in the figure. 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 q E= 4 0 r 2 1 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 <a href="/keyword/electric-field-vector/" >electric field vector</a> s generated b each charge. h f h l i fi ld d by hh In the example shown in the figure, E = E1 + E2 + E3 . Here E1 , E2 , and E3 are th <a href="/keyword/electric-field-vector/" >electric field vector</a> s generated by q1 , q1 , and q3 , H d the l t i fi ld t t db d respectively. Note: E1 , E2 , and E3 must be added as vectors: Ex = E1x + E2 x + E3 x , E y = E1 y + E2 y + E3 y , Ez = E1z + E2 z + E3 z 6 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, hi d i i h l i fi ld Th l i f hh k b ll id and the ball attains equilibrium when the string makes an angle of 35 with the vertical. The mass of the ball is 3x10-5 kg, and the charge on the ball is 4x10-7 C. What is the magnitude of the electric field? g X 5.1x102 N/C 7 8 v2 = vo2 + 2ax (if xo=0 at to=0) m/s2 / 9 10 <a href="/keyword/electric-dipole/" >electric dipole</a> A system of two equal charges of opposite sign +q d -q +q/2 ( q ) placed at a distance d apart is known as an &quot;electric <a href="/keyword/electric-dipole/" >electric dipole</a> .&quot; For every <a href="/keyword/electric-dipole/" >electric dipole</a> we associate a vector known as &quot;the <a href="/keyword/electric-dipole/" >electric dipole</a> moment&quot; (symbol p ) defined as follows: The magnitude p = qd +q/2 Th direction of p is along the line that connects The di i fil h li h the two charges and points from - q to + q. y p Many molecules have a built-in <a href="/keyword/electric-dipole/" >electric dipole</a> moment. An example is the water molecule (H 2 O). -q The bonding between the O atom and the two H atoms involves the sharing of 10 valence electrons (8 from O and 1 from each H atom). The 10 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 2 O molecule. 11 Electric Field Generated by an <a href="/keyword/electric-dipole/" >electric dipole</a> We will determine the electric field E generated by the <a href="/keyword/electric-dipole/" >electric dipole</a> shown in the figure using the principle of superposition. The positive charge generates at P an electric 1q field whose magnitude E( + ) = . The negative charge 2 4 0 r+ creates an electric field with magnitude E( ) t l t i fi ld ith it d The net electric field at P is E = E( + ) E( ) . q q 1 q q 1 E= 2 2 = 2 ( z d / 2 )2 r 4 0 4 0 r+ ( z + d / 2) 2 2 q d d d 1 1 + We assume: E= 2 4 0 z 2 z 2z 2z E= qd 1p d d = 1 + 1 = 4 0 z 2 z z 2 0 z 3 2 0 z 3 q 12 q = . 2 4 0 r 1 1 (1 + x ) 2 1 2x Electric Field Generated by a Continuous Charge Distribution P r dV dE Consider the continuous charge distribution shown in the r dq figure. We assume that we know the volume density of dq ( Units: C/m3 ). the electric charge. This is defined as = 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 &quot;elements&quot; of volume dV . Each element has charge dq = dV . We assume that point P is at a distance r from dq. 2. 2 Determine the electric field dE generated by dq at point P. dq The magnitude dE of dE is given by the equation dE = . 2 4 0 r dVr 1 3. Sum all the contributions: E = r2 . 4 0 13 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 figure. between the element and point P is r = z 2 + R 2 . g q generates at P an electric field of magnitude g The charge dq g dE that points outward along the line AP: dq dE = . The z -component of dE is given by p g y 2 4 0 r A dq C dEz = dE cos . From triangle PAC we have: cos = z / r zdq zdq dEz = = . 3 2 2 3/2 4 0 r 4 0 ( z + R ) Ez = 4 0 ( z 2 + R 2 ) z 3/2 Ez = dEz dq d = 4 0 ( z 2 + R 2 ) zq q 3/2 14 Example 2 : Determine the electric field E generated at point P by a uniformly positively charged disk of radius R and charge density . 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 g g contributions of all the rings. For a ring with radius r and radial width dr, dq = dA = (2 rdr ). We have already solved the problem of the electric field due to a ring of charge: dE = dqz d 4 o ( z + r ) 2 2 3 2 = z 2 rdr d 4 o ( z + r ) 2 2 3 2 = 2rdr d 4 o ( z 2 + r 2 ) 3 2 We can now find E by integrating this eq. over the surface y g g q of the disk, that is , by integrating with respect to r , from r = 0 to r = R, while z = const. 15 z R 2 2 3 2 E = dE = 0 ( z + r ) (2r )dr 4 o To l thi i t T solve this integral, we cast it in the f l t i th form X m dX b setting by tti 3 X = ( z 2 + r 2 ), m = , and dX = (2r )dr 2 2 2 1 2 m +1 z z X z (z + r ) = E= X m dX = 1 4 o 4 o m + 1 4 o 2 0 z E= 1 2 2 o z + R2 If we let R (infinite sheet) while keeping z finite, the seconf term in the parentheses approaches zero and E= R (infinite sheet) 2 o This is the electic field produced by an infinite sheet of uniform charge located on one side of a nonconductor such as plastic. 16 <a href="/keyword/electric-field-lines/" >electric field lines</a> . In the 19th century Michael Faraday introduced the concept of <a href="/keyword/electric-field-lines/" >electric field lines</a> , which help visualize the <a href="/keyword/electric-field-vector/" >electric field vector</a> E without using mathematics. For the relation between the <a href="/keyword/electric-field-lines/" >electric field lines</a> and E : 1. At any point P the <a href="/keyword/electric-field-vector/" >electric field vector</a> E is tangent to the <a href="/keyword/electric-field-lines/" >electric field lines</a> . EP P electric field line 2. The magnitude of the <a href="/keyword/electric-field-vector/" >electric field vector</a> E is proportional to the density of the <a href="/keyword/electric-field-lines/" >electric field lines</a> . lines EP &gt; EQ EP Q EQ <a href="/keyword/electric-field-lines/" >electric field lines</a> 17 P Two graphical methods of graphical representation of electric gp gp p field: a) <a href="/keyword/electric-field-vector/" >electric field vector</a> s; b) <a href="/keyword/electric-field-lines/" >electric field lines</a> (directed radially outward from the positive charge +q) 18 <a href="/keyword/electric-field-lines/" >electric field lines</a> extend away from positive charges (where they originate) and toward negative charges (where they terminate). Example 1 : <a href="/keyword/electric-field-lines/" >electric field lines</a> of a negative point charge - q : E= 1 q -The <a href="/keyword/electric-field-lines/" >electric field lines</a> 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. di t f d Note : In the case of a positive point charge the <a href="/keyword/electric-field-lines/" >electric field lines</a> have the same form but they point outward. q 4 0 r 2 q 19 The electric field can be represented graphically by field lines 20 Major properties of the field lines 21 Example 2 : <a href="/keyword/electric-field-lines/" >electric field lines</a> 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 h the form shown i fig. b. l i fi ld db hl has h f h in fi 1. The electric field on either side of the plane has a constant magnitude. pp gp 2. The <a href="/keyword/electric-field-vector/" >electric field vector</a> is perpendicular to the charge plane. 3. The <a href="/keyword/electric-field-vector/" >electric field vector</a> E points away from the plane. The corresponding <a href="/keyword/electric-field-lines/" >electric field lines</a> are given in fig. c. Note : For a negatively charged plane the <a href="/keyword/electric-field-lines/" >electric field lines</a> point inward. 22 The concept of the field lines for the computation of the electric field of a large, flat, charged sheet 23 The Electric Field of the Two Charged Sheets 24 ED=(1.1x105N/C)[-1-1+1] = -1.1x105 N/C 25 Example 3 : <a href="/keyword/electric-field-lines/" >electric field lines</a> generated by an <a href="/keyword/electric-dipole/" >electric dipole</a> (a positive and a negative point charge of the same size but of opposite sign) Example 4 : <a href="/keyword/electric-field-lines/" >electric field lines</a> generated by two equal positive point charges 26 F+ Forces and Torques Exerted on <a href="/keyword/electric-dipole/" >electric dipole</a> s by a <a href="/keyword/uniform-electric-field/" >uniform electric field</a> Consider the <a href="/keyword/electric-dipole/" >electric dipole</a> shown in the figure in the presence of a uniform (constant magnitude and direction) electric field E along the x-axis. axis The electric field exerts a force F+ = qE on the Fx-axis i p positive charge and a force F = qE on the g q negative charge. The net force on the dipole is Fnet = qE qE = 0. The net torque generated by F+ and F about the dipole center is = + + = F+ d d sin F sin = qEd sin = pE sin q p 2 2 In vector form: = p E p The <a href="/keyword/electric-dipole/" >electric dipole</a> in a <a href="/keyword/uniform-electric-field/" >uniform electric field</a> does not move but can rotate about its center. Fnet = 0 = p E 27 Potential Energy of an <a href="/keyword/electric-dipole/" >electric dipole</a> in a <a href="/keyword/uniform-electric-field/" >uniform electric field</a> The dipole has its least potential energy, when it is in equilibrium p ) orientation ( p is lined up with the E ; = p E = 0). To rotate th e 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 <a href="/keyword/electric-dipole/" >electric dipole</a> in an external electric field is simplest if we choose it to be zero when angle =90 . We can find U at any other value of by calculating the work W done by the field on the dipole w hen the dipole is rotated to that value from 90 . U = pE cos U = p E B U 180 U = d = pE sin d p 90 90 U = pE sin d = pE cos = p E 90 90 A At point A ( = 0), U has a minimum At point B ( = 180 ), U has a maximum value U max = + pE. value U min = pE. It is a position of stable equilibrium. It is a position of unstable equilibrium. p E p E 28 p i Fig. a Work Done by an External Agent to Rotate an Electric E Dipole in a <a href="/keyword/uniform-electric-field/" >uniform electric field</a> Consider the <a href="/keyword/electric-dipole/" >electric dipole</a> in fig. a. It has an <a href="/keyword/electric-dipole/" >electric dipole</a> moment p and is positioned so that p is at an angle p f Fig. b g i with respect to a <a href="/keyword/uniform-electric-field/" >uniform electric field</a> E. An external agent rotates the <a href="/keyword/electric-dipole/" >electric dipole</a> and brings it to its final position shown in fig. b. In this position fig E p is at an angle f with respect to E. 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 f U i = pE cos f ( pE cos i ) W = pE ( cos i cos f ) 29