21.5 Exercises 723 Fig 21.28 See Exercise (8) S2 S1 r r q R λ Fig 21.29 See Exercise (9) S1 S4 q −q S3 S2 −2 q 2q (10) A point charge q = 25 µC is located at the center of a sphere of radius R = 25 cm A circular cut of radius r = cm is removed from the surface of the sphere, see Fig 21.30 (a) Find the electric flux that passes through that cut (b) Repeat when the cut has a radius r = 25 cm Is the answer q/2 ◦ ? Fig 21.30 See Exercise (10) r q+ R (11) A point charge q = 53.1 nC is located at the center of a cube of side a = cm, see Fig 21.31 (a) Find the electric flux through each face of the cube (b) Find the flux through the four slanted surfaces of a pyramid formed from a vertex on the center of the cube and one of its six square faces 724 21 Gauss’s Law Fig 21.31 See Exercise (11) y a q + x a z a (12) At an altitude h1 = 700 m above the ground, the electric field in a particular region is E1 = 95 N/C downwards At an altitude h2 = 800 m, the electric field is E2 = 80 N/C downwards Construct a Gaussian surface as a box of horizontal area A and height lying between h1 and h2 , to find the average volume-charge density in the layer of air between these two elevations (13) A point P is at a distance a = 10 cm from an infinite rod, charged with a uniform charge per unit length λ = nC/m (a) Find the electric flux through a sphere of radius r = cm centered at P, see left of Fig 21.32 (b) Find the electric flux through a sphere of radius r = 15 cm centered at P, see right of Fig 21.32 r - + + P r a λ + + + + + P a - + + + + + + λ + Fig 21.32 See Exercise (13) (14) A point charge q is located at a distance δ just above the center of the flat face of a hemisphere of radius R as shown in Fig 21.33 (a) When δ is very small, use the argument of symmetry to find an approximate value for the electric flux through the curved surface of the hemisphere (b) When δ is very small, use Gauss’s law to find an approximate value of the electric flux flat through curved the flat surface of the hemisphere 21.5 Exercises 725 Fig 21.33 See Exercise (14) q δ R Section 21.3 Applications of Gauss’s Law (15) An infinite horizontal sheet of charge has a charge per unit area σ = 8.85 µC/m2 Find the electric field just above the sheet (16) A nonconductive wall carries a uniform charge density σ = 8.85 µC/cm2 Find the electric field cm away from the wall Does your result change as the distance from the wall increases such that it is much less than the wall’s dimensions? (17) Two infinitely long, nonconductive charged sheets are parallel to each other Each sheet has a fixed uniform charge The surface charge density on the left sheet is σ while on the right sheet is −σ, see Fig 21.34 Use the superposition principle to find the electric field: (a) to the left of the sheets, (b) between the sheets, and (c) to the right of the sheets Fig 21.34 See Exercise (17) + σ + + + + + L −σ B R (18) Repeat the calculations for Exercise 17 when: (i) both the sheets have positive uniform surface charge densities σ, and (ii) both the sheets have negative uniform surface charge densities −σ (19) A thin neutral conducting square plate of side a = 80 cm lies in the xy-plane, see Fig 21.35 A total charge q = nC is placed on the plate Assuming that 726 21 Gauss’s Law the charge density is uniform, find: (a) the surface charge density on the plate, (b) the electric field just above the plate, and (c) the electric field just below the plate Fig 21.35 See Exercise (19) z a a y x (20) A long filament has a charge per unit length λ = −80 µC/m Find the electric field at: (a) 10 cm, (b) 20 cm, and (c) 100 cm from the filament, where distances are measured perpendicular to the length of the filament (21) A uniformly charged straight wire of length L = 1.5 m has a total charge Q = µC A thin uncharged nonconductive cylinder of height = cm and radius r = 10 cm surrounds the wire at its central axis, see Fig 21.36 Using reasonable approximations, find: (a) the electric field at the surface of the cylinder and (b) the total electric flux through the cylinder Fig 21.36 See Exercise (21) Q + + + + r L + + + (22) A thin nonconductive cylindrical shell of radius R = 10 cm and length L = 2.5 m has a uniform charge Q distributed on its curved surface, see Fig 21.37 The radial outward electric field has a magnitude × 104 N/C at a distance r = 20 cm from its axis (measured from the midpoint of the shell) 21.5 Exercises 727 Find: (a) the net charge on the shell, and (b) the electric field at a point r = cm from its axis Fig 21.37 See Exercise (22) R Q E r L (23) A long non-conducting cylinder of radius R has a uniform charge distribution of density ρ throughout its volume Find the electric field at a distance r from its axis where r < R? (24) A thin spherical shell of radius R = 15 cm has a total positive charge Q = 30 µC distributed uniformly over its surface, see Fig 21.38 Find the electric field at: (a) 10 cm and (b) 20 cm from the center of the charge distribution Fig 21.38 See Exercise (24) Q + + + R Spherical shell + + + + + (25) Two concentric thin spherical shells have radii R1 = cm and R2 = 10 cm The two shells have charges of the same magnitude Q = µC, but different in sign, see Fig 21.39 Use the shown three Gaussian surfaces S1 , S2 , and S3 to find the electric field in the three regions: (a) r < R1 , (b) R1 < r < R2 , and (c) r > R2 (26) A particle with a charge q = −60 nC is located at the center of a non-conducting spherical shell of volume V = 3.19 × 10−2 m3 , see Fig 21.40 The spherical shell carries over its interior volume a uniform negative charge Q of volume density ρ = −1.33 µC/m3 A proton moves outside the spherical shell in a circular orbit of radius r = 25 cm Calculate the speed of the proton 728 21 Gauss’s Law Fig 21.39 See Exercise (25) S3 −Q S2 Q R2 R1 S1 Spherical shells Fig 21.40 See Exercise (26) Q – – – q – – Spherical shell – – – – – – – – – – – – – r – – – Point charge – – +e + – Proton – (27) A solid non-conducting sphere is cm in radius and carries a 7.5 µC charge that is uniformly distributed throughout its interior volume Calculate the charge enclosed by a spherical surface, concentric with the sphere, of radius (a) r = cm and (b) r = cm (28) A solid non-conducting sphere of radius R = 20 cm has a total positive charge Q = 30 µC that is uniformly distributed throughout its volume Calculate the magnitude of the electric field at: (a) cm, (b) 10 cm, (c) 20 cm, (d) 30 cm, and (e) 60 cm from the center of the sphere (29) If the electric field in air exceeds the threshold value Ethre = × 106 N/C, sparks will occur What is the largest charge Q can a metal sphere of radius 0.5 cm hold without sparks occurring? (30) The charge density inside a non-conducting sphere of radius R varies as ρ = α r (C/m3 ), where r is the radial distance from the center of the sphere Use Gauss’s law to find the electric field inside and outside the sphere (31) A solid sphere of radius R with a center at point C1 has a uniform volume charge density ρ A spherical cavity of radius R/2 with a center at point C2 is then scooped out and left empty, see Fig 21.41 Point A is at the surface of the big sphere and collinear with points C1 and C2 What is the magnitude and direction of the electric field at points C1 and A? 21.5 Exercises 729 Fig 21.41 See Exercise (31) A C1 R C2 R /2 Section 21.4 Conductors in Electrostatic Equilibrium (32) A non-conducting sphere of radius R and charge +Q uniformly distributed throughout its volume is concentric with a spherical conducting shell of inner radius R1 and outer radius R2 The shell has a net charge −Q, see Fig 21.42 Find an expression for the electric field as a function of the radius r when: (a) r < R (within the sphere) (b) R < r < R1 (between the sphere and the shell) (c) R1 < r < R2 (inside the shell) (d) r > R2 (outside the shell) (e) What are the charges on inner and outer surfaces of the conducting shell? Fig 21.42 See Exercise (32) −Q Q R R1 R2 c (33) A large, thin, copper plate of area A has a total charge Q uniformly distributed over its surfaces The same charge Q is uniformly distributed over the upper surface of a glass plate, which is identical to the copper plate, see Fig 21.43 (a) Find the surface charge density on each face of the two plates (b) Compare the electric fields just above the center of the upper surface of each plate 730 21 Gauss’s Law Q Copper plate Q Glass plate Fig 21.43 See Exercise (33) (34) A thin, long, straight wire carries a charge per unit length λ The wire lies along the axis of a long conducting cylinder carrying a charge per unit length 3λ The cylinder has an inner radius R1 and an outer radius R2 , see Fig 21.44 (a) Use a Gaussian surface inside the conducting cylinder to find the charge per unit length on its inner and outer surfaces (b) Use Gauss’s law to find the electric field E outside the wire (c) Sketch the electric field E as a function of the distance r from the wire’s axis Fig 21.44 See Exercise (34) 3λ λ R1 R2 (35) An uncharged solid conducting sphere of radius R contains two cavities A point charge q1 is placed within the first cavity, and a point charge q2 is placed within the second one Find the magnitude of the electric field for r > R, where r is measured from the center of the sphere 22 Electric Potential Newton’s law of gravity and Coulomb’s law of electrostatics are mathematically identical Therefore, the general features of the gravitational force introduced in Chap apply to electrostatic forces In particular, electrostatic forces are conservative Consequently, it is more convenient to assign an electric potential energy U to describe any system of two or more charged particles This idea allows us to define a scalar quantity known as the electric potential It turns out that this concept is of great practical value when dealing with devices such as capacitors, resistors, inductors, batteries, etc, and when dealing with the flow of currents in electric circuits 22.1 Electric Potential Energy → The electric force that acts on a test charge q placed in an electric field E (created by → → a source charge distribution) is defined by F = qE For an infinitesimal displacement d→ s , the work done by the conservative electric field is: → → dW = F • d → s = qE • d → s (22.1) According to Eq 6.39, this amount of work corresponds to a change in the potential energy of the charge-field system given by: → → dU = −dW = −F • d → s = −qE • d → s (22.2) For a finite displacement of the charge from an initial point A to a final point B, the change in electric potential energy U = UB − UA of the charge-field system When dealing with electric and magnetic fields, it is common to use this notation to represent an infinitesimal displacement vector that is tangent to a path through space H A Radi and J O Rasmussen, Principles of Physics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_22, © Springer-Verlag Berlin Heidelberg 2013 731 732 22 Electric Potential is given by integrating along any path that the charge can take between these two points Thus: B U = UB − UA = −WAB = −q → E • d→ s (22.3) A This integral does not depend on the path taken from A to B because the electric force is conservative For convenience, we usually take the reference configuration of the charge-field system when the charges are infinitely separated Moreover, we usually set the corresponding reference potential energy to be zero Therefore, we assume that the charge-field system comes together from an initial infinite separation state at ∞ with U∞ = to a final state B with UB We also let W∞B represent the work done by the electrostatic force during the movement of the charge Thus: B UB = −W∞B = −q → E • d→ s (22.4) ∞ Although the electric potential energy at a particular point UB (or simply U) is associated with the charge-field system, one can say that the charge in the electric field has an electric potential energy at a particular point UB You should always keep in mind the fact that the electric potential energy is actually associated with the → charge plus the source charge distribution that establishes the electric field E Moreover, when defining the work done on a certain charged particle by the electric field, you are actually defining the work done on that certain particle by the → electric force due to all the other charges that actually created the electric field E in which that certain particle moved Example 22.1 In the air above a particular region near Earth’s surface, the electric field is uniform, directed downwards, and has a value 100 N/C, see Fig 22.1 Find the change in the electric potential energy of an electron released at point A such that the electrostatic force due to the electric field causes it to move up a distance s = 50 m Solution: The change in the electric potential energy of the electron is related to the work done on the electron by the electric field given by Eq 22.3 Since the 738 22 Electric Potential Arbitrary Displacement in the Field Now, let us calculate the potential difference between two points A and B when the → displacement → s has an arbitrary direction with respect to the uniform field E , see Fig 22.4a Equation 22.6 gives: B V = VB − VA = − → → → B E • d s = −E • A d→ s (22.20) A Path of constant potential A s Equipotential surfaces d B C E V1 V2 V3 E (a) (b) Fig 22.4 (a) When the uniform electric field lines point down, the electric potential at point B is lower than that at point A Points B and C are at the same electric potential (b) Portions of equipotential surfaces → that are perpendicular to the electric field E Integrating d → s along the path gives → sB − → sA = → s Thus: → V = VB − VA = −E • → s (22.21) When a charge q moves from A to B, the change in the electric potential energy U of the charge-field system is given according to Eq 22.6 by U = q V Then, using Eq 22.21 we get: → U = q V = −qE • → s (From A to B) (22.22) → From Fig 22.4a, we notice that E • → s = Es cos θ = Ed Thus, any point in a plane perpendicular to a uniform electric field has the same electric potential We can see this in Fig 22.4a, where the potential difference VB − VA is equal to the potential difference VC − VA ; that is VC = VB All points that have the same electric potential form what is called an equipotential surface, see Fig 22.4b 22.3 Electric Potential in a Uniform Electric Field 739 Example 22.2 The electric potential difference between two opposing parallel plates is 12 V, while the separation between the plates is d = 0.4 cm Find the magnitude of the electric field between the plates Solution: We use Eq 22.14, namely V = VB −VA = −Ed, to find the magnitude of the uniform electric field as follows: E= |VB − VA | 12 V = = 3,000 V/m (or N/C) d 0.4 × 10−2 m Example 22.3 Figure 22.5 shows two oppositely charged parallel plates that are separated by a distance d = cm The electric field between the plates is uniform and has a magnitude E = 10 kV/m A proton is released from rest at the positive plate, see Fig 22.5 (a) Find the potential difference between the two plates (b) Find the change in potential energy of the proton just before striking the opposite plate (c) Calculate the speed of the proton when it strikes the plate Fig 22.5 + + + + Proton E A + + A B B d Solution: (a) Since the potential must be lower in the direction of the field, then VA > VB in Fig 22.5 According to Eq 22.14 the potential difference between the two plates along the field is: V = VB − VA = −Ed = −(10 × 103 V/m)(5 × 10−2 m) = −500 V (b) Equation 22.15 gives the change in potential energy as: U = q V = e V = (1.6 × 10−19 C)(−500 V) = −8 × 10−17 J The negative sign indicates that the potential energy of the proton decreases as the proton moves in the direction of the field 740 22 Electric Potential (c) Using Eq 22.16, we can find for the proton that: K =− U 2 mp vB vB = −2 U = mp −0=− U −2(−8 × 10−17 J) = 3.1 × 105 m/s 1.67 × 10−27 kg Example 22.4 Redo Example 22.3, but this time with an electron that is released from rest at the negative plate, see Fig 22.6 Fig 22.6 + E + + + B A - Electron - A B d Solution: (a) Since the potential must be lower in the direction of the field, then VB > VA in Fig 22.6 Then, according to Eq 22.17 the potential difference between the two plates is: V = VB − VA = +Ed = +(10 × 103 V/m)(5 × 10−2 m) = +500 V This means that VB = VA + 500 V as expected, since point B lies on the positive plate (b) From Eq 22.18, we find for the electron that: U = q V = −e V = −(1.6 × 10−19 C)(500 V) = −8 × 10−17 J The negative sign indicates that the potential energy of the electron decreases as the electron moves in the opposite direction of the field (c) From Eq 22.19, we find for the electron that: K =− U 22.3 Electric Potential in a Uniform Electric Field 2 me vB vB = −2 U = me 741 −0=− U −2(−8 × 10−17 J) = 1.33 × 107 m/s 9.11 × 10−31 kg Although we are using the same constants of Example 22.3, with the same change in potential energy of the electron as for the proton, the speed of the electron is much greater than the speed of the proton because the electron’s mass is much smaller 22.4 Electric Potential Due to a Point Charge To find the electric potential at a point located at a distance r from an isolated positive point charge q, we begin with the general expression for potential difference: B VB − VA = − → E • d→ s (22.23) A where A and B are two points having position vectors → r A and → r B , respectively, see Fig 22.7 In this representation, the origin is at q According to Eq 20.4, the electric → field at a distance r from the point charge is E = kq → rˆ /r = Er → rˆ , where → rˆ is a unit → → vector directed from the charge to point P The quantity E • d s , can be written as: → E • d→ s = Er → rˆ • d → s =k q → rˆ • d → s r2 (22.24) Fig 22.7 The potential difference between points A ds E and B due to a point charge q rB depends only on the radial coordinates rA and rB B dr r + q A rA → From Fig 22.7, we see that → rˆ • d → s = × ds × cos θ = dr Then, E • d → s = Er dr, and Eq 22.23 can be written as: 742 22 Electric Potential rB VB − VA = − rB Er dr = −kq rA Therefore: r −2 dr = −kq −r −1 rA VB − VA = kq rB rA 1 − rB rA (22.25) This result does not depend on the path between A and B, but depends only on the initial and final radial coordinates rA and rB It is common to choose a reference point where VA = at rA = ∞ With this reference choice, the electric potential at any arbitrary distance r from a point charge q will be given by: V =k q r (22.26) The sign of V depends on q If r → 0, then V → +∞ or V → −∞ depending on q Figure 22.8 shows a plot for V in the xy-plane Fig 22.8 A computer- V generated plot for the electric potential V(r) of a single point-charge in the xy-plane The predicted infinite value of V(r) is not plotted y x When many point charges are involved, we can get the resulting electric potential at point P from the superposition principle That is: V =k n qn , rn (n = 1, 2, 3, ) (22.27) where rn is the distance from the point P to the charge qn Electric Potential Energy of a System of Point Charges The electric potential energy of a system of two point charges q1 and q2 can be obtained first by having both charges placed at rest and set infinitely apart Then, by 22.4 Electric Potential Due to a Point Charge 743 bringing q1 by itself from infinity and putting it in place as shown in Fig 22.9a, we no work for such a move The electric potential at point P which is at a distance r12 from q1 is given by Eq 22.26 as VP = kq1 /r12 Later, by bringing q2 without acceleration from infinity to point P at a distance r12 from q1 , as shown in Fig 22.9b, we must work W∞P (app) for such a move, since q1 exerts an electrostatic force on q2 during the move P VP k r12 q1 r12 q2 r12 U k q1 q1 (a) q q2 r12 (b) Fig 22.9 (a) The potential VP at a distance r12 from a point charge q1 (b) The potential energy of two point charges is U = kq1 q2 /r12 We can calculate W∞P (app) by using K = W∞P (app) + W∞P = 0, i.e W∞P (app) = −W∞P When we replace q2 by the general charge q in Eq 22.6, we find that: VP − V∞ = UP − U ∞ W∞P =− q2 q2 (22.28) Setting V∞ as well as U∞ to zero (our reference point at ∞), we get: VP = UP W∞P (app) = q2 q2 ⇒ UP = W∞P (app) = q2 VP (22.29) Substituting with VP = kq1 /r12 , we can generalize the electric potential energy of a system of two point charges q1 and q2 separated by a distance r12 as follows: U=k q1 q2 r12 (22.30) If the charges have the same sign, we have to positive work to overcome their mutual repulsion, and then U is positive If the charges have opposite signs, we have 744 22 Electric Potential to negative work against their mutual attraction to keep them stationary, and then U is negative When the system consists of more than two charges, we calculate the potential energy of each pair and add them algebraically For instance, the total potential energy of three charges q1 , q2 , and q3 is: q1 q2 q1 q3 q2 q3 + + r12 r13 r23 U=k (22.31) Example 22.5 Two charges q1 = µC and q2 = − µC are fixed in their positions and separated by a distance d = 10 cm; see Fig 22.10a (a) Find the total electric potential at the point P in Fig 22.10a (b) Find the change in potential energy of the two charges when a third charge q3 = µC is brought from ∞ to P, see Fig 22.10b (c) What is the total electric potential energy of the three charges? P + d q1 + d d d d - q2 q1 + (a) q3 d - q2 (b) Fig 22.10 Solution: (a) For two charges, Eq 22.27 gives: VP = k × 10−6 C −4 × 10−6 C q1 q2 + = (9 × 109 N.m2 /C2 ) + d d 0.1 m 0.1 m = −1.8 × 105 V (b) When we replace q3 by the charge q of Eq 22.6, and bring q3 from infinity to point P, this equation gives: U = q3 (VP − V∞ ) = (6 × 10−6 C)(−1.8 × 105 V − 0) = −1.08 J 22.4 Electric Potential Due to a Point Charge 745 (c) The total electric potential energy of the three charges is: q1 q3 q2 q3 q1 q2 + + d d d [(2)(−4) + (2)(6) + (−4)(6)] × 10−12 C2 = (9 × 109 N.m2 /C2 ) = −1.8 J 0.1 m U=k 22.5 Electric Potential Due to a Dipole As introduced in Chap 20, an electric dipole consists of a positive charge q+ = +q and an equal-but-opposite negative charge q− = −q separated by a distance 2a Let us find the electric potential at a point P in the xy-plane as shown in Fig 22.11a We assume that V+ is the electric potential produced at P by the positive charge, and that V− is the electric potential produced at P by the negative charge y y P(x,y) r r r r q - ( a,0) r r r y q + O (a,0) xa x a q x q - + O x 2a (b) (a) Fig 22.11 (a) The electric potential V at point P(x, y) due to an electric dipole located along the x-axis with a length 2a Point P is at a distance r from the midpoint O of the dipole, where OP makes an angle θ with the dipole’s x-axis (b) When P is very far, the lines of length r+ and r− are approximately parallel to the line OP The total electric potential at P is thus: V = V+ + V− = k q+ q− 1 +k = kq − r+ r− r+ r− = kq r − − r+ r+ r− (22.32) = (x − a)2 + y2 and r = (x + a)2 + From the geometry of Fig 4.11a, we find that r+ − y2 Accordingly, Eq 22.32 becomes: V = kq (x − a)2 + y2 − (x + a)2 + y2 (22.33) 746 22 Electric Potential Note that V = ∞ at P(x = a, y = 0) and V = −∞ at P(x = −a, y = 0) Figure 22.12 shows a plot of the general shape of V in the xy-plane Fig 22.12 A computer- V y generated plot of the electric potential V in the xy-plane for an electric dipole The predicted infinite values of V are not plotted x Because naturally occurring dipoles have very small lengths, such as those possessed by many molecules, we are usually interested only in points far away from the dipole, i.e r 2a Considering these conditions, we find from Fig 22.11b that: r− − r+ 2a cos θ, and r − r+ r2 (22.34) When substituting these approximate quantities in Eq 22.32, we find that: V = kq 2a cos θ r2 (r 2a) (22.35) As introduced in Chap 20, the product of the positive charge q and the length of the dipole 2a is called the magnitude of the electric dipole moment, p = 2aq The direction of → p is taken to be from the negative charge to the positive charge of the → dipole, i.e → p = p i This indicates that the angle θ is measured from the direction of → p Using this definition, we have: V =k p cos θ r2 (r 2a) (22.36) Example 22.6 Find the electric potential along the axis of the electric dipole at the four points A, B, C, and D in Fig 22.13 22.5 Electric Potential Due to a Dipole 747 y D -q C - B +q A + x O (-a,0) (a,0) Fig 22.13 Solution: We use Eq 22.32 with r+ and r− as the distance from each point to the positive and negative charges, respectively: (1) For point A in Fig 22.13, we have x > a Therefore, r+ = x − a and r− = a + x The electric potential VA is: 1 1 − = kq − r+ r− x−a a+x 2k qa (x a) x2 VA = kq = 2k qa x − a2 (VA positive) (2) For point B in Fig 22.13, we have < x < a Therefore r+ = a − x and r− = a + x The electric potential VB is: VB = kq 1 − r+ r− = kq 1 − a−x a+x = 2kq x − x2 a2 (VB positive) (3) For point C in Fig 22.13, we have −a < x < Therefore r+ = a − x and r− = a + x The electric potential VC is: VC = kq 1 − r+ r− = kq 1 − a−x a+x = 2kq x − x2 a2 (VC negative) (4) For point D in Fig 22.13, we have x < −a Therefore r+ = a − x and r− = −x − a The electric potential VD is: 1 − r+ r− 2kqa − (x x VD = kq 22.6 = kq 1 + a−x a+x =− 2kqa x − a2 −a) (VD negative) Electric Dipole in an External Electric Field Consider an electric dipole of electric dipole moment → p is placed in a uniform → external electric field E , as shown in Fig 22.14 Do not get confused between the 748 22 Electric Potential field produced by the dipole and this external field In addition, we assume that the → vector → p makes an angle θ with the external field E +q + 2a p o −F θ F E p ⊗ θ E - -q (b) (a) Fig 22.14 (a) An electric dipole has an electric dipole moment → p in an external uniform electric → → field E The angle between → p and E is θ The line connecting the two charges represents their rigid connection and their center of mass is assumed to be midway between them (b) Representing the electric → dipole by a vector → p in the external electric field E and showing the direction of the torque → τ into the page by the symbol ⊗ → → Figure 22.14 shows a force F , of magnitude qE in the direction of E , is exerted on → the positive charge, and a force −F , of the same magnitude but in opposite direction, is exerted on the negative charge The resultant force on the dipole is zero, but since the two forces not have the same line of action, they establish a clockwise torque → τ about the center of mass of the two charges at o The magnitude of this torque about o is: τ = (2a sin θ )F (22.37) Using F = qE and p = 2aq, we can write this torque as: τ = (2aq sin θ )E = pE sin θ (22.38) The vector torque τ→ on the dipole is therefore the cross product of the vectors → p → and E Thus: → τ→ = → p ×E (22.39) The effect of this torque is to rotate the dipole until the dipole moment → p is aligned → with the electric field E 22.6 Electric Dipole in an External Electric Field 749 Potential energy can be associated with the orientation of an electric dipole in an electric field The dipole has the least potential energy when it is in the equilibrium → orientation, which occurs when → p is along E On the other hand, the dipole has the → greatest potential energy when → p is antiparallel to E We chose the zero-potential→ energy configuration when the angle between → p and E is 90◦ According to Eq 22.3, U = UB − UA = −WAB , we can find the electric potential energy of the dipole by calculating the work done by the field from the initial orientation θ = 90◦ , where UA ≡ U(90◦ ) = 0, to any orientation θ, where UB ≡ U(θ ) In addition, we use the relation W = τ dθ, to find U(θ ) as follows: U(θ ) − U(90◦ ) = −W90◦ →θ = − θ τ dθ (22.40) 90◦ Letting U(θ ) ≡ U, U(90◦ ) = 0, τ = pE sin θ, and integrating we get: U = −pE cos θ (22.41) This relation can be written in vector form as follows: → U = −→ p •E (22.42) Equation 22.42 shows the least and greatest value of U as follows: 180° p τ max p 90° τ E p 22.7 E E Electric Potential Due to a Charged Rod For a Point on the Extension of the Rod Figure 22.15 shows a rod of length L with a uniform positive charge density λ and a total charge Q In this figure, the rod lies along the x-axis and point P is taken to be at the origin of this axis, located at a distance a from the left end When we consider a segment dx on the rod, the charge on this segment will be dq = λ dx 750 22 Electric Potential y P dx x + + + + + + + + + + + + dq L a x Fig 22.15 The electric potential V at point P due to a uniformly charged rod lying along the x-axis The electric potential due to a segment of charge dq at a distance x from P is k dq/x The total electric potential is the algebraic sum of all the segments of the rod The electric potential dV at P due to this segment is given by: dV = k λ dx dq =k x x (22.43) We obtain the total electric potential at P due to all the segments of the rod by integrating from one end of the rod (x = a) to the other (x = a + L) as follows: a+L V = dV = a λ dx = kλ k x a+L x −1 dx = kλ| ln x|a+L = kλ {ln(a + L) − ln a} a a Therefore: V = kλ ln a+L a = kQ a+L ln L a (22.44) For a Point on the Perpendicular Bisector of the Rod A rod of length L has a uniform positive charge density λ and a total charge Q The rod is placed along the x-axis as shown in Fig 22.16 Assuming that P is a point on the perpendicular bisector of the rod and is located a distance a from the origin of the x-axis, then the charge of a segment dx on the rod will be dq = λ dx The electric potential dV at P due to this segment is: dV = k λ dx dq =k r r (22.45) The total electric potential at P due to all segments of the rod is given by two times the integral of dV from the middle of the rod (x = 0) to one of its ends (x = L/2) Thus: 22.7 Electric Potential Due to a Charged Rod 751 Fig 22.16 A rod of length L y with a uniform positive charge P density λ and an electric potential dV at point P due to a a charge segment r dq + + + + + + + + + ++ + x dx L x=L/2 V =2 L/2 dV = 2kλ x=0 dx r x (22.46) To perform the integration of this expression, we relate the variables x and r From √ the geometry of Fig 22.16, we use the fact that r = x + a2 Therefore, Eq 22.46 becomes: L/2 V = 2kλ (x dx + a2 )1/2 (22.47) From the table of integrals in Appendix B, we find that: dx = ln(x + (x + a2 )1/2 V = 2kλ ln(x + x + a2 ) Thus: = 2kλ ln(L/2 + Therefore: V = 2kλ ln x + a2 ) (22.48) L/2 (L/2)2 + a2 ) − ln(a) L/2 + (L/2)2 + a2 a (22.49) When we use the fact that the total charge Q = λL, we get: V = 2kQ L/2 + ln L (L/2)2 + a2 a (22.50) 752 22 Electric Potential For a Point Above One End of the Rod When a point P is located at a distance a from one of the rod’s ends, see Fig 22.17, we can perform similar calculations to find that: V = L+ kQ ln L √ L + a2 a Fig 22.17 A setup similar to y Fig 22.16 except P is above P one end a 22.8 (22.51) r dq ++++ +++++ +++ x dx L x Electric Potential Due to a Uniformly Charged Arc Assume that a rod has a uniformly distributed total positive charge Q Assume now that the rod is bent into an arc of radius R and central angle φ rad, see Fig 22.18a To find the electric potential at the center P of this arc, we first let λ represent the linear charge density of this arc, which has a length Rφ Thus: λ= Q Rφ (22.52) For an arc element ds subtending an angle dθ at P, we have: ds = R dθ (22.53) Therefore, the charge dq on this arc element will be given by: dq = λ ds = λ R dθ (22.54) To find the electric field at P, we first calculate the differential electric potential dV at P due to the element ds of charge dq, see Fig 22.18b, as follows: dV = k dq = kλ dθ R (22.55)