21.5 Exercises 723 Fig 21.28 See Exercise (8) S2 r q S1 r 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 22.1 Electric Potential Energy 733 electron’s displacement is upwards and the electric field is directed downwards, i.e θ = 180◦ , then we have: B U = −q → E • d→ s = −(−e) A B E ds cos 180◦ A B = −eE ds = −eE(sB − sA ) = −eEs A = −(1.6 × 10−19 C)(100 N/C)(50 m) = −8 × 10−16 J Fig 22.1 - s F e B - A E Notice that the sign of the electron’s charge is used in this calculation Thus, during the 50 m ascent of the electron, the electric potential of the electron decreases by × 10−16 J Also, from Eq 22.3, the work done by the electrostatic force on the electron is: WAB = − U = × 10−16 J 22.2 Electric Potential The electric potential energy depends on the charge q However, the electric potential energy per unit charge U/q has a unique value at any point and depends only on the electric field (or alternatively on the source-charge distribution) This quantity is called the electric potential V (or simply the potential) Thus: V = U q (22.5) 734 22 Electric Potential This equation implies that the electric potential is a scalar quantity Spotlight Electric potential is a scalar quantity that characterizes an electric field and is independent of any charge that may be placed in the field The electric potential difference V (or simply the potential difference) between two points A and B in an electric field is defined as the difference in the electric potential energy per unit charge between the two points Thus, dividing Eq 22.3 by q leads to: V = VB − VA = WAB U =− =− q q B → E • d→ s (22.6) A It is clear that the electric potential difference between A and B depends only on the source-charge distribution, and is equal to the negative of the work done by the electrostatic force per unit charge The SI unit of both the electric potential and the electric potential difference is joule per coulomb, or volt (abbreviated by V) That is: V = J/C (22.7) From this unit, we see that the joule is one coulomb times V (J = CV) Additionally, the volt unit allows us to adopt a more convenient unit for the electric field From Eq 22.6, the volt unit also has the units of electric field times distance Then we have: N/C = V/m (22.8) Spotlight Electric field can be expressed as the rate of change of electric potential with position From now on, we shall express values of electric fields in volts per meter (V/m) rather than newtons per coulomb (N/C) 22.2 Electric Potential 735 Suppose we move a particle of an arbitrary charge q from point A to point B by applying a force on it and changing its kinetic energy The applied force must perform work WAB (app), while the electric field does work WAB From Eq 6.54, the change in the kinetic energy of the particle is: K = WAB (app) + WAB (22.9) Now suppose the kinetic energy of the particle does not change during the move Then, Eq 22.9 reduces to: WAB (app) = −WAB (22.10) Since Eq 22.6 gives q V = −WAB , then we have: WAB (app) = q V (22.11) In atomic and nuclear physics, we use a convenient unit of energy called electronvolt (eV) One electron-volt (1 eV) is the energy equal to the work required to move a single elementary charge e, such as that of an electron or a proton, through an electric potential difference of exactly one volt (1 V) To find the value of this unit, one can use Eq 22.11 where q = 1.6 × 10−19 C and V = V, Thus: eV = (1.6 × 10−19 C)(1 V) = (1.6 × 10−19 C)(1 J/C) = 1.6 × 10−19 J (22.12) An electron that strikes the screen of a typical TV set may have a speed of × 107 m/s This corresponds to a kinetic energy of 7.28 × 10−16 J, which is equivalent to 4.55 keV In other words, such an electron has to be accelerated from rest through a potential difference of 4.55 kV to reach a speed of × 107 m/s Equations 22.3 and 22.6 hold true for both discrete and continuous source distributions, and for both uniform and varying fields In the following sections, we calculate the electric potential for various cases 22.3 Electric Potential in a Uniform Electric Field Displacement Parallel to a Field Let us calculate the potential difference between two points A and B separated by a → distance |→ s | = d, where → s is a displacement along a uniform field E , see Fig 22.2 Then, Eq 22.6 gives: 736 22 Electric Potential Fig 22.2 A uniform electric A field pointing downwards and → d→ s is along E The electric ds d potential at point B is lower than that at point A B E B V = VB − VA = − B → E • d→ s =− A B (E cos 0◦ )ds = −E A ds (22.13) A Integrating ds along a straight line gives (sB − sA ) = d Thus: V = VB − VA = −Ed (Along the field) (22.14) The negative sign means that the electric potential at point B is lower than that at point A, i.e VB < VA In fact, electric field lines always point in the direction of decreasing electric potential If a charge q moves along the field from point A to point B, then according to Eq 22.6, the change in the electric potential energy U of the charge-field system is given by U = q V Then, substituting with Eq 22.14, we get the following: U = q V = −qEd (Along the field) (22.15) (1) If q is positive (i.e q = +|q|) and the charge moves in the direction of the field from A to B, then U = −|q|Ed is negative This means that the positive chargefield system loses electric potential energy Also, since WAB = − U, this means that the electric field does positive work WAB on the positive charge during this motion Furthermore, if this positive charge is released from rest at point A, it accelerates downwards, gaining kinetic energy K = WAB when it reaches point B Thus: K = − U = +|q|Ed (Along the field for positive q) (22.16) (2) If q is negative (i.e q = −|q|), and the charge moves in the direction of the field from A to B, then U = +|q|Ed is positive This means that the negative chargefield system gains electric potential energy In order for the negative charge to move along the field, an external agent must apply a force and positive work WAB (app) during that motion For motion with zero acceleration from A to B, 22.3 Electric Potential in a Uniform Electric Field 737 this positive work compensates the negative work WAB done on the negative charge by the field Similar steps can be performed when the displacement → s is opposite to the field from point A to point B, see Fig 22.3, which leads to: V = VB − VA = Ed (Opposite the field) (22.17) Fig 22.3 Same as Fig 22.2 except that d → s is opposite B → to E d ds A E If a charge q moves against the field from A to B, then according to Eq 22.6, the change in the electric potential energy U of the charge-field system is given by U = q V By substitution with Eq 22.17 we get the following: U = q V = qEd (Opposite the field) (22.18) (1) If q is positive (i.e q = +|q|), and the charge moves against the field from A to B, then U = +|q|Ed is positive This means that the positive charge-field system gains electric potential energy In order for the positive charge to move against the field, an external agent must apply a force and positive work WAB (app) during that motion For motion with zero acceleration from A to B, this positive work compensates for the negative work WAB done on the positive charge by the electric field (2) If q is negative (i.e q = −|q|) and is released from rest at A, it will accelerate upwards to B In this case V = VB −VA = +Ed, and U = −|q| V = −|q|Ed (loss of electric potential energy) Since WAB = − U, then WAB = +|q|Ed, i.e the electric field does positive work on the negative charge during its motion from A to B The gain in kinetic energy K = − U = +|q|Ed K = WAB is thus: (Opposite the field for negative q) This gain is the same as for a positive charge moving along the field (22.19) 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 = − → → E • d→ s = −E • A B 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= 12 V |VB − VA | = = 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: → q ˆ → E • d→ s = Er → rˆ • d → s =k → r • ds r (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 [...]... difference of 4.55 kV to reach a speed of 4 × 107 m/s Equations 22 .3 and 22 .6 hold true for both discrete and continuous source distributions, and for both uniform and varying fields In the following sections, we calculate the electric potential for various cases 22 .3 Electric Potential in a Uniform Electric Field Displacement Parallel to a Field Let us calculate the potential difference between two points A. .. is positive This means that the negative chargefield system gains electric potential energy In order for the negative charge to move along the field, an external agent must apply a force and do positive work WAB (app) during that motion For motion with zero acceleration from A to B, 22 .3 Electric Potential in a Uniform Electric Field 737 this positive work compensates the negative work WAB done on the... external agent must apply a force and do positive work WAB (app) during that motion For motion with zero acceleration from A to B, this positive work compensates for the negative work WAB done on the positive charge by the electric field (2) If q is negative (i.e q = −|q|) and is released from rest at A, it will accelerate upwards to B In this case V = VB −VA = +Ed, and U = −|q| V = −|q|Ed (loss of electric... 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... A and B separated by a → distance |→ s | = d, where → s is a displacement along a uniform field E , see Fig 22 .2 Then, Eq 22 .6 gives: 736 22 Electric Potential Fig 22 .2 A uniform electric A field pointing downwards and → d→ s is along E The electric ds d potential at point B is lower than that at point A B E B V = VB − VA = − B → E • d→ s =− A B (E cos 0◦ )ds = −E A ds (22 .13) A Integrating ds along... Eq 22 .6 gives q V = −WAB , then we have: WAB (app) = q V (22 .11) In atomic and nuclear physics, we use a convenient unit of energy called electronvolt (eV) One electron-volt (1 eV) is the energy equal to the work required to move a single elementary charge e, such as that of an electron or a proton, through an electric potential difference of exactly one volt (1 V) To find the value of this unit, one... distribution, and is equal to the negative of the work done by the electrostatic force per unit charge The SI unit of both the electric potential and the electric potential difference is joule per coulomb, or volt (abbreviated by V) That is: 1 V = 1 J/C (22 .7) From this unit, we see that the joule is one coulomb times 1 V (J = CV) Additionally, the volt unit allows us to adopt a more convenient unit for the... to find the magnitude of the uniform electric field as follows: E= 12 V |VB − VA | = = 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 = 5 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... From Eq 22 .6, the volt unit also has the units of electric field times distance Then we have: 1 N/C = 1 V/m (22 .8) Spotlight Electric field can be expressed as the rate of change of electric potential with position From now on, we shall express values of electric fields in volts per meter (V/m) rather than newtons per coulomb (N/C) 22 .2 Electric Potential 735 Suppose we move a particle of an arbitrary... coordinates rA and rB B dr r + q A rA → From Fig 22 .7, we see that → rˆ • d → s = 1 × ds × cos θ = dr Then, E • d → s = Er dr, and Eq 22 .23 can be written as: 7 42 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 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