1. Electric Potential Chapter 24

2. 24-2 Electric Potential Energy When an electrostatic force acts between two or more charged particles within a system of particles, we can assign an electric potential energy U to the system. If the system changes its configuration from an initial state i to a different final state f, the electrostatic force does work W on the particles. We then know that the resulting change U in the potential energy of the system is: 1436-1437

3. 24-2 Electric Potential Energy Suppose a charged particle within the system moves from point i to point f while an electrostatic force between it and the rest of the system acts on it. The work W done by the force on the particle is the same for all paths between points i and f. The work done by the electrostatic force is path independent. Suppose that several charged particles come together from initially infinite separations (state i) to form a system of neighboring particles (state f ). Let the initial potential energy U i be zero, and let W∞ represent the work done by the electrostatic forces between the particles during the move in from infinity. Then the final potential energy U of the system is 1436-1437

4. CHECKPOINT 1 (a) negative; (b) increase 1436-1437

5. Sample Problem (p629) 1436-1437

6. 24-3 Electric Potential The potential energy of a charged particle in an electric field depends on the charge magnitude. However, the potential energy per unit charge has a unique value at any point in an electric field. The potential energy per unit charge at a point in an electric field is called the electric potential V (or simply the potential) at that point. Note that electric potential is a scalar, not a vector 1436-1437

7. 24-3 Electric Potential electric potential difference The electric potential difference V between any two points i and f in an electric field is equal to the difference in potential energy per unit charge between the two points Work by the Field. The work W done by the electric force as the particle moves from i to f: 1436-1437

8. 24-3 Electric Potential electric potential difference The electric potential V at a point P in the electric field of a charged object is where W ∞ is the work that would be done by the electric force on a positive test charge q 0 were it brought from an infinite distance to P, and U is the electric potential energy that would then be stored in the test charge–object system. (a)A test charge has been brought in from infinity to point P in the electric field of the rod. (b)We define an electric potential V at P based on the potential energy of the configuration in (a). If a particle with charge q is placed at a point where the electric potential of a charged object is V, the electric potential energy U of the particle–object system is 1436-1437

9. 24-3 Electric Potential SI unit for potential The SI unit for potential is the joule per coulomb. This combination occurs so often that a special unit, the volt (abbreviated V), is used to represent it. Thus, 1 volt = 1 joule per coulomb. This new unit allows us to adopt a more conventional unit for the electric field which we have measured up to now in newtons per coulomb. With two unit conversions, we obtain 1436-1437

10. 24-3 Electric Potential Finally, we can now define an energy unit that is a convenient one for energy measurements in the atomic and subatomic domain: One electron-volt (eV) is the energy equal to the work required to move a single elementary charge e, such as that of the electron or the proton, through a potential difference of exactly one volt 1436-1437

11. 24-3 Electric Potential: Work Done by an Applied Force Suppose we move a particle of charge q from point i to point f in an electric field by applying a force to it. During the move, our applied force does work W app on the charge while the electric field does work W on it. By the work–kinetic energy theorem, the change ΔK in the kinetic energy of the particle is Now suppose the particle is stationary before and after the move. Then K f and K i are both zero In words, the work W app done by our applied force during the move is equal to the negative of the work W done by the electric field—provided there is no change in kinetic energy. 1436-1437

12. 24-3 Electric Potential Work Done by an Applied Force W app can be positive, negative, or zero depending on the signs and magnitudes of q and V. 1436-1437

13. CHECKPOINT 2 (a) positive; (b) higher 1436-1437

14. 24-4 Equipotential Surfaces Adjacent points that have the same electric potential form an equipotential surface, which can be either an imaginary surface or a real, physical surface. No net work W is done on a charged particle by an electric field when the particle moves between two points i and f on the same equipotential surface Figure shows a family of equipotential surfaces associated with the electric field due to some distribution of charges. The work done by the electric field on a charged particle as the particle moves from one end to the other of paths I and II is zero because each of these paths begins and ends on the same equipotential surface and thus there is no net change in potential. The work done as the charged particle moves from one end to the other of paths III and IV is not zero but has the same value for both these paths because the initial and final potentials are identical for the two paths; that is, paths III and IV connect the same pair of equipotential surfaces. 1436-1437

15. 24-4 Equipotential Surfaces E( the magnitude of the electric filed) = constant For Equipotential surface V(the electric potential ) = constant W (the work done) = zero 1436-1437

16. 24-5 Calculating the Potential from the Field A test charge q 0 moves from point i to point f along the path shown in a non-uniform electric field. During a displacement ds, an electric force q 0 E acts on the test charge. This force points in the direction of the field line at the location of the test charge. We can calculate the potential difference between any two points i and f in an electric field if we know the electric field vector all along any path connecting those points. To make the calculation, we find the work done on a positive test charge by the field as the charge moves from i to f, and then use 1436-1437

17. 24-5 Calculating the Potential from the Field A test charge q 0 moves from point i to point f along the path shown in a non-uniform electric field. During a displacement ds, an electric force q 0 E acts on the test charge. This force points in the direction of the field line at the location of the test charge. 1436-1437

18. 24-5 Calculating the Potential from the Field The electric potential difference between two points i and f is A test charge q 0 moves from point i to point f along the path shown in a non-uniform electric field. During a displacement ds, an electric force q 0 E acts on the test charge. This force points in the direction of the field line at the location of the test charge. If we choose V i = 0, we have, for the potential at a particular point, Thus, the potential difference V f -V i between any two points i and f in an electric field is equal to the negative of the line integral (meaning the integral along a particular path) of from i to f. However, because the electrostatic force is conservative, all paths (whether easy or difficult to use) yield the same result. 1436-1437

19. CHECKPOINT (a) rightward; (b) 1, 2, 3, 5: positive; 4, negative; (c) 3, then 1, 2, and 5 tie, then 4 1436-1437

23. Sample Problem (p634) 1436-1437

24. Sample Problem (P634) 1436-1437

25. 24-6 Potential due to a Charged Particle Consider a point P at distance R from a fixed particle of positive charge q. We imagine that we move a positive test charge q 0 from point P to infinity. Because the path we take does not matter, let us choose the simplest one a line that extends radially from the fixed particle through P to infinity 1436-1437

26. 24-6 Potential due to a Charged Particle We know that the electric potential difference between two points i and f is 1436-1437

27. 24-6 Potential due to a Charged Particle the electric potential V due to a particle of charge q at any radial distance r from the particle. Solving for V and switching R to r, we get We set V f =0 (at ∞) and V i =V (at R), For the magnitude of the electric field at the site of the test charge 1436-1437

28. 24-6 Potential due to a Charged Particle Equation 24-26 also gives the electric potential either outside or on the external surface of a spherically symmetric charge distribution. We can prove this by using one of the shell theorems of Sections 21-4 and 23- 9 to replace the actual spherical charge distribution with an equal charge concentrated at its center. Then the derivation leading to Eq. 24-26 follows, provided we do not consider a point within the actual distribution. 1436-1437

29. 24- 7 Potential due to a group of Charged Particles Thus, the potential is the algebraic sum of the individual potentials, with no consideration of directions. t The potential due to a collection of charged particles is 1436-1437

30. CHECKPOINT 4 Answer: Same net potential (a)=(b)=(c) 1436-1437

31. Sample Problem (P 636) 1436-1437

32. Sample Problem (P637) 1436-1437

33. 24-9 Potential due to a Continuous Charge Distribution For a continuous distribution of charge (over an extended object), the potential is found by (1)dividing the distribution into charge elements dq that can be treated as particles and then (2) summing the potential due to each element by integrating over the full distribution: We now examine two continuous charge distributions, a line and a disk. 1436-1437

34. 24-9 Potential due to a Continuous Charge Distribution Line of Charge Fig. a has a thin conducting rod of length L. As shown in fig. b the element of the rod has a differential charge of This element produces an electric potential dV at point P (fig c) given by We now find the total potential V produced by the rod at point P by integrating dV along the length of the rod, from x = 0 to x = L (Figs.d and e) 1436-1437

36. 24-9 Potential due to a Continuous Charge Distribution Charged Disk In figure, consider a differential element consisting of a flat ring of radius R’ and radial width dR’. Its charge has magnitude in which (2πR’)(dR’) is the upper surface area of the ring. The contribution of this ring to the electric potential at P is We find the net potential at P by adding (via integration) the contributions of all the rings from R’=0 to R’=R: A plastic disk of radius R, charged on its top surface to a uniform surface charge density σ. We wish to find the potential V at point P on the central axis of the disk. 1436-1437

37. Note that the variable in the second integral is R’ and not z 1436-1437

38. 24-10 Calculating the Field from the Potential Suppose that a positive test charge q 0 moves through a displacement ds from one equipotential surface to the adjacent surface. The work the electric field does on the test charge during the move is – q 0 dV. On the other hand the work done by the electric field may also be written as the scalar product (q 0 E)  ds. Equating these two expressions for the work yields or, Since E cosθ is the component of E in the direction of ds, we get, A test charge q 0 moves a distance ds from one equipotential surface to another. (The separation between the surfaces has been exaggerated for clarity.) The displacement ds makes an angle θ with the direction of the electric field E. 1436-1437

39. 24-10 Calculating the Field from the Potential 1436-1437

40. Sample Problem (P642) 1436-1437

41. 24-11 Electric Potential Energy of a System of Charged Particles In Section 24-2, we discussed the electric potential energy of a charged particle as an electrostatic force does work on it. In that section, we assumed that the charges that produced the force were fixed in place, so that neither the force nor the corresponding electric field could be influenced by the presence of the test charge 1436-1437

42. Two charges held a fixed distance r apart. The figure shows two point charges q 1 and q 2, separated by a distance r. To find the electric potential energy of this two-charge system, we must mentally build the system, starting with both charges infinitely far away and at rest. When we bring q 1 in from infinity and put it in place, we do no work because no electrostatic force acts on q 1. However, when we next bring q 2 in from infinity and put it in place, we must do work because q 1 exerts an electrostatic force on q 2 during the move. We can calculate that work by dropping the minus sign (so that the equation gives the work we do rather than the field’s work) and substituting q 2 for the general charge q. Our work is then equal to q 2 V where V is the potential that has been set up by q 1 at the point where we put q 2. 24-11 Electric Potential Energy of a System of Charged Particles 1436-1437

43. The electric potential energy of a system of charged particles is equal to the work needed to assemble the system with the particles initially at rest and infinitely distant from each other. For two particles at separation r, Two charges held a fixed distance r apart. 1436-1437

44. Sample Problem 1436-1437

45. 24 Summary Electric Potential The electric potential V at point P in the electric field of a charged object: Electric Potential Energy Electric potential energy U of the particle-object system: If the particle moves through potential ΔV: Eq. 24-2 Eq. 24-3 Eq. 24-4 Mechanical Energy Applying the conservation of mechanical energy gives the change in kinetic energy: In case of an applied force in a particle In a special case when ΔK=0: Eq. 24-9 Eq. 24-11 Eq. 24-12 Finding V from E The electric potential difference between two point I and f is: Eq. 24-18 1436-1437

46. 24 Summary Potential due to a Charged Particle due to a single charged particle at a distance r from that particle : due to a collection of charged particles Eq. 24-26 Eq. 24-27 Potential due to a Continuous Charge Distribution For a continuous distribution of charge: Eq. 24-32 Eq. 24-40 Calculating E from V The component of E in any direction is: Eq. 24-46 Electric Potential Energy of a System of Charged Particle For two particles at separation r: 1436-1437

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