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25 CHAPTER ELECTRIC POTENTIAL ActivPhysics can help with these problems: Activities 11.9, 11.10 Section 25-2: Potential Difference Problem 1. How much work does it take to move a 50- C charge against a 12-V potential difference? Solution The potential difference and the work per unit charge, done by an external agent, are equal in magnitude, so W = q V = (50 C)(12 V) = 600 J. (Note: Since only magnitudes are needed in this problem, we omitted the subscripts A and B.) Problem 2. The potential difference between the two sides of an ordinary electrical outlet is 120 V. How much energy does an electron gain when it moves from one side to the other? Solution Moving from the negative to the positive side (i.e., opposite to the electric field), an electron gains e V = 120 eV = (1.6 10 19 C)(120 V ) = 1.92 10 17 J of energy. Problem 3. It takes 45 J to move a 15-mC charge from point A to point B. What is the potential difference VAB ? Solution The work done by an external agent equals the potential energy change, U AB = 45 J = q VAB , hence VAB = 45 J= mC = 3 kV. (Since the work required to move the charge from A to B is positive, VB > VA and VAB is positive.) 15 Problem 4. Show that 1 V/m is the same as 1 N/C. Solution Problem 5. Since a volt = joule/coulomb = newton-meter/coulomb, it follows that a V/m = N/C. Find the magnitude of the potential difference between two points located 1.4 m apart in a uniform 650 N/C electric field, if a line between the points is parallel to the field. Solution For in the direction of a uniform electric field, Equation 25-2b gives V = E = (650 N/C)(1.4 m ) = 910 V. (See note in solution to Problem 1. Since dV = E d, the potential always decreases in the direction of the electric field.) Problem 6. . A charge of 31 C moves from the positive to the negative terminal of a 9.0-V battery. How much energy does the battery impart to the charge? Solution UAB = q VAB = (31 C)(9.0 V) = 27.9 J. . 2 CHAPTER 25 Problem 7. Two points A and B lie 15 cm apart in a uniform electric field, with the path AB parallel to the field. If the potential difference VAB is 840 V, what is the field strength? Solution 0.15 m = 5.60 kV/m. (See notes in solutions to Problems 1 Equation 25-2b for a uniform field gives E = V= = 840 V= and 5.) Problem 8. Figure 25-37 shows a uniform electric field of magnitude E. Find expressions for (a) the potential difference VAB and (b) VBC . (c) Use your result to determine VAC . FIGURE 25-37 Problem 8. Solution (a) On the line A to B, d is antiparallel to E, so Equation 25-2 gives VB VA = B E d = E B d = Ed . (b) The line A A B to C makes an angle of 45 with E, so VC VB = E C cos 45 d = Ed= 2 . (c) Addition yields VC VA = B VC VB + VB VA = Ed (1 1= 2 ) = 0.293 Ed . z z z Problem 9. A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a potential difference of 100 V. Find the energy each gains. The energy gained is q V (see Example 25-1). The proton and singly-ionized helium atom have charge e, so they gain 100 eV = (1.6 10 19 C)(100 V) = 1.6 10 17 J, while the -particle has charge 2e and gains twice this energy. Solution Problem 10. The electric field within the membrane separating the inside and outside of a biological cell is approximately 8.0 MV/m, and is essentially uniform. If the membrane is 10 nm thick, what is the potential difference across the membrane? Solution For a uniform electric field parallel to the thickness of the membrane, Equation 25-2b gives a potential difference of magnitude E = (8.0 MV/m )(10 nm ) = 80 mV. Problem 11. What is the potential difference between the terminals of a battery that can impart 7.2 10 19 J to each electron that moves between the terminals? Solution q 1 The magnitude of the potential difference is W= = 7.2 10 19 J=.6 10 19 C = 4.5 V. (The energy imparted per electron 4.5 eV.) is CHAPTER 1 L 3 Problem 12. Electrons in a TV tube are accelerated from rest through a 25-kV potential difference. With what speed do they hit the TV screen? Solution The work done on an electron equals the change in its kinetic energy, W = e V = v= 2e V= = m 1 2 mv2 (if it starts from rest). Thus, 2(1.6 10 19 C)(25 10 3 V) = 9.37 10 7 m/s. 31 (9.11 10 kg) Problem 13. A 12-V car battery stores 2.8 MJ of energy. How much charge can move between the battery terminals before it is totally discharged? Assume the potential difference remains at 12 V, an assumption that is not realistic. Solution A charge q, moving through a potential difference V, is equivalent to electrostatic potential energy U = q V , stored in the battery. Thus, q = 2.8 MJ= V = 2.33 10 5 C. 12 Problem 14. What is the charge on an ion that gains 1.6 10 15 J when it moves through a potential difference of 2500 V? Solution The energy gained is 1.6 10 15 J = 10 4 eV, so the charge (really its magnitude) on the ion is q = W= V = 4 10 eV= 2500 V = 4e. Problem 15. Two large, flat metal plates are a distance d apart, where d is small compared with the plate size. If the plates carry surface charge densities , show that the potential difference between them is V = d= 0 . Solution The electric field between the plates is uniform, with E = = 0 , directed from the positive to the negative plate (see last paragraph of Section 24-6 and Fig. 24-35). Then Equation 25-2b gives V = V+ V = (= 0 )( d ) = d= 0 (the displacement from the negative to the positive plate is opposite to the field direction). Problem 16. An electron passes point A moving at 6.5 Mm/s. At point B the electron has come to a complete stop. Find the potential difference VAB . Solution The work-energy theorem (for an electron under the influence of just an electric force) gives WAB = q VAB = K = K A , where WAB is the work done by the electric field (also equal to U AB ), and K B = 0. Thus, VAB = m v2 KA (9.11 10 31 kg)(6.5 10 6 m/s) 2 = eA= = 120 V. q 2( e ) 2( 1.6 10 19 C) (To stop an electron, a negative potential difference must be applied.) Problem 17. A 5.0-g object carries a net charge of 3.8 C. It acquires a speed v when accelerated from rest through a potential difference V. A 2.0-g object acquires twice the speed under the same circumstances. What is its charge? 4 CHAPTER 25 Solution The speed acquired by a charge q, starting from rest at point A and moving through a potential difference of V, is given by 2 1 2V (q= ). (This is the work-energy theorem for the electric force. A positive charge is m 2 mvB = q(V A VB ) = qV , or vB = accelerated in the direction of decreasing potential.) If the second object acquires twice the speed of the first object, moving m through the same potential difference, it must have four times the charge to mass ratio, q= . Thus, q2 = 4(q1= 1 )m2 = 4(3.8 C)(2g= g) = 6.08 C. m 5 Section 25-3: Calculating Potential Difference Problem 18. An electric field is given by E = E0 , where E0 is a constant. Find the potential as a function of position, taking j V = 0 at y = 0. Solution Problem Since this is a uniform field, Equations 25-2a or b give V y af V (0) = V ( y) = E = ( E j) (r 0) = E y. 0 0 19. The classical picture of the hydrogen atom has a single electron in orbit a distance 0.0529 nm from the proton. Calculate the electric potential associated with the protons electric field at this distance. Solution The potential of the proton, at the position of the electron (both of which may be regarded as point-charge atomic a constituents) is (Equation 25-4) V = ke= 0 , where a0 is the Bohr radius. Numerically, V = (9 10 9 N m 2 /C 2 ) 19 11 (1.6 10 C)=5.29 10 m ) = 27.2 V. (The energy of an electron in a classical, circular orbit, around a stationary ( proton, is one half its potential energy, or 1 U = 1 ( e)V = 13.6 eV. The excellent agreement with the ionization energy of 2 2 hydrogen was one of the successes of the Bohr model.) Problem 20. Earth carries an electric charge of 4.3 10 5 C, distributed essentially uniformly over its surface. What is the potential difference between Earths surface and the base of the ionosphere, about 80 km above the surface? Solution Outside a uniformly charged sphere, Equation 25-4 can be used: 1 1 Vsurf ion = kQ(rion rsurf ) = (9 GN m 2 /C 2 )( 4.3 10 5 C)[(6450 km ) 1 (6370 km ) 1 ] = 7.54 MV. Problem 21. Points A and B lie 20 cm apart on a line extending radially from a point charge Q, and the potentials at these points are VA = 280 V, VB = 130 V. Find Q and the distance r between A and the charge. Solution r r V 130 Since VA = kQ=A and VB = kQ=B , division yields rB = (VA= B )rA = (280= )rA = 2.15rA . But rB rA = 20 cm, so 1 rA = (20 cm )(2.15 1) = 17.3 cm. Then Q = VA rA= = (280 V)(17.3 cm )=9 10 9 N m 2 /C 2 ) = 5.39 nC. k ( Problem 22. What is the maximum potential allowable on a 5.0-cm-diameter metal sphere if the electric field at the spheres surface is not to exceed the 3 MV/m breakdown field in air? Solution R For an isolated metal sphere, the potential at the surface is V = kQ= , while the electric field strength at the surface is CHAPTER 1 L 5 R kQ= 2 = V= . Thus, V= 3 MV/m implies V (3 MV/m )( 1 0.05 m ) = 75 kV. R R 2 Problem 23. A 3.5-cm-diameter isolated metal sphere carries a net charge of 0.86 C. (a) What is the potential at the spheres surface? (b) If a proton were released from rest at the spheres surface, what would be its speed far from the sphere? Solution (a) An isolated metal sphere has a uniform surface charge density, so Equation 25-4 gives the potential at its surface, Vsurf = kQ= = (9 GN m 2 /C 2 )(0.86 C)=1 3.5 cm ) = 442 kV. (b) The work done by the repulsive electrostatic field R (2 (the negative of the change in the protons potential energy) equals the protons kinetic energy at infinity, W = 2 e(V Vsurf ) = eVsurf = 1 mv2 . Then v = 2eVsurf = = [2(1.6 10 19 C)( 442 kV)=1.67 10 27 kg)]1= = 9.21 Mm/s. m ( 2 Problem 24. A sphere of radius R carries a negative charge of magnitude Q, distributed in a spherically symmetric way. Find the escape speed for a proton at the spheres surfacethat is, the speed that would enable the proton to escape to arbitrarily large distances. Solution The work done by the electric field, when a proton escapes from the surface to an infinite distance, equals the change in kinetic energy, or e(V Vsurf ) = eVsurf = K Ksurf = 1 mv2 . (We assumed zero kinetic energy for the proton at 2 R mR infinity, and that the sphere is stationary.) For a uniformly negatively charged sphere, Vsurf = kQ= , so v = 2keQ= . Problem 25. A thin spherical shell of charge has radius R and total charge Q distributed uniformly over its surface. What is the potential at its center? Solution R The potential at the surface of the shell is kQ= (as in Example 25-3). The electric field inside a uniformly charged shell is zero, so the potential anywhere inside is a constant, equal, therefore, to its value at the surface. Problem 26. A solid sphere of radius R carries a net charge Q distributed uniformly throughout its volume. Find the potential difference from the spheres surface to its center. Hint: Consult Example 24-1. Solution The electric field inside a uniformly charged sphere is radially symmetric with strength E = kQr= 3 . Then V ( R) V (0) = R R z( kQr= 3 )dr = kQ= R. (The potential is higher at the center if Q is positive.) R 2 0 Problem 27. Find the potential as a function of position in an electric field given by E = ax , where a is a constant and where V = 0 at x = 0. Solution r r x Since V (0) = 0, V (r) = z E dr = z ax dr = z ax dx = 1 ax 2 . 0 0 0 2 Problem 28. A coaxial cable consists of a 2.0-mm-diameter inner conductor and an outer conductor of diameter 1.6 cm and negligible thickness (Fig. 25-38). If the conductors carry line charge densities 0.56 nC/m, what is the magnitude of the potential difference between them? 6 CHAPTER 25 Solution If the length of the cable is long compared to its outer diameter, the electric field inside, away from the ends, will be due to the inner conductor only (apply cylindrical symmetry and Gausss law). The potential difference is given by Equation 25-5. V = r ln 2 2 0 r1 F I = 2(9 10 GJ HK 9 N m 2 /C 2 )(0.56 10 9 C/m ) ln 1 F.6 I = 21.0 V. H2 K 0. Problem 29. The potential difference between the surface of a 3.0-cm-diameter power line and a point 1.0 m distant is 3.9 kV. What is the line charge density on the power line? Solution If we approximate the potential from the line by that from an infinitely long charged wire, Equation 25-5 can be used to find : = 2 0 VAB= rA=B ) = (3.9 kV)[2(9 10 9 N m 2 /C 2 ) ln(100=.5)]1 = 51.6 nC/m. (Note: VAB = VB VA so ln( r 1 B is at the surface of the wire and A is 100 cm distant.) Problem 30. Three equal charges q form an equilateral triangle of side a. Find the potential at the center of the triangle. Solution 2 The center is equidistant from each vertex, and r = a= cos 30 = a= 3 . Each charge contributes equally to the potential, so V = 3kq= = 3 3kq=. r a Problem 30 Solution. Problem 31. A charge +Q lies at the origin, and 3Q at x = a. Find two points on the x-axis where V = 0. Solution r The potential on the x-axis, kqi =i = kQ=x + k ( 3Q)=x a (from Equation 25-6), is zero when 3 x = x a . For x < 0, this implies 3 x = a x, or x = a= . Note that the absolute value of a negative number is minus the number, and 2 0 < x < a, the condition is 3 x = a x, or x = a= . For x > a, there are no 4 we assume that a is a positive constant. For 2 2 solutions. (The same results follow from the quadratic 8 x + 2 ax a = 0, which results from the square of the above condition.) Problem 32. Two identical charges q lie on the x-axis at a. (a) Find an expression for the potential at all points in the x-y plane. (b) Show that your result reduces to the potential of a point charge for distances large compared with a. Solution 2 ] = kq{[( x a) 2 + y 2 ]1= + 2 r [( x + a) 2 + y 2 ]1= }. (b) If r 2 = x 2 + y 2 a, a can be neglected relative to x or y, so V (r ) 2 kq=, which is the potential of a point charge of magnitude 2q. (a) Equation 25-6 and some geometry give V (r) = kqi =i = kq[ r a r 1 + r + a 1 CHAPTER 1 L 7 Problem 33. Find the potential 10 cm from a dipole of moment p = 2.9 nC m (a) on the dipole axis, (b) at 45 to the axis, and (c) on the perpendicular bisector. The dipole separation is much less than 10 cm. Solution Equation 25-7 gives the potential from a point dipole as a function of distance and angle from the dipole axis. For the dipole moment and distance given, V (r, ) = kp cos = 2 = (9 GN m 2 /C 2 )(2.9 nC-m ) cos =10 cm ) 2 = (2.61 kV)cos . For the r ( three given angles, (a) V = (2.61 kV) cos 0 = 2.61 kV; (b) V = (2.61 kV) cos 45 = 1.85 kV; and (c) V = (2.61 kV) cos 90 = 0. Problem 34. Two points A and B lie 55 cm from a dipole of moment p = 6.4 nC m, whose charge separation is much less than 55 cm. A line from the dipole to A makes a 20 angle with the dipole axis, and a line to B makes a 50 angle. Find the potential difference VB VA . Solution From Equation 25-7, VA = (9 GN m 2 /C 2 )(6.4 nC m ) cos 20=55 cm ) 2 = 179 V, and VB = VA cos 50= 20 = 122 V. ( cos Therefore, VAB = VB VA = 56.5 V. Problem 35. A hollow, spherical conducting shell of inner radius b and outer radius c surrounds, and is concentric with, a solid conducting sphere of radius a, as shown in Fig. 25-39. The sphere carries a net charge Q and the shell carries a net charge +3Q. Both conductors are in electrostatic equilibrium. Find an expression for the potential difference from infinity to the surface of the sphere. FIGURE 25-39 Problem 35. Solution The electric field between the solid sphere and the shell is like that due to a point charge Q located at their common center (the origin), so the potential difference between the sphere and the shell is Vsph Vshell = kQ( a 1 b 1 ). The electric field outside the shell is like that due to a point charge 2Q at the origin (the total charge is 3Q Q), so the potential difference between the shell and infinity is Vshell V = 2kQc 1 . (See Examples 24-3 and 25-3/) The entire shell is at one potential, as is the entire sphere, because each is a conductor in electrostatic equilibrium. Therefore, the potential difference between the sphere and infinity is Vsph V = Vsph Vshell + Vshell V = kQ(2c 1 + b 1 a 1 ). Problem 36. A thin plastic rod 20 cm long carries 3.2 nC distributed uniformly over its length. (a) If the rod is bent into a circular ring, find the potential at its center. (b) If the rod is bent into a semicircle, find the potential at the center (i.e., at the 8 CHAPTER 25 center of the circle of which the semicircle is part). Solution (a) The potential at the center of a uniformly charged ring, of radius a = 2, is V = kQ= = (9 GN m 2 /C 2 )(3.2 nC) = a (20 cm= ) = 905 V (see Equation 25-9). (b) Since each charge element, dq, of a circular arc, in Fig. 25-15, is the same 2 distance from the center of the arc, Example 25-6 also gives the potential at the center of a uniformly charged semicircular ring. However, the radius is now 20 cm/, or twice the value in part (a). Thus, V = 1 905 V = 452 V. 2 Problem 37. A thin ring of radius R carries a charge 3Q distributed uniformly over three-fourths of its circumference, and Q over the rest. What is the potential at the center of the ring? Solution The result in Example 25-6 did not depend on the ring being uniformly charged. For a point on the axis of the ring, the 2 geometrical factors are the same, and ring dq = Qtot for any arbitrary charge distribution, so V = kQtot ( x 2 + a 2 ) 1= still holds. Thus, at the center ( x = 0 ) of a ring of total charge Qtot = 3Q Q = 2Q, and radius a = R, the potential is V = 2 kQ= . R z Problem 38. The potential at the center of a uniformly charged ring is 45 kV, and 15 cm along the ring axis the potential is 33 kV. Find the rings radius and its total charge. Solution 2 45 The result of Example 25-6 and the given data imply 45 kV = kQ= and 33 kV = kQ[(15 cm ) 2 + a 2 ]1= . Thus, (33= ) = a 2 2 1= 22 2 2 2 1= or a = (33= )(15 cm )[1 (33= ) ] 45 45 = 16.2 cm, and Q = ( 45 kV)(16.2 cm )=9 GN m /C ) = ( a[(15 cm ) + a ] 809 nC. Problem 39. The annulus shown in Fig. 25-40 carries a uniform surface charge density . Find an expression for the potential at an arbitrary point P on its axis. FIGURE 25-40 Problem 39. Solution The annulus can be considered to be composed of thin rings of radius r ( a r b) and charge dq = 2 r dr (see Example 25-7 and Fig.s 25-15 and 16). The element of potential from a ring on its axis, a distance x from the center, is dV = k dq= x 2 + r 2 (see Example 25-6) so the potential from the whole annulus is: V= z dV = 2 k z b a r dr x +r 2 2 = 2 k x2 + r2 b a = 2 k ex 2 + b2 x 2 + a2 . j (Note: This reduces to the potential on the axis of a uniformly charged disk if a 0.) CHAPTER 1 L 9 Problem 40. A thin rod of length carries a charge Q distributed uniformly over its length. (a) Show that the potential in the plane that perpendicularly bisects the rod is given by V (r ) = 2 kQ 2 ln + 1+ 2 , 2r 4r L M M N O P P Q where r is the distance from the rod center. (b) Show that this expression reduces to an expected result when r . Hint: See Appendix A for a series expansion of the logarithm. Solution (a) The potential produced by an element of charge dq = dz = (Q=) dz, at a point in the perpendicular bisecting plane, is 2 2 dV = k dq= z + r , as shown in the sketch. Thus, V= kQ z 2 = 2 = dz z +r 2 2 = 2 kQ ln( z + z2 + r2 ) 2 = 0 = 2 kQ 2 ln + 1+ 2 , 2r 4r F G H I J K where we used the symmetry of the rod about the bisecting plane ( z = 0 ). (b) For 2r 1, ln[( 2 r ) + 1 + ( 2r ) 2 ] = = = ln[( 2r ) + 1] 2 r, so V (2 kQ=)( 2 r ) = kQ=, as expected for a point charge. = = = r Problem 40 Solution. Problem 41. (a) Find the potential as a function of position in the electric field E = E0 ( + ), where E0 = 150 V/m. Take the zero j x = 2.0 m, y = 1.0 m to the point x = 3.5 m, of potential at the origin. (b) Find the potential difference from the point y = 1.5 m. Solution (a) Equation 25-2b gives the potential for a uniform field. Take the zero of potential at the origin (point A in Equation 25-2b) and let = r = x + y + zk be the vector from the origin to the field point (point B in Equation 25-2b). j Then VAB = VB VA = V (r) 0 = V ( x, y) = E0 ( + ) r = E0 ( x + y). (The potential is independent of z, so we j wrote V (r) = V ( x, y). ) (b) V (3.5 m, 1.5 m) V (2.0 m, 1.0 m) = (150 V/m)(3.5 m 1.5 m 2.0 m 1.0 m) = 150 V. Section 25-4: Potential Difference and the Electric Field Problem 42. In a uniform electric field, equipotential planes that differ by apart. What is the field strength? Solution For a uniform field, Equation 25-10 can be written as V = E x, where the x-axis is in the direction of the field 2 (V decreases in the direction of the field). Thus, E = V= x = (1 V= .5 cm ) = 40 V/m. 10 CHAPTER 25 Problem 43. Figure 25-41 shows a plot of potential versus position along the x-axis. Make a plot of the x component of the electric field for this situation. FIGURE 25-41 Problem 43. Solution dx Figure 25-24 illustrates the relation E x = dV= , which may be used to estimate E x for the seven straight-line segments ( shown in Fig. 25-41. Thus, for x = 0 to 2 m, E x = 0, for x = 2 m to 4 m, E x = ( 2 V 2 V)=4 m 2 m ) = 2 V/m; etc., as sketched below. Problem 43 Solution. Problem 44. Figure 25-42 shows some equipotentials in the x-y plane. (a) In what region is the electric field strongest? What are (b) the direction and (c) the magnitude of the field in this region? FIGURE 35-42 Problem 44 Solution. Solution (a) The equipotentials in Fig. 25-37 are most closely spaced along the x-axis between x = 2 m and x = 5 m. (b) The potential decreases in the direction of the electric field, which, for 2 m x 5 m, is in the negative x direction. (c) The potential drops by 10 V/m, which is the field strength in this region. CHAPTER 1 11 L Problem 45. The potential in a certain region is given by V = axy, where a is a constant. (a) Determine the electric field in the region. (b) Sketch some equipotentials and field lines. Solution (a) The x and y components of the electric field can be found from Equation 25-10: E x = V= x = = x ( axy) = ay, and E y = V= y = ax. Thus E = a( y + x ). (The field has no z component.) (b) See sketch below. The field lines (dashed) j are perpendicular to the equipotentials (solid) in the direction of decreasing potential (arrows for a > 0, in this case). These equipotentials and field lines are confocal hyperbolas, proportional to xy and 1 ( x 2 y 2 ) respectively, and are sketched 2 only for x and y in the first quadrant. Problem 45 Solution. Problem 46. Sketch some equipotentials and field lines for a distribution consisting of two equal point charges. Problem 46 Solution. Solution The equipotential surfaces for two equal point charges, q, located on the x-axis at a, are given by 12 CHAPTER 25 F I V ( x, y, z) = 4 Gq J HK 0 1 ( x a) + y + z 2 2 2 + 1 ( x + a) + y 2 + z 2 2 = constant Lines of force are perpendicular to the equipotentials at every point. We sketched the field in the x-y plane, without adhering to any consistent numerical mapping convention, but in sufficient detail to display its general shape, in the vicinity of the charges. Standard calculus techniques and a personal computer help in preparing such pictures. A three-dimensional plot, as in Fig. 25-14, has also been included; caveatthe lines on the three-dimensional plot are not equipotentials. Problem 47. The electric potential in a region of space is given by V = 2 xy 3zx + 5 y 2 , with V in volts and the coordinates in meters. If point P is at x = 1 m, y = 1 m, z = 1 m, find (a) the potential at P and (b) the x, y, and z components of the electric field at P. Solution (a) Direct substitution gives V ( P) = 2(1)(1) 3(1)(1) + 5(1) 2 = 4 V. (b) Use of Equation 25-10 gives E x = V= x = 2 y + 3z, E y = V= y = 2 x 10 y, and Ez = V= z = 3 x. At P( x = y = z = 1), E x = 1 V/m, E y = 12 V/m, and Ez = 3 V/m. Problem 48. Use Equation 25-7 to calculate the electric field on the perpendicular bisector of a point dipole, and show that your result is equivalent to Equation 23-7a. Solution Equations 25-7 and 10 give V = kp cos = 2 , Er = V= r = 2 kp cos = 3, and E = (1=) V= = kp sin = 3 . On the r r r r = (kp= 3 ) . To compare with Equation 23-7a, take the origin at the bisecting center plane, = 90 so Er = 0 and E = E r (p = p ), and the y-axis up in Fig. 25-13, so = sin 90 + of the dipole, the dipole moment along the x-axis cos 90 = and r = 0 2 + y 2 = y on the bisecting plane. Then E = kp= 3. (Of course, instead of using polar j y 2 coordinates, one could first express V in terms of x and y, V ( x, y) = kpx ( x 2 + y 2 ) 3= , and then differentiate, E x = V= x and E y = V= y. The result is the same.) Problem 49. Use the result of Example 25-6 to determine the on-axis field of a charged ring, and verify that your answer agrees with the result of Example 23-8. Solution On the axis of a uniformly charged ring (the x-axis), V = kQ= x 2 + a 2 (Equation 25-9), and the electric field only has 2 an x component (by symmetry). Then E = ( dV= ) = kQx ( x 2 + a 2 ) 3= , in accord with Example 23-8. (In general, one dx needs to know the potential in a 3-dimensional region in order to calculate the field from its partial derivatives.) Problem 50. A charge +4q is located at the origin and a charge q is on the x-axis at x = a. (a) Write an expression for the potential on the x-axis for x > a. (b) Find a point in this region where V = 0. (c) Use the result of (a) to find the electric field on the x-axis for x > a and (d) find a point where E = 0. Solution (a) From Equation 25-6 on the x-axis, for x > a, V ( x ) = kq[4 x 1 ( x a) 1 ]. (b) V ( x ) = 0 when 4( x a) = x, or x = 4 a=. (c) For points on the x-axis, the electric field only has an x component, so Equation 25-10 gives E = ( dV= ) = 3 dx kq[4 x 2 ( x a) 2 ] (see note to solution of previous problem). (d) E = 0 when x = 2( x a), or x = 2 a (remember x > a when taking square roots). CHAPTER 1 13 L Problem R 51. The electric potential in a region is given by V = V0 (r= ), where V0 and R are constants, r is the radial distance from the origin, and where the zero of potential is taken at r = 0. Find the magnitude and direction of the electric field in this region. Solution Since depends only on r, the field is spherically symmetric and its direction is radial. From Equation 25-10 (which applies dr R R unmodified for the radial coordinate), Er = dV= = V0= , and E = (V0 = )r (r is a unit vector radially outward). Section 25-5: Potentials of Charged Conductors Problem 52. (a) What is the maximum potential (measured from infinity) for the sphere of Example 25-3 before dielectric breakdown of air occurs at the spheres surface? (Breakdown of air occurs at a field strength of 3 MV/m.) (b) What is the charge on the sphere when its at this potential? Solution (b) Dielectric breakdown in the air occurs if the field at the surface, E = = 0 , exceeds 3 10 6 V/m. Therefore, the 2 charge (for a uniformly charged sphere) must not be greater than q = 4 R = 4 0 ER 2 = (3 10 6 V/m )(2.3 m ) 2 R (9 10 9 N m 2 /C 2 ) = 1.76 mC. (a) From Equation 25-11, V = kq= = RE = 6.9 MV. Problem 53. The spark plug in an automobile engine has a center electrode made from wire 2.0 mm in diameter. The electrode is worn to a hemispherical shape, so it behaves approximately like a charged sphere. What is the minimum potential on this electrode that will ensure the plug sparks in air? Neglect the presence of the second electrode. Solution If we can treat the field from the central electrode as that from an isolated sphere, then E = kq= 2 and V = kq= , so that R R 6 V = RE. Dielectric breakdown in air would occur for potentials exceeding V = (1 mm )(3 10 V/m ) = 3 kV. Problem 54. A large metal sphere has three times the diameter of a smaller sphere and carries three times as much charge. Both spheres are isolated, so their surface charge densities are uniform. Compare (a) the potentials and (b) the electric field strengths at their surfaces. Solution R (a) The potential of an isolated metal sphere, with charge Q and radius R, is kQ= , so a sphere with charge 3Q and radius 3R has the same potential. (b) However, the electric field at the surface of the smaller sphere is = 0 = kQ= 2 , so tripling R 1 Q and R reduces the surface field by a factor of 3 . Problem 55. Two metal spheres each 1.0 cm in radius are far apart. One sphere carries 38 nC of charge, the other 10 nC. (a) What is the potential on each? (b) If the spheres are connected by a thin wire, what will be the potential on each once equilibrium is reached? (c) How much charge must move between the spheres in order to achieve equilibrium? Solution (a) Since the spheres are far apart (approximately isolated), we can use Equation 25-11 to find their potentials: V1 = kQ1= 1 = (9 GN m 2 /C 2 )(38 nC)=1 cm ) = 34.2 kV and V2 = kQ2 = 2 = 9 kV. (b) When connected by a thin wire, R ( R R R the spheres reach electrostatic equilibrium with the same potential, so V = kQ1= 1 = kQ2= 2 . Since the radii are equal, so must be the charges, Q1 = Q2 . The total charge is 38 nC 10 nC = 28 nC = Q1 + Q2 = 2Q1 (if we assume that the wire is Q1 = Q2 = 14 nC. Then V = k (14 nC)=1 cm ) = 12.6 kV. (c) In this process, the ( so thin that it has a negligible charge), so first sphere loses 38 14 = 24 nC to the second. 14 CHAPTER 25 Problem 56. Sketch some equipotentials and field lines for the isolated, charged conductor shown in Fig. 25-43. Solution The description of nonspherical conductors in Section 25-5 outlines a qualitative, field mapping procedure, which will produce a sketch very similar to Fig. 25-28. Problem 57. Two conducting spheres are each 5.0 cm in diameter and each carries 0.12 C. They are 8.0 m apart. Determine (a) the potential on each sphere; (b) the field strength at the surface of each sphere; (c) the potential midway between the spheres; (d) the potential difference between the spheres. Solution Since the spheres are small and widely separated, at small distances ( 8 m ) they behave like isolated spheres, and at large distances ( 5 cm ) they behave like two point charges. (a) At the surface of either sphere, Equation 25-11 gives Vsurf = kq= = (9 10 9 N m 2 /C 2 )(1.2 10 7 C)=0.025 m ) = 43.2 kV. (b) Then Esurf = = 0 = kq= 2 = Vsurf = = R ( R R 6 ( 43.2 kV)=2.5 cm) = 1.73 10 V/m. (c) Midway between the spheres, the potential from each one is the same, so ( Vmid - pt. = 2 kq= = 2(9 10 9 V m/C)(1.2 10 7 C)=4 m ) = 540 V. (d) The spheres are at the same potential, so the r ( difference is zero. Problem 58. Two small metal spheres are located 2.0 m apart. One has radius 0.50 cm and carries 0.20 C. The other has radius 1.0 cm and carries 0.080 C. (a) What is the potential difference between the spheres? (b) If they were connected by a thin wire, how much change would move along it, and in which direction? Solution (a) Since the spheres are approximately isolated, their potentials are V1 = kq1= 1 = (9 10 9 V m/C)(2 10 7 C) R 3 5 9 8 2 (5 10 m ) = 3.6 10 V, and V2 = (9 10 V m /C)(8 10 C)=10 m ) = 0.72 10 5 V. The difference is ( 5 0.28 C = q1 + q2 , where primes refer to the spheres V1 V2 = 2.88 10 V. (b) The total charge on the spheres is after they are connected. (We assume that the wire has a negligible effect, other than bringing both spheres to the same R R potential.) Since V1 = kq1= 1 = V2 = kq2 = 2 , R2 q1 R1q2 = 0, or 2q1 q2 = 0. Solving for q1 , we find q1 = 1 (0.28 C) = 0.0933 C, therefore a charge of q1 q1 = 0.107 C was transferred from the first sphere to the 3 second. Paired Problems Problem 59. Three 50-pC charges sit at the vertices of an equilateral triangle 1.5 mm on a side. How much work would it take to bring a proton from very far away to the midpoint of one of the triangles sides? Solution Two of the charges are at distances of 1 (1.5 mm ) = 0.75 mm, and the third is at 3 (0.75 mm ) from the midpoint of one 2 ( side. Therefore, the potential at this point (Equation 25-6) is V ( P) = k (50 pC)(1 + 1 + 1= 3 )=0.75 mm ) = 1.55 kV. The work it would take to move a proton of charge e from infinity (where the potential is zero) to this point is eV = 1.55 keV = (1.6 10 19 C)(1.55 kV) = 2.47 10 16 J. CHAPTER 1 15 L Problem 59 Solution. Problem 60. Repeat the preceding problem for the case when one of the charges is 50 pC and the proton is brought to the midpoint of the side between the two positive charges. Solution The positive charges are still 0.75 mm from the midpoint, but the charge 3 (0.75 mm ) away is 50 pC. Thus, eV = ek (50 pC)(1 + 1 1= 3 )=0.75 mm ) = 854 eV = 1.37 10 16 J. (See solution to previous problem.) ( Problem 61. A pair of equal charges q lies on the x-axis at x = a. (a) Find expressions for the potential at points on the x-axis for which x > a and (b) show that your result reduces to a point-charge potential for x a. Solution 1 1 (a) From Equation 25-6, the potential on the x-axis is V ( x ) = kq( x a + x + a ). For x > a , this becomes V ( x ) = x kq[( x a) 1 + ( x + a) 1 ] = 2 kqx ( x 2 a 2 ) 1 . (b) For x a, V ( x ) 2 kq= , the potential of a point charge 2q at a distance x. Problem 62. (a) For the charge distribution of the preceding problem, find an expression for the potential at all points on the y-axis. (b) Show that your result reduces to a point-charge potential for y a. Solution (a) For points on the y-axis, which are equidistant from x a, the potential is V ( y) = 2 kq= y 2 + a 2 . (b) For y a, V ( y) 2 kq= as expected. (See previous solution.) y Problem 73. A 2.0-cm-radius metal sphere carries 75 nC and is surrounded by a concentric spherical conducting shell of radius 10 cm carrying 75 nC. (a) Find the potential difference between the shell and the sphere. (b) How would your answer change if the shell charge were changed to +150 nC? Solution (a) The electric field outside the sphere ( radius R1 = 2 cm ), but inside the shell (inner radius R2 = 10 cm ), is only due to the charge q1 on the sphere (recall Gausss law), and equals kq1= 2 radially outward. The potential difference is V1 V2 = r z kq dr= r R1 R2 1 2 = kq1 ( R11 R2 1 ) = (9 GN m 2/C 2 )(75 nC)[(2 cm ) 1 (10 cm ) 1 ] = 27.0 kV. (We used Equation 25-2a, with A at R2 and B at R1. Note that E dr = kq1 dr= 2 regardless of the choice of points A and B.) (b) Adding more charge r to the conducting shell does not affect the field inside, nor the potential difference between points inside, like V1 V2 in part (a). The field outside the shell and the potential relative to infinity would change. 16 CHAPTER 25 Problem 64. A coaxial cable consists of a 2.0-mm-radius central wire carrying 75 nC/m, and a concentric outer conductor of radius 10 mm carrying 75 nC/m. (a) Find the potential difference between the outer and inner conductor. (b) How would your answer change if the outer conductor were charged to +150 nC/m? Solution (a) For a long straight cable, the field between the two conductors is approximately cylindrically symmetric and depends only on the charge density of the wire, but not on that of the outer conductor (hence no change gor part (b)). In fact, 2 E = 2 k wire = (radially outward) for 2 mm r 10 mm (see Example 24-4). Then V = 10mm 2 k wire dr= = r r mm 2 k wire ln(10 mm= mm) = 2(9 75 V) ln 5 = 2.17 kV. 2 z Problem 65. On the x-axis, the electric field of a certain charge distribution is given by E = a= 4 , where a = 55 V m 3 . Find the x potential difference from the point x = 1.3 m to the point x = 2.8 m. Solution Direct application of Equation 25-2a yields: Problem 66. A sphere of radius R carries a nonuniform but spherically symmetric volume charge density that results in an electric field in the sphere given by E = E0 (r= ) 2 r, where E0 is a constant. Find the potential difference from the spheres R surface to its center. Solution In this case, Equation 25-2a gives V (0 ) V ( R) = zE dr = z( E =R )r dr = E R=3. R 0 r R 0 0 2 2 0 Problem 67. The potential as a function of position in a certain region is given by V ( x ) = 3 x 2 x 2 x 3 , with x in meters and V in volts. Find (a) all points on the x-axis where V = 0, (b) an expression for the electric field, and (c) all points on the x-axis where E = 0. Solution (a) The expression for the potential can be factored into V ( x ) = x ( x + 3)(1 x ), so V ( x ) = 0 at x = 0, 1 m, and 3 m. (b) V is independent of y and z, hence E = E x has only an x component, and E x = dV= = 3 x 2 + 4 x 3. (c) E x = 0 dx 3 for x = ( 2 4 + 9 )= = 0.535 m and 1.87 m. Problem 68. The potential in a certain region is given by V ( x, y) = [2 x 2 + ( y 1) 2 1]. Find (a) a point where V = 0, (b) an expression for the electric field, and (c) a point in the x-y plane where E = 0. Solution (a) V ( x, y) = 0 if 2 x 2 + ( y 1) 2 = 1. This represents an ellipse centered at (0, 1), with major axis of 2 along the y-axis and minor axis of 2 along the x-axis. (The ellipse is the intersection of the zero valued equipotential surface with the x-y plane.) Thus, four easily found points where V ( x, y) = 0 are (0, 0 ), (0, 2), ( 1 2, 1), and ( 1 2, 1). (b) E x = 2 2 V= x = 4 x, E = V= y = 2( y 1), and E = 0, hence E = 4 x + 2( y 1). (c) E = 0 is a vector equation equivalent j to E x = 4 x = 0 and E y = 2( y 1) = 0. The only solution is (0, 1). (Note: Units are not specified in this problem.) y z CHAPTER 1 17 L Supplementary Problems Problem 69. A conducting sphere 5.0 cm in radius carries 60 nC. It is surrounded by a concentric spherical conducting shell of radius 15 cm carrying 60 nC. (a) Find the potential at the spheres surface, taking the zero of potential at infinity. (b) Repeat for the case when the shell also carries +60 nC. Solution The potential difference between the sphere and the shell depends only on the electric field inside the shell (which is due to 1 1 qsphere only). For a spherically symmetric configuration, this was found to be Vsphere Vshell = kqsphere ( Rsphere Rshell ) in Problem 63. (Here, Rshell is the inner radius of the shell.) Outside the shell, the potential depends on the total charge, which, R at the surface of a spherical distribution, was found to be Vshell V = k ( qsphere + qshell )= shell in Example 25-3. (Here, Rshell is the outer radius of the shell, the conducting material of which is assumed to be an equipotential region in 1 1 electrostatic equilibrium, and V = 0.) Thus, Vsphere V = Vsphere Vshell + Vshell V = kqsphere ( Rsphere Rshell ) + 1 1 1 k ( qsphere + qshell ) Rshell = kqsphere Rsphere + kqshell Rshell . (For a thin shell, the inner and outer radii are approximately equal.) (a) In this problem, qsphere = qshell = 60 nC, and Rsphere = 1 Rshell = 5 cm, so Vsphere = (9 6 )(1 1 ) kV = 7.2 kV. (b) If 3 5 3 qshell is changed to equal + qsphere , Vsphere = (9 6 )(1 + 1 ) kV = 14.4 kV. 5 3 Problem 70. Show that the result of Example 25-9 approaches the field of a point charge for x a. Hint: You will need to apply the binomial theorem to the quantity 1= x 2 + a 2 . Solution 2 x For a= 1, the field in Example 25-9 is approximately (2 kQ= 2 )[1 (1 + a 2 = 2 ) 1= ] = (2 kQ= 2 ) a x a 2 2 2 2 2 2 [1 (1 a = x + . . . )] = (2 kQ= )( a = x + . . . ) kQ= , like that of a point charge. 2 a 2 x Problem 71. The potential on the axis of a uniformly charged disk at 5.0 cm from the disk center is 150 V; the potential 10 cm from the disk center is 110 V. Find the disk radius and its total charge. Solution Combining the given data with the potential in Exercise 25-7, we find 150 V = 2 kQ= 2 ( (5 cm ) 2 + a 2 5 cm ) and a 110 V = 2 kQ= 2 ( (10 cm ) 2 + a 2 10 cm). The charge can be eliminated by division, a 150 F I= HK 110 1 + (a= cm ) 2 1 5 4 + ( a= cm ) 2 2 5 . 52 Several lines of algebra to remove the square roots finally yields a = (5 cm ) 105 209= = 14.2 cm. We can now solve 2 2 2 1 for Q from either of the first two equations, Q = (Va = k )[ x + a x ] = 1.67 nC. 2 Problem 72. A uranium nucleus (mass 238 u, charge 92e) decays, emitting an alpha particle (mass 4 u, charge 2e) and leaving a thorium nucleus (mass 234 u, charge 90e). At the instant the alpha particle leaves the nucleus, the centers of the two are 7.4 fm apart and are essentially at rest. Find their speeds when they are a great distance apart. Treat each particle as a spherical charge distribution. Solution The work done by the Coulomb repulsion when the thorium nucleus and the alpha particle separate is U R = q VTh,R = q (kqTh = ), where R = 7.4 fm is the initial separation and the final separation is essentially infinite. (See R also Equation 26-1.) With the neglect of other interactions, the work energy theorem requires that this equal the change in 18 CHAPTER 25 2 2 the kinetic energy of the two particles, or 1 m v + 1 mTh vTh (since they start from rest). The conservation of momentum 2 2 (under the same assumptions) requires also that 0 = m v + mTh v Th (since the total momentum is zero initially), so 2 2 234 R2 that v and vTh can be determined. Thus k (90e)(2e)= = 1 (234 u vTh + 4 u v ) and vTh = ( 4= )v . Then 2 ke 2 (180)= (1 u) = ( 4 + 16= )v , or v = 4.07 10 7 m/s and vTh = 6.96 10 5 m/s, where values from the inside front R 234 2 cover were used. Problem 73. A power line consists of two parallel wires 3.0 cm in diameter spaced 2.0 m apart. If the potential difference between the wires is 4.0 kV, what is the charge per unit length on each wire? The wires carry equal but opposite charges. Hint: The wires are far enough apart that they dont greatly affect each others fields. Solution Since the radius of the wires ( a = 1.5 cm) is much smaller than their separation (b = 200 cm ), their potential difference can be found from the superposition of the potentials of two (approximately infinite) line charges (as calculated in Example 25-4). Thus, for any two points between the wires, V2 V1 = (V2+ V1+ ) + (V2 V1 ) = r ln 1 2 0 r2 F I + ( ) ln F r I = ln F (b r ) I. b r GJ 2 G r J 2 G(b r ) J b r HK HK HK 1 2 1 2 0 0 2 1 To obtain the potential difference between the wires, we let r2 = a (the positive wire is at a higher potential) and r1 = b a. Then V = (b a)(b a) ba ln = ln . 2 0 a(b b + a) 0 a FI HK I K From the given numerical values, we can solve for : = 0 V 4000 V = = 22.7 nC/m. ba 200 ln (36 10 9 V m/C) ln 1 a 1.5 FI HK F H ln( a (Incidentally, we have found the capacitance per unit length of a bifilar transmission line, C= = 0= b 1), in the limit a b; see Section 26-4.) Problem 73 Solution. Problem 74. For the dipole of Example 25-5, show that the electric field at an arbitrary point far from the dipole can be written CHAPTER 1 19 L E= kp [(3 cos 2 1) + 3 sin cos ]. j 3 r Solution In the solution to Problem 48, we found Er = 2kp cos = 3 and E = kp sin = 3 . In Fig. 25-10 (in which a r for a r r sin and = sin + cos . ( is in the point dipole) p is parallel to the x-axis, so the unit vectors are r = cos + j j 3 r j j direction of increasing . ) Thus, E dip = Er r + E = (kp= )[2 cos ( cos + sin ) + sin ( sin + cos )] = 3 2 2 ], where sin + cos 2 = 1 was used to express the x component. ( kp= )[(3 cos 1) + 3 sin cos j r Problem 75. A thin rod of length lies on the x-axis with its center at the origin. It carries a line charge density given by = = 0 ( x=) 2 , where 0 is a constant. (a) Find an expression for the potential on the x-axis for x > 2. (b) Integrate the charge density to find the total charge on the rod. (c) Show that your answer for (a) reduces to the potential of a point charge whose charge is the answer to (b), for x . Solution = (a) For points P on the x-axis at x > 2, the potential from a charge element dq = dx at x along the rod ( 2 x 2) is dV = k dq=x x = k dx =x x ). (We used x for the variable position of dq, and took the = = ( potential relative to zero at infinity.) The potential due to the entire rod, for = 0 ( x = ) 2 , is V( x) = z 2 = 2 = dV = k 0 2 z 2 = k x 2 dx x2 = 20 x 2 ln( x x ) xx 2 ( x x ) = 2 2 = = 2 = k 0 2 x + 2 = x ln x . 2 x 2 = LF MG M NH IO JP KQ P (Use partial fractions, x 2=x x ) = x x + x 2=x x ), or standard tables to evaluate the integral.) (b) The total charge ( ( 2 = 2 2 3= = on the rod is Q = dq = = ( 0 = ) x dx = ( 0= 2 )( 2 )( 2) 3 = 0 12. (c) For x , we can expand the logarithms, 2 zz 1 2 ln(1 2 x ) = ( 2 x ) = = ( 2 x ) 2 = 1 3 ( 2 x ) 3 . . . . .Therefore: = V ( x ) = (k 0 = 2 )[ x 2 ln(1 + 2 x ) x 2 ln(1 2 x ) x] = = = k F G H k F =G H 0 2 0 2 IL F 1 F I + 1 F I . . .I x F 1 F I 1 F I . . .I x O JMGx 2 Hx K 3 Hx K J G2 x 2 Hx K 3 Hx K J P Kx H 2 2 2 2 KH KQ M P N2 IL F + 2 F I + . . .I x O k = kQ=x, = JMG 3 Hx K J P 12 x Kx H 2 KP Mx N Q 2 2 3 2 2 3 2 3 0 as expected. Problem 75 Solution. Problem 76. Repeat the preceding problem for the case = 0 ( x=). Why is your answer for x different? Hint: What does this charge distribution resemble at large distances? Solution (a) If = 0 ( x=) for the rod in the preceding problem, 20 CHAPTER 25 V( x) = k 0 z 2 = 2 = k 0 k 0 x + 2 = x dx 2 = = x ln( x x ) x = = x ln . 2 ( x x ) x 2 = (b) Now, however, the total charge is Q = ( 0 =) 2 2 z LF MG M NH IO JP KQ P 2 = 2 = = x d x = 0. (c) For x, the potential is x times the preceding limit, or V ( x ) = k 0 = x . This is like the potential on the axis of a point dipole, with dipole moment p = 0 2= . In 12 12 fact, p = x dq = z z 2 = 2 2 ( 0 =) x = dx. Problem 77. For the situation of Example 25-10, find an equation for the equipotential with V = 0 in the x-y plane. Plot the equipotential, and show that it passes through the points described in Example 25-10 and its exercise. Solution The potential in the x-y plane, for the charges in Example 25-10, is V ( x, y) = k (2 q ) k ( q ) + = kq r+ r L2 Mx + a) M N( 2 + y2 1 ( x a) 2 + y 2 O P , P Q where r+ , r are the distances from a point ( x, y) to the charges +2q at ( a, 0 ) and q at ( a, 0) respectively. The potential is zero when r+ = 2 r , or ( x + a) 2 + y 2 = 4[( x a) 2 + y 2 ]. This equation simplifies to y 2 + x 2 10 ax= + a 2 = 0, or 3 2 3 2 3 after completion of the square in x, to y + ( x 5a= ) = ( 4 a= ) . This is the equation of a circle centered at (5a=, 0) with 3 3 3 3 3 3 radius 4 a=. The intercepts of the circle on the x-axis are at 5a= 4 a= = 3a or a=, as found in Example 25-10 and the following exercise. Problem 77 Solution. Problem a 78. A disk of radius a carries a nonuniform surface charge density given by = 0 (r= ), where 0 is a constant. (a) Find the potential at an arbitrary point on the disk axis, a distance x from the disk center. (b) Use the result of (a) to find the electric field on the disk axis, and (c) show that the field reduces to an expected form for x a. Solution (a) As in Example 25-7, V ( x ) = integral tables gives V( x) = zk dq= x a 0 2 a + r 2 , where dq = 2 r dr , but now = 0 r=. Reference to standard 2 k 0 a a 2 2 k 0 a z a 0 r 2 dr x2 + r2 = Lx M M N 2 + a2 x2 a+ ln 2 F G H x 2 + a2 x IO JP P K Q CHAPTER 1 21 L = k 0 a 1 + ( x= ) 2 ( x= ) 2 ln( a= + 1 + ( a= ) 2 ) . a a x x dx (b) As in Example 25-9, E x = dV= results in E x = k 0 Lx lnF + 2 a MG aH M N F+ a G x H x 2 + a2 x I J K x x 2 + a2 + x2 a F G+ H a x x 2 + a2 1= 2 IF 1 JGx + a K H 2 2 a+ x 2 + a2 x2 IO JP P K Q = (2 k 0 x= ) ln( a= + 1 + ( a= ) 2 ) ( a= )(1 + ( a= ) 2 ) a x x x x . (c) The logarithm has to be expanded carefully, up to order ( a= )3 , to evaluate E x for x a. Thus, x ln 1+ a F I I = lnF+ a + a + . . .I 1 J HKJ G x 2 x xKH K F a I 1 F + a I + 1 F + a I + a a a G+ x x x H 2x J 2 G 2x J 3 G 2x J KH K H K 2 2 2 2 2 2 2 3 2 2 2 . . . a a3 3. x 6x Also, a a2 1+ 2 x x Then, Ex This is like a point charge field for Q = FI GJ HK 1= 2 a a2 1 x 2x2 F G H I= a a J x 2x K O P Q 3 3 . 2 k 0 x a 2 k 0 a 2 a3 a a3 + = . a x 6x 3 x 2x 3 3x 2 a a2 0 (2 0 = )r z L M N dr = 2 0 a 2 =. 3 Problem 79. An open-ended cylinder of radius a and length 2a carries charge q spread uniformly over its surface. Find the potential on the cylinder axis at its center. Hint: Treat the cylinder as a stack of charged rings, and integrate. Solution 2 The cylinder can be considered to be composed of rings of radius a, width dx, and charge dq = (q= a ) dx. The potential at the center of the cylinder (which we take as the origin, with x-axis along the cylinder axis) due to a ring at x ( a x a) is dV = k dq= x 2 + a 2 (see Example 25-6). The whole potential at the center follows from integration: V= zF IK H a a kq 2a dx x +a 2 2 = kq a + 2a kq kq ln = ln(1 + 2 ) = 0.881 . 2a a + 2a a a F G H I J K ... View Full Document

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