F5ch25 - CHAPTER 25 ELECTRIC POTENTIAL 681 CHAPTER 25 Answer to Checkpoint Questions 1 2 3 4 5 6 7(a negative(b increase(a positive(b higher(a

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Unformatted text preview: CHAPTER 25 ELECTRIC POTENTIAL 681 CHAPTER 25 Answer to Checkpoint Questions 1. 2. 3. 4. 5. 6. 7. (a) negative; (b) increase (a) positive; (b) higher (a) rightward; (b) 1, 2, 3, 5: positive; 4: negative; (c) 3, then 1, 2, and 5 tie, then 4 all tie a, c (zero), b (a) 2, then 1 and 3 tie; (b) 3; (c) accelerate leftward closer (half of 9:23 fm) Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. (a) higher; (b) positive; (c) negative; (d) all tie (a) left; (b) 50 V; (c) positive; (d) negative (a) 1 and 2; (b) none; (c) no; (d) 1 and 2 yes, 3 and 4 no 4q=40 d (b), then (a), (c) and (d) tie all tie (a) negative; (b) zero (a){(c) Q=40 R; (d) a, b, c (a) 1, then 2 and 3 tie; (b) 3 Ez , Ey , Ex left (a) 2, 4, and then a tie of 1, 3, and 5 (where E = 0); (b) negative x direction; (c) positive x direction (a), (b), (c) (a) 3 and 4 tie, then 1 and 2 tie; (b) 1 and 2, increase; 3 and 4, decrease 682 CHAPTER 25 ELECTRIC POTENTIAL 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. (a) c, b, a; (b) zero (a) { (d) zero (a) positive; (b) positive; (c) negative; (d) all tie farther (a) no; (b) yes no no (a particle at the intersection would have two dierent potential energies) (a) { (c) C , B , A; (d) all tie (a) { (b) all tie; (c) C , B , A; (d) all tie situation 2 Solutions to Exercises & Problems The magnitude is U = eV = 1:2 109 eV = 1:2 GeV: (a) An ampere is a coulomb per second, so 84 A h = (84 C h=s) (3600 s=h) = 3:0 105 C : (b) The change in potential energy is U = q V = (3:0 105 C)(12 V) = 3:6 106 J. (a) When charge q moves through a potential dierence V its potential energy changes by U = q V . In this case, U = (30 C)(1:0 109 V) = 3:0 1010 J. (b) Equate the nal kinetic energy of the automobile to the energy released by the lightning: U = 1 mv2 , where m is the mass of the automobile and v is its nal speed. Thus 2 3P 2E 1E v= r 2 U = m s 2(3:0 1010 J) = 7:7 103 m/s : 1000 kg (c) Equate the energy required to melt mass m of ice to the energy released by the lightning: U = mLF , where LF is the heat of fusion for water. Thus 10 J m = U = 3:33:0 10 J/kg = 9:0 104 kg : LF 105 CHAPTER 25 ELECTRIC POTENTIAL 683 4E + - z axis 5P -2 -2 + + some electric field lines + some equipotential cross sections + (a) VB VA = U=( e) = 3:94 10 19 J=( 1:60 10 19 C) = 2:46 V: (b) VC VA = VB VA = 2:46 V: (c) VC VB = 0 (Since C and B are on the same equipotential line). 6E 684 CHAPTER 25 ELECTRIC POTENTIAL V = E s = (1:92 105 N/C)(0:0150 m) = 2:90 103 V: (a) E = F=e = 3:9 10 15 N=(1:60 10 19 C) = 2:4 104 N/C: (b) V = E s = (2:4 104 N/C)(0:12 m) = 2:9 103 V: 9E 8E 7E The electric eld produced by an in nite sheet of charge has magnitude E = =20 , where is the surface charge density. The eld is normal to the sheet and is uniform. Place the origin of a coordinate system at the sheet and take the x axis to be parallel to the eld and positive in the direction of the eld. Then the electric potential is V = Vs Z x where Vs is the potential at the sheet. The equipotential surfaces are surfaces of constant x; that is, they are planes that are parallel to the plane of charge. If two surfaces are separated by x then their potentials dier by V = E x = (=20 )x. Thus 12 2 2 x = 20 V = 2(8:85 10 C 6/N m2)(50 V) = 8:8 10 3 m : 0:10 10 C/m (a) 10P 0 E dx = Vs Ex ; q0 Z z dz = q0 z : W = q0 E ds = 2 20 0 0 i (b) Since V V0 = W=q0 = z=20 , z V = V0 2 : 0 Z 11P f The potential dierence between the wire and cylinder is given, not the linear charge density on the wire. Use Gauss' law to nd an expression for the electric eld a distance r from the center of the wire, between the wire and the cylinder, in terms of the linear charge density. Then integrate with respect to r to nd an expression for the potential dierence between the wire and cylinder in terms of the linear charge density. Use this result to obtain an expression for the linear charge density in terms of the potential dierence and substitute the result into the equation for the electric eld. This will give the electric eld in terms of the potential dierence and will allow you to compute numerical values for the eld at the wire and at the cylinder. CHAPTER 25 ELECTRIC POTENTIAL 685 For the Gaussian surface use a cylinder of radius r and length `, concentric with the wire and cylinder. The electric eld is normal to the rounded portion of the cylinder's surface and is uniform over that surface. This means the electric ux through the Gaussian surface is given by 2r`E , where E is the magnitude of the electric eld. The charge enclosed by the Gaussian surface is q = `, where is the linear charge density on the wire. Gauss' law yields 20 r`E = `. Thus E = 2 r : 0 Since the eld is radial, the dierence in the potential Vc of the cylinder and the potential Vw of the wire is and r V = Vw Vc = E dr = 2 r dr = 2 ln r c ; 0 0 w rc rw where rw is the radius of the wire and rc is the radius of the cylinder. This means that 2 = ln(r0=r V) Z Z rw rc c w (a) Substitute rw for r to obtain the eld at the surface of the wire: V E = r ln(rV =r ) = (0:65 10 6 m) ln [(1:085010 2 m)=(0:65 10 6 m)] w c w = 1:36 108 V/m : (b) Substitute rc for r to nd the eld at the surface of the cylinder: E = r ln(rV =r ) = (1:0 10 2 m) ln [(1:0 850 V 2 m)=(0:65 10 6 m)] 10 w c w 3 V/m : = 8:82 10 (a) 12P E = 2 r = r ln(r V=r ) : 0 c w qr dr = qr2 : E (r)dr = 0 V (r) = V (0) 80 R3 0 0 40 R3 (b) V = V (0) V (R) = q=80 R: (c) Since V = V (0) V (R) > 0; the potential at the center of the sphere is higher. Z Z 13P r r (a) Use Gauss' law to nd expressions for the electric eld inside and outside the spherical charge distribution. Since the eld is radial the electric potential can be written as an integral of the eld along a sphere radius, extended to in nity. Since dierent expressions for the eld apply in dierent regions the integral must be split into two parts, one from in nity to the surface of the distribution and one from the surface to a point inside. 686 CHAPTER 25 ELECTRIC POTENTIAL Outside the charge distribution the magnitude of the eld is E = q=40 r2 and the potential is V = q=40 r, where r is the distance from the center of the distribution. To nd an expression for the magnitude of the eld inside the charge distribution use a Gaussian surface in the form of a sphere with radius r, concentric with the distribution. The eld is normal to the Gaussian surface and its magnitude is uniform over it, so the electric ux through the surface is 4r2 E . The charge enclosed is qr3 =R3 . Gauss' law becomes 3 40 r2 E = qr3 ; R so qr E = 4 R3 : 0 If Vs is the potential at the surface of the distribution (r = R) then the potential at a point inside, a distance r from the center, is q Z r r dr = V qr2 q E dr = Vs 4 R3 V = Vs s 8 R3 + 8 R : 0 0 0 R R The potential at the surface can be found by replacing r with R in the expression for the potential at points outside the distribution. It is Vs = q=40 R. Thus q 1 r2 + 1 = q (3R2 r2 ) : V = 4 R 2R3 2R 80 R3 0 Z r (b) In 12P the electric potential was taken to be zero at the center of the sphere. In this problem it is zero at in nity. According to the expression derived in part (a) the potential at the center of the sphere is Vc = 3q=80 R. Thus V Vc = qr2 =80 R3 . This is the result of 12P. (c) The potential dierence is 2 3 q V = Vs Vc = 8q R 8q R = 8 R : 0 0 0 The same value as is given by the expression obtained in 12P. (d) Only potential dierences have physical signi cance, not the value of the potential at any particular point. The same value can be added to the potential at every point without changing the electric eld, for example. Changing the reference point from the center of the distribution to in nity changes the value of the potential at every point but it does not change any potential dierences. 14P (a) For r > r2 the eld is like that of a point charge and 1 V = 4 Q ; 0 r where the zero of potential was taken to be at in nity. CHAPTER 25 ELECTRIC POTENTIAL 687 and the magnitude of the electric eld is (b) To nd the potential in the region r1 < r < r2 , rst use Gauss's law to nd an expression for the electric eld, then integrate along a radial path from r2 to r. The Gaussian surface is a sphere of radius r, concentric with the shell. The eld is radial and therefore normal to the surface. Its magnitude is uniform over the surface, so the ux through the surface 3 3 is = 4r2 E . The volume of the shell is (4=3)(r2 r1 ), so the charge density is = 4(r3Q r3 ) 3 2 1 and the charge enclosed by the Gaussian surface is 3 4 (r3 r3 ) = Q r3 r1 : q= 3 1 3 3 r2 r1 Gauss' law yields 3 r3 1 2E = Q r 40 r 3 r3 r 2 1 3 3 Q E = 4 r2rr3 r1 3 ) : 0 ( 2 r1 If Vs is the electric potential at the outer surface of the shell (r = r2 ) then the potential a distance r from the center is given by Z r 3 Q 1 Z r r r1 dr E dr = Vs 4 r3 r3 V = Vs 2 0 2 r2 1 r2 r 2 2 3 3 Q = Vs 4 r3 1 r3 r2 r22 + rr1 r1 : r2 0 2 1 The potential at the outer surface is found by placing r = r2 in the expression found in part (a). It is Vs = Q=40 r2 . Make this substitution and collect terms to nd r2 Q V = 4 r3 1 r3 322 0 2 1 r2 2 3 r1 : r 3 3 Since = 3Q=4(r2 r1 ) this can also be written 2 (c) The electric eld vanishes in the cavity, so the potential is everywhere the same inside and has the same value as at a point on the inside surface of the shell. Put r = r1 in the result of part (b). After collecting terms the result is 2 2 Q V = 4 3(r2 r1 ) ; 3 3 0 2(r2 r1 ) r2 V = 3 322 0 r2 3 r1 : r 688 CHAPTER 25 ELECTRIC POTENTIAL 2 2 or in terms of the charge density V = (=20 )(r2 r1 ). (d) The solutions agree at r = r1 and at r = r2 . 15E Let V = 0 at r ! 1. Then V (r) = q=40 r: Thus (a) VA VB = 4q r q 0 A 40 rB 1 1 = 4500 V: 6 C)(8:99 109 N m2 =C2 ) = (1:0 10 2:0 m 1:0 m (b) Since V (r) depends only on the magnitude of r, the result is unchanged. (a) From symmetry consideration, the equipotential surface with V = 30 V must be a sphere (of radius R) centered at q, where V = q=40 R. Solve for R: 8 9 2 2 q :99 R = 4 V = (1:5 10 C)(830 V 10 N m =C ) = 4:5 m : 0 16E (b) Since R = (q=40 )(1=V ), R is not proportional to V (but rather to V 1 ), so the surfaces are not evenly spaced. Imagine moving all the charges on the surface of the sphere to the center of the the sphere. Using Gauss' law you can see that this would not change the electric eld outside the sphere. The magnitude of the electric eld E of the uniformly charged sphere as a function of r, the distance from the center of the sphere, is thus given by E (r) = q=(40 r2 ) for r > R. Here R is the radius of the sphere. Thus the potential V at the surface of the sphere (where r = R) is given by 17E V (R) = V q dr = q 40 R R 1 40 r2 r=1 9 2 2 8 = (8:99 10 N m :=C )(1:50 10 C) = 8:43 102 V : 16 0 cm + Z 1 E (r) dr = Z R 18E (10 10 q = 40 RV = 8:99 m)(9 N :mV)C2 = 1:1 10 9 C : 2= 10 CHAPTER 25 ELECTRIC POTENTIAL 689 If the electric potential is zero at in nity then at the surface of a uniformly charged sphere it is V = q=40 R, where q is the charge on the sphere and R is the sphere radius. Thus q = 40 RV and the number of electrons is 19E N = jqj = 40 RjV j = e e (1:0 10 6 m)(400 V) (8:99 109 N m2 /C2 )(1:60 10 19 C) = 2:8 105 : 20E (a) +q +2q some electric field lines (b) +q + 2q some equipotential cross sections First, convince yourself that V (x) cannot be equal to zero for x > d. In fact V (x) is always negative for x > d. Now consider the two remaining regions on the x axis: x < 0 and 21E 690 CHAPTER 25 ELECTRIC POTENTIAL 0 < x < d. For x < 0 the separation between q1 and a point on the x axis whose coordinate is x is given by d1 = x; while the corresponding separation for q2 is d2 = d x. Now set q 1 q q V (x) = 4 d1 + d2 = 4 0 1 2 0 1 + 3 =0 x d x to obtain x = d=2. Similarly, for 0 < x < d we have d1 = x and d2 = d x. Let q 1 1 q q V (x) = 4 d1 + d2 = 4 x + d 3x = 0 0 1 2 0 and solve for x: x = d=4. (a) Since both charges are positive, so is the electric potential created by each of them. Thus the net potential cannot possibly be zero anywhere except at in nity. (b) y q1 0 E2 E=0 x E1 q2 x d 22E In the gure above, let E (x) = E1 E2 = 0, we obtain 1 q 1 E1 (x) = 4 x1 = E2 (x) = 4 (d q2x)2 : 2 0 0 Combine this equation with q1 = +q and q2 = +2q and solve for x: x= 1+ 2 m dp = 1:0 p = 0:41 m : 1+ 2 Since according to the problem statement there is a point in between the two charges on the x axis where the net electric eld is zero, the elds at that point due to q1 and q2 must be directed opposite to each other. This means that q1 and q2 must have the same sign (i.e., either both are positive or both negative). Thus the potentials due to either of them must be of the same sign. Therefore the net electric potential cannot possibly be zero anywhere except at in nity. 23E CHAPTER 25 ELECTRIC POTENTIAL 691 (a) 24E (b) The eld just outside the sphere would be 6 2 2 q 99 9 V = 4 R = (4:0 10 C)(8::10 m 10 N m =C ) = 3:6 105 V : 0 0 5 E = 4q R2 = V = 3:60 10 V = 3:6 106 V/m ; R :10 m 0 which would have exceeded 3:0 MV/m. So this situation cannot occur. (a) (b) 25E (200 V)(0: q = 40 V R = 8:99 109 N15 m)C2 = 3:3 10 9 C : m2 = 3 10 9 q = 4R2 = 3:(0:15 m)C = 1:2 10 8 C/m2 : 2 4 26P (a) The electric potential V at the surface of the drop, the charge q on the drop, and the radius R of the drop are related by V = q=40 R. Thus q = (8:99 109 N m2 /C2 )(30 10 R = 4 V 500 V 0 12 C) = 5:4 10 4 m : (b) After the drops combine the total volume is twice the volume of an original drop, so the radius R0 of the combined drop is given by (R0 )3 = 2R3 and R0 = 21=3 R. The charge is twice the charge of original drop: q0 = 2q. Thus 1 q0 1 V 0 = 4 R0 = 4 212qR = 221V3 = 2(5003V) = 790 V : = 21= 0 0 =3 27P Assume the charge on the Earth is distributed with spherical symmetry. If the electric potential is zero at in nity then at the surface of the Earth it is V = q=40 R, where q is the charge on the Earth and R (= 6:37106 m) is the radius of the Earth. The magnitude of the electric eld at the surface is E = q=40 R2 , so V = ER = (100 V/m)(6:37 106 m) = 6:4 108 V. 692 CHAPTER 25 ELECTRIC POTENTIAL A charge 5q is a distance 2d from point P , a charge 5q is a distance d from P , and two charges +5q are each a distance d from P , so the electric potential at P is 5 5 + 5 + 5 = 5q : q V = 4 2d d d d 80 0 The zero of the electric potential was taken to be at in nity. Use q = 137; 000 C from Sample Problem 22-4 to nd V : 9 2 2 V = 4q R = (137; 000 C)(8:99 10 N m =C ) = 1:93 108 V : 6:37 106 m 0 e The net electric potential at point P is the sum of those due to the six charges: 30P 29P 28P VP = qi i=1 i=1 40 ri 1 20 = 4 p 2 5:0q 2 + d=:2q + p 2 3:0q 2 d + (d=2) d + (d=2) 0 3:0q 2:0q + p 5:0q +p 2 + d + (d=2)2 d=2 d2 + (d=2)2 = 40:94q : d X 6 VPi = X 6 0 (a) and (b). Consider the two points A and B , with (Ax ; Ay ; Az ) = (R + xc ; 0; 0) and (Bx ; By ; Bz ) = (R xc ; 0; 0), where the circle intersects the x axis. We have 31P q1 40 VA = R + x + x (q2 + x ) = 0 ; R c c 2 q1 + q2 > : 40 VB = R x x + (R x ) = 0 : 8 > < Solve for R and xc : c 2 c q2 (6 0e 2 xc = q2 1 x2 2 = (6:0:e)2) (8(:6 nm) 2 = 4:8 nm ; 10e) 1 q2 > q1 q2 x2 = (6:0e)( 10e)(8:6 nm) = 8:1 nm : > : R = 2 2 q1 q2 (6:0e)2 ( 10e)2 8 > > < CHAPTER 25 ELECTRIC POTENTIAL 693 (c) No. In fact the V = 0 one is the only circular one. (a) The net charge carried by the sphere after a time t is q = 1 aet, where a is the activity 2 of the nickel coating. Thus from V = q=40 R = aet=80 R we solve for t: 0V t = 8ae R = (8:99 109 N m22(1000 V)(0:0108m) :60 10 =C2 )(3:70 10 =s)(1 19 C) 32P = 38 s : (b) The time t0 required satis es 1 at0 = cT , where = 100 keV, c = 14:3 J/K, and 2 T = 5:0 C . Thus 2(14:3 J/K)(5:0 C ) 0 = 2cT = t a (3:70 108 =s)(100 103 1:6 10 19 J) = 2:4 109 s = 2:8 102 da : 33E Use Eq. 25-20: 9 2 = 2 :47 3 34 1 V = 4 rp2 = (8:99 10 N m(52C0 )(110 9 m):2 10 : 0 30 C m) = 1:63 10 5 V : A positive charge q is a distance r d from point P , another positive charge q is a distance r from P , and a negative charge q is a distance r + d from P . Sum the individual electric potentials created at P to nd the total: 1 q V = 4 r 1 d + 1 r + d : r 0 Use the binomial theorem to approximate 1=(r d) for r much larger than d: 1 = (r d) 1 (r) 1 (r) 2 ( d) = 1 + d : r d r r2 34E 1 1 d: r + d r r2 Only the rst two terms of each expansion were retained. Thus Similarly, q q q V 4 1 + rd2 + 1 1 + rd2 = 4 1 + 2d = 4 r 1 + 2rd : 2 r r 0 r 0 r r 0 694 CHAPTER 25 ELECTRIC POTENTIAL (a) From Eq. 25-35 35E (b) V = 0 due to superposition. 36E 2= 2 1=2 V = 2 4 ln L=2 + (L d4 + d ) : 0 1 Z dq = 1 Z dq = Q : VP = 4 40 R rod 40 R 0 rod R (a) All the charge is the same distance R from C , so the electric potential at C is 1 Q 6Q = 5Q ; V = 4 R R 40 R 0 where the zero was taken to be at in nity. p (b) All the charge is the same distance from P . That distance is R2 + z 2 , so the electric potential at P is 1 p Q 6Q 5Q V = 4 2 + z 2 pR2 + z 2 = 40 pR2 + z 2 : R 0 The disk is uniformly charged. This means that when the full disk is present each quadrant contributes equally to the electric potential at P , so the potential at P due to a single quadrant is one-fourth the potential due to the entire disk. First nd an expression for the potential at P due to the entire disk. Consider a ring of charge with radius r and width dr. p area is 2r dr and it contains Its charge dq = 2r dr. All the charge in it is a distance r2 + z 2 from P , so the potential it produces at P is 1 2 r dV = 4 pr dr2 = p 2dr 2 : 2+z 20 r + z 0 r The total potential at P is 38E 37E Z R p r dr = pr2 + z2 R = pR2 + z2 z : V = 2 2 + z 2 20 20 0 r 0 0 The potential Vsq at P due to a single quadrant is V = pR2 + z2 z : Vsq = 4 8 0 CHAPTER 25 ELECTRIC POTENTIAL 695 Consider an in nitesimal segment of the ring, located between r0 and r + dr0 from the center of the ring. The area of this segment is dA = 2r0 dr0 , and the charge it carries is dq = dA = 2r0 dr0 . The separation between any point on this segment of the ring p and point P is a = r0 2 + z 2 . Thus the contribution to VP from this segment is given by dVP = dq=(40 a). Integrate over the entire ring to obtain VP : Z 1 Z dq = 1 Z R p2r0 dr0 : VP = dVP = 4 40 r r0 2 + z 2 0 ring a ring Upon making a change of variable u = r0 2 the integral above can be readily evaluated to yield the result R p p VP = 2 [ r02 + z2 ] = 2 ( R2 + z2 0 0 r For z = 2:00R and r = 0:200R this gives VP = 0:113R=0 . p 39P r2 + z 2 ) : 1 2 4 = (2:20 10 2 m)2 [1:50 10 6 C/m2 + 3(8:00 10 7 C/m2 )] 4 = 1:48 10 9 C : (b) Use Eq. 25-36: Z 1 Z R=2 p R0 )dR0 + Z R p R0 )dR0 1 (2 2 (2 dV = 4 V (z) = z2 + R02 0 0 disk R=2 z 2 + R02 disk 0 (a) Denote the surface charge density of the disk as 1 for 0 < r < R=2, and as 2 for R=2 < r < R. Thus the total charge on the disk is given by Z Z R=2 Z R q= dq = 2 rdr + 2 rdr = R2 ( + 3 ) 1 40P R=2 2 2 2 p z2 + R z + 22 z2 + R2 z2 + R : 4 4 0 Plug in thenumerical values of 1 , 2 , R and z to obtain V (z ) = 7:95 102 V. = 21 0 r r (a) Consider an in nitesimal segment of the rod, located between x and x + dx. It has length dx and contains charge dq = dx, where = Q=L is the linear charge density of the rod. Its distance from P1 is d + x and the potential it creates at P1 is 1 dx 1 dV = 4 d dq x = 4 d+ x : 0 + 0 41P 696 CHAPTER 25 ELECTRIC POTENTIAL To nd the total potential at P1 , integrate over the rod: Z L dx = ln(d + x) L = Q ln 1 + L : V = 4 d + x 40 40 L d 0 0 0 42P (a) Similar to the last problem, let us consider an in nitesimal segment of the rod, located between x and x + dx. It has length dx and contains charge dq = dx = cxdx. Its distance from P1 is d + x and the potential it creates at P1 is 1 cx dx 1 dV = 4 d dq x = 4 d + x : 0 + 0 To nd the total potential at P1 , integrate over the rod: c Z L x dx = c [x d ln(x + d)] L = c L d ln 1 + L V = 4 40 d 0 0 d + x 40 0 : The magnitede of the electric eld is given by jE j = V = 2(5:0 V) = 6:7 102 V/m : s 0:015 m E points from the positively charged plate to the negatively charged one. Use Ex = dV=dx, where dV=dx is the local slope of the V vs. x curve depicted in Fig. 25-48. The results are: Ex (ab) = 6:0 V/m, Ex (bc) = 0, Ex (cd) = Ex (de) = 3:0 V/m, Ex (ef ) = 15 V/m, Ex (fg) = 0, Ex (gh) = 3:0 V/m. Ex (V/m) 15 12 44E 43E e f 6 c b -5 d 0 e f g 5 c x (m) h g a b -6 CHAPTER 25 ELECTRIC POTENTIAL 697 On the dipole axis = 0 or so j cos j = 1. The magnitude of the electric eld is thus given by @V = p d 1 = p : jE (r)j = @r 4 dr r2 2 r3 0 0 46E 45E Use Eq. 25-41: @ Ex (x; y) = @V = @x [(2:0 V/m2 )x2 (3:0 V/m2 )y2 ] = 2(2:0 V/m2 )x ; @x @V = @ [(2:0 V/m2 )x2 (3:0 V/m2 )y2 ] = 2(3:0 V/m2 )y : Ey (x; y) = @y @y 2 2 Now plug in x = 3:0 m and y = 2:0 m to obtain the magnitude of E: E = Ex + Ey = 17 V/m. E makes an angle with the positive x axis, where q = tan 1 Ey = 135 : Ex 47E E = d = dx (1500x2 ) i = ( 3000x) i = ( 3000 V/m2 )(0:0130 m) i = ( 39 V/m) i : i dV dx (a) The E - eld corresponding to V is 2 d E = dV = dr 4Ze 1 23 + 2r 3 dr 0 r R R r = 4Ze r12 R3 : 0 48E (b) The expression for V (r) is valid inside the atom only and cannot be extended to r ! 1: 698 CHAPTER 25 ELECTRIC POTENTIAL (a) The charge on every part of the ring is the same distance from any point P on the axis. p This distance is r = z 2 + R2 , where R is the radius of the ring and z is the distance from the center of the ring to P . The electric potential at P is Z 1 Z dq = 1 Z p dq = 1 p 1 1 V = 4 dq = 4 p 2 q 2 : r 40 z2 + R2 40 z2 + R2 0 0 z +R 49P (b) The electric eld is along the axis and its component is given by q d E = @V = 4 dz (z2 + R2 ) 1=2 @z 0 q 1 q = 4 2 (z 2 + R2 ) 3=2 (2z ) = 4 2 z 2 3=2 : 0 0 (z + R ) This agrees with the result of Section 23-6. 50P Replace x = d with a general value x to obtain From 41P the electric potential for a point on the x axis whose x coordinate is given by x = d < 0 is Q ln 1 + L : V (x = d) = 4 L d 0 Q V (x) = 4 L ln 1 L : x 0 Thus d Q Ex (x = d) = dV = dx 4 L ln 1 L dx x= d x 0 1 = 4 d(LQ d) : + 0 x= d Note that here Ex < 0, indicating that E is in the negative x direction, as expected. (b) From symmetry it is obvious that the electric eld does not have any y component, i.e., Ey = 0. (a) Consider an in nitesimal segment of the rod from x to x + dx. Its contribution to the potential at point P is 51P cx 1 1 dVP = 4 p (x)dx 2 = 4 p 2 2 dx: 0 x2 + y 0 x +y CHAPTER 25 ELECTRIC POTENTIAL 699 Thus (b) c Z L p x dx = c (pL2 + y2 y) : VP = dVP = 4 40 0 0 x2 + y2 rod Z : L2 + y2 (c) All we obtained above for the potential is its value at any point P on the y-axis. In order to obtain Ex (x; y) we need to rst calculate V (x; y); i.e., the potential for an arbitrary point located at (x; y). Then Ex (x; y) can be obtained from Ex (x; y) = @V (x; y)[email protected] p 52E c c d p EP;y = @VP = 4 dy ( L2 + y2 y) = 4 1 @y 0 0 y (a) Use Eq. 25-43 with q1 = q2 = e and r = 2:00 nm: 9 2 2 )(1 19 2 1 1 2 U = 4 q1rq2 = 4 er = (8:99 10 N2:m =C10 9:60 10 C) = 1:15 10 00 m 0 0 (b) Since U > 0 and U / r 1 the potential energy U decreases as r increases. 53E 19 J : (a) The charges are equal and are the same distance from C . Use the Pythagorean theorem p p to nd the distance r = (d=2)2 + (d=2)2 = d= 2. The electric potential at C is the sum of the potential due to the individual charges but since they produce the same potential, it is twice that of either one: p p 2q 2 = 2 2q V = 4 d 4 d 0 0 9 N m2 /C2 )(2)p2(2:0 10 6 C) = (8:99 10 = 2:5 106 V : 0:020 m (b) As you move the charge into position from far away the potential energy changes from zero to qV , where V is the electric potential at the nal location of the charge. The change in the potential energy equals the work you must do to bring the charge in: W = qV = (2:0 10 6 C)(2:5 106 V) = 5:1 J. (c) The work calculated in part (b) represents the potential energy of the interactions between the charge brought in from in nity and the other two charges. To nd the total potential energy of the three-charge system you must add the potential energy of the interaction between the xed charges. Their separation is d so this potential energy is q2 =40 d. The total potential energy is q2 U = W + 4 d 0 (8:99 109 N m2 /C2 )(2:0 10 6 C)2 = 6:9 J : = 5:1 J + 0:020 m 700 CHAPTER 25 ELECTRIC POTENTIAL The potential energy of the two-charge system is 1 p q1 q2 U = 4 (x1 x2 )2 + (y1 y2 )2 0 9N 2 2 6 6 = (8:99 10p m =C )(3:0 10 C)( 24:0 10 C) = 1:9 J : (3:5 + 2:0)2 + (0:50 1:5) cm Thus 1:9 J of work is needed. (a) Denote the side length of the triangle as a. Then the electric potential energy is 9 2 = 2 )(1 19 2 e= 2 U = 3(3) = (8:99 10 N m82 C 10 :60) 10 C) 4 0a 3(2: 31 = 2:72 10 14 J : (b) U=c2 = 2:72 10 14 J=(3:00 108 m/s)2 = 3:02 10 31 kg: This is about a third of the 55E 54E accepted value of the electron mass. 56E Choose the zero of electric potential to be at in nity. The initial electric potential energy Ui of the system before the particles are brought together is therefore zero. After the system is set up the nal potential energy is 1 1 + p1 q2 1 1 + p1 Uf = 4 a a 2a a a 2a 0 2 1 2 21 2q = 4 a p 2 = 0: aq : 2 0 0 Thus the amount of work required to set up the system is given by W = U = Uf Ui = 0:21q2 =(0 a). Let the quark-quark separation be a. Then (a) 1 4e2 Uup-up = 4 (2e=3)(2e=3) = 4 a a 0 0 9 N m2 =C2 )(1:60 10 = 4(8:99 109(1:32 10 15 m) = 4:84 105 eV = 0:484 MeV: 19 C)e 57E CHAPTER 25 ELECTRIC POTENTIAL 701 (b) 1 U = 4 (2e=3)(2e=3) a 0 2(2e=3)(2e=3) = 0 : a 58E 1 X qi qj = 1 q q + q q + q q + q q + qpq4 + qpq3 1 2 U = 4 1 2 1 3 2 4 3 4 40 d 2 2 0 i6=j rij (8:99 109 N m2 =C2 ) (12)( 24) + (12)(31) + ( 24)(17) + (31)(17) = 1:3 m (12)(17) + ( 24)(31 (10 19 C)2 p + p 2 2 6 J: = 1:2 10 Let q = 0:12 C and a = 1:7 m. The change in electric potential energy of the three-charge system as one of the charges is moved as described in the problem is 2q2 q2 2 2 U = 4 a=2 a = 4 a 0 0 2 (8:99 109 N m2 =C2 ) = 1:5 108 J: = 2(0:12 C) 1:7 m 59P Thus the number of days required would be 8 n = (0:83 (1:53 10 J) s/d) = 2:1 d : 10 W)(86400 (a) Let ` (= 0:15 m) be the length of the rectangle and w (= 0:050 m) be its width. Charge q1 is a distance ` from point A and charge q2 is a distance w, so the electric potential at A is 1 q1 + q2 VA = 4 ` w 0 6 6 9 N m2 /C2 ) 5:0 10 C + 2:0 10 C = (8:99 10 0:15 m 0:050 m 4 V: = 6:0 10 60P 702 CHAPTER 25 ELECTRIC POTENTIAL (b) Charge q1 is a distance w from point b and charge q2 is a distance `, so the electric potential at B is 1 q1 + q2 VB = 4 w ` 0 6 6 9 N m2 /C2 ) 5:0 10 C + 2:0 10 C = (8:99 10 0:050 m 0:15 m 5 V: = 7:8 10 (c) Since the kinetic energy is zero at the beginning and end of the trip, the work done by an external agent equals the change in the potential energy of the system. The potential energy is the product of the charge q3 and the electric potential. If UA is the potential energy when q3 is at A and UB is the potential energy when q3 is at B , then the work done in moving the charge from B to A is W = UA UB = q3 (VA VB ) = (3:0 10 6 C)(6:0 104 V + 7:8 105 V) = 2:5 J. (d) The work done by the external agent is positive, so the energy of the three-charge system increases. (e) and (f) The electrostatic force is conservative, so the work is the same no matter what the path. The work required is )(5 1 W = U = 4 (4q2d q) + (5q)(d 2q) = 0 : 0 61P The particle with charge q has both potential and kinetic energy and both these change when the radius of the orbit is changed. Find an expression for the total energy in terms of the orbit radius. Q provides the centripetal force required for q to move in uniform circular motion. The magnitude of the force is F = Qq=40 r2 , where r is the orbit radius. The acceleration of q is v2 =r, where v is its speed. Newton's second law yields Qq=40 r2 = mv2 =r, so mv2 = Qq=40 r and the kinetic energy is K = 1 mv2 = Qq=80 r. 2 The potential energy is U = Qq=40 r and the total energy is 62P Qq = Qq : 40 r 80 r 0 When the orbit radius is r1 the energy is E1 = Qq=80 r1 and when it is r2 the energy is E2 = Qq=80 r2 . The dierence E2 E1 is the work W done by an external agent to change the radius: Qq E = K + U = 8 r W = E2 E1 = 8Qq r1 0 2 1 = Qq 1 r1 80 r1 1 : r 2 CHAPTER 25 ELECTRIC POTENTIAL 703 9 m2 = 2 19 1 e V (r) = 4 r = (8:99 10 N:29 C )(1:60 10 C) = 27:2 V: 5 10 11 m 0 (b) U = eV (r) = 27:2 eV: (c) Since me v2 =r = e2 =40 r2 ; 1 mv2 = 1 e2 = 1 V (r) = 27:2 eV = 13:6 eV: K=2 2 40 r 2 2 (a) 63P (d) The energy required is E = 0 [V (r) + K ] = 0 ( 27:2 eV + 13:6 eV) = 13:6 eV: 64P Use the conservation of energy principle. The initial potential energy is Ui = q2 =40 r1 , the initial kinetic energy is Ki = 0, the nal potential energy is Uf = q2 =40 r2 , and the nal kinetic energy is Kf = 1 mv2 , where v is the nal speed of the particle. Conservation 2 of energy yields q2 = q2 + 1 mv2 : 4 r 4 r 2 The solution for v is s 2q2 v = 4 m r1 0 1 s 0 1 0 2 r2 1 9 m2 /C2 6 2 = (8:99 10 N 20 10)(2)(3:1 10 C) 0:90 1 3 m 6 kg 10 = 2:5 103 m/s : 1 2:5 10 3 m Let r = 1:5 m, x = 3:0 m, q1 = 9:0 nC, and q2 = 6:0 pC. The work done is given by q1 q2 1 p 1 W = U = 4 r r2 + x2 0 = ( 9:0 10 9 C)( 6:0 10 12 C)(8:99 109 N m2 =C2 ) # " 1 1 1:5 m p (1:5 m)2 + (3:0 m)2 = 1:8 10 10 J : 65P 704 CHAPTER 25 ELECTRIC POTENTIAL 66P (a) The potential energy is relative to the potential energy at in nite separation. (b) Each sphere repels the other with a force that has magnitude 9 2 /C2 )(5 6 2 q2 F = 4 d2 = (8:99 10 N m :00 m)2 :0 10 C) = 0:225 N : (1 0 According to Newton's second law the acceleration of each sphere is the force divided by the mass of the sphere. Let mA and mB be the masses of the spheres. The acceleration of sphere A is F : aA = m = 5:00225 N kg = 45:0 m/s2 10 3 and the acceleration of sphere B is A 9 2 /C2 )(5 6 2 q2 U = 4 d = (8:99 10 N m1:00 m :0 10 C) = 0:225 J ; 0 part (a). The initial kinetic energy is zero since the spheres start from rest. The nal potential energy is zero since the spheres are then far apart. The nal kinetic energy is 1 mA v 2 + 1 mB v 2 , where vA and vB are the nal velocities. Thus A 2 B 2 1 2 1 2 U = 2 mA vA + 2 mB vB : Momentum is also conserved, so 0 = mA vA + mB vB : Solve these equations simultaneously for vA and vB . Substitute vB = (mA =mB )vA , from the momentum equation, into the energy equation 2 and collect terms. You should obtain U = 1 (mA =mB )(mA + mB )vA . Thus 2 F : aB = m = 100225 N)kg = 22:5 m/s2 : 10 3 B (c) Energy is conserved. The initial potential energy is U = 0:225 J, as calculated in vA = m (2UmBm ) A mA + B s 3 = (5:0 10 32(0:2250J)(10 310 +kg) 10 3 kg) = 7:75 m/s : kg)(5: 10 kg 10 Now calculate vB : mA v = 5:0 10 3 kg (7:75 m/s) = 3:87 m/s : vB = m A 10 10 3 kg B s CHAPTER 25 ELECTRIC POTENTIAL 705 The initial speed vi of the electron satis es Ki = 1 me vi2 = eV , which gives 2 67P vi = r 2eV = m e s 2(1:60 10 19 J)(625 V) = 1:48 107 m/s : 9:11 10 31 kg (a) At the smallest center-to-center separation rmin the initial kinetic energy Ki of the proton is entirely converted to the electric potential energy between the proton and the neucleus. Thus 1 82 2 Ki = 4 eq lead = 4 e : r r Thus 0 min 0 min 68P 82e2 = 82e2 82(1:6 10 19 C)(8:99 109 N m2 =C2 ) rmin = 4 K 4 (4:80 106 e) V 4:80 106 V 0 i 0 = 2:5 10 14 m = 25 fm : Note that 1 eV = 1e 1 V. (b) In this case 1 q q lead = 2 82e2 82 2 = 4 e ; Ki = 4 r0 0 40 rmin 0 min 0 rmin 0 so the new minimum separation is rmin = 2rmin = 50 fm: 69P The idea for solving this problem is the same as that for the last one. In this case 1 K = 4 rqQ ; 0 min which gives rmin = qQ=40 K . The change in electric potential energy of the electron-shell system as the electron starts from its initial position and just reaches the shell is U = ( e)( V ) = eV: Thus from U = K = 1 me vi2 we nd the initial electron speed to be 2 70P vi = r 2U = m e r 2eV : m e 706 CHAPTER 25 ELECTRIC POTENTIAL Use the conservation of energy principle. Take the potential energy to be zero when the moving electron is far away from the xed electrons. The nal potential energy is then Uf = 2e2 =40 d, where d is the half the distance between the xed electrons. The initial kinetic energy is Ki = 1 me v2 , where me is the mass of an electron and v is the initial 2 speed of the moving electron. The nal kinetic energy is zero. Thus Ki = Uf or 1 me v2 = 2 2e2 =40 d. Hence s 71P v = 4 dm = 0 e 72P 4e2 s (8:99 109 N m2 /C2 )(4)(1:60 10 (0:010 m)(9:11 10 31 kg) 19 C)2 = 3:2 102 m=s : The potential dierence between the surface of the sphere of charge Q and radius r and a point in nitely far from it is V = Q=40 r. Thus the escape speed vesc of the electron of 2 mass me satis es K = 1 me vesc = eV , or 2 19 C)(1 60 15 2 :99 109 2 2 vesc = 4eQ r = 2(1:60 10 (9:11 :10 10 :C)(810 m) N m =C ) 31 kg)(1 0 2 0 me = 2:2 104 m/s : r s Let the distance in question be r. The initial kinetic energy of the electron is Ki = 1 me vi2 , 2 where vi = 3:2 105 m/s. As the speed doubles, K becomes 4Ki . Thus e2 = K = (4K K ) = 3K = 3 m v2 ; U = 4 r i i i 2 ei 0 or 19 2 (8:99 109 m2 = 2 22 r = 3(4 e)m v2 = 2(1:6 10 10 C) kg)(3: 10N m/s)C ) 5 2 3(9:11 19 2 0 e i = 1:6 10 9 m : Since the electric potential throughout the entire conductor is a constant, the electric potential at its center is also +400 V. (a) and (b) For r < R = 20 cm the electric eld is zero since this region is inside the conductor. For r > R we have E (r) = q=(40 r2 ), where q = +3:0 C. In particular, just outside the sphere E = q=(40 R2 ) = E0 = 6:8 105 N/C. 75E 74E 73P CHAPTER 25 ELECTRIC POTENTIAL R 707 Use V (r) = 01 E (r)dr to calculate V (r). For r < R = 20 cm the electric potential is a constant since this region is inside the conductor, an equipotential body. For r R we have V (r) = q=(40 r). In particular, at r = R V = q=(40 R) = V0 = 1:4 105 V. This is also the value of V (r) for r < R. Both E and V as functions of r are plotted below. E V E0 1/r 2 V0 1/r r 20 cm 20 cm r V0=1.4x105 V. 76E If the electric potential is zero at in nity, then the potential at the surface of the sphere is given by V = q=40 r, where q is the charge on the sphere and r is its radius. Thus q = 40 rV = (0:15 m)(15002V) 2 = 2:5 10 8 C : 8:99 109 N m /C 77E (a) Since the two conductors are connected V1 and V2 must be the same. (b) Let V1 = q1 =40 R1 = V2 = q2 =40 R2 and note that q1 + q2 = q and R2 = 2R1 . Solve for q1 and q2 : q1 = q=3, q2 = 2q=3. (c) 2 1 = q1 =4R1 = q1 R2 2 = 2 : q =4R2 q R 2 2 2 2 1 78P In sketching the electric eld lines and the equipotential lines pay attention to the following: (1) The electric eld lines and the equipotential lines and always perpendicular to each other wherever they intersect. (2) The led lines are perpenduluar to the surface of the conductor (which is an equipotential surface); while the equipotential lines just outside 708 CHAPTER 25 ELECTRIC POTENTIAL the conductor follow its surface countour. (3) The greater the curvature of the surface of the conductor, the higher the concentration of the surface charge and hence the greater the electric eld outside the surface. This means that the electric eld just outside the conductor is the strongest near its narrower end; and the weakest in between the two ends, where the surface is at. The eld is of course zero inside the conductor, where the potential is a constant. axis some equipotential lines and electric field lines (denoted with arrows) (a) The potential would be 2 Qe = 4Re e Ve = 4 R 4 R 0 e 0 e 6 m)(1:0 electron= m2 )( 1:6 10 = 4(6:37 10 = 0:12 V : 19 C= electron)(8:99 109 N m2 =C2 ) 79P (b) where the minus sign indicates that E is radially inward. 80P V 0 12 V E = e = Re = 6:37 : 106 m = 1:8 10 8 N/C ; 0 e (a) The electric potential is the sum of the contributions of the individual spheres. Let q1 be the charge on one and q2 be the charge on the other. The point halfway between them CHAPTER 25 ELECTRIC POTENTIAL 709 is the same distance d (= 1:0 m) from the center of each sphere, so the potential at the halfway point is 9 2 2 8 8 1+ V = q4 q2 = (8:99 10 N m /C )(1::0 10 C 3:0 10 C) = 1:80 102 V : 1 0m 0d (b) The distance from the center of one sphere to the surface of the other is d R, where R is the radius of either sphere. The potential of either one of the spheres is due to the charge on that sphere and the charge on the other sphere. The potential at the surface of sphere 1 is 1 V1 = 4 q1 + d q2 R 0 R = (8:99 109 N m2 /C2 ) = 2:7 103 V : 1:0 10 8 C 0:030 m 3:0 10 8 C 1:0 m 0:030 m The potential at the surface of sphere 2 is 1 V2 = 4 d q1 R + q2 R 0 = (8:99 109 N m2 /C2 ) = 8:9 103 V : 1:0 10 8 C 1:0 m 0:030 m 3:0 10 8 C 0:030 m (a) 81P 8 :99 9 2 2 E = = 4q R2 = (3:0 10 C)(815 m)210 N m =C ) = 1:2 104 N/C : (0: 0 0 (b) V = RE = (0:15 m)(1:2 104 N/C) = 1:8 103 V: (c) Let the distance be x. Then q 1 1 V = V (x) V = 4 R + x R = 500 V ; 0 which gives R :15 m)( x = V V V = (01800 V +500 V) = 5:8 10 2 m : 500 V 710 CHAPTER 25 ELECTRIC POTENTIAL Since the charge distribution is spherically symmetric we may write 1 E (r) = 4 qencl ; 0 r where qR 1 is the charge enclosed in a sphere of radius r centered at the origin. Also encl V (r) = r E (r) dr. The results are as follows: For r > R2 > R1 8 q1 + q2 ; > V (r ) = < 40 r q1 + q2 ; > : E (r ) = 4 r2 for R2 > r > R1 0 8 > > < 82P and for R2 > R1 > r q 1 V (r) = 4 qr1 + R2 ; 0 2 q1 ; > > : E (r ) = 4 r2 0 E = 0; 1 q1 + q2 : : V (r ) = 40 R1 R2 8 < E(r) (10 4 N/C) 8.0 6.0 4.0 V(r) (10 4 V) 5.0 4.0 3.0 2.0 1.0 2.0 0.50 1.0 2.0 3.0 4.0 r(m) 0.50 1.0 2.0 3.0 4.0 r(m) (c) 4:24 V (a) for N < 12 con guration 1 is less energetic, for N 12 con guration 2 is less energetic; (b) 12; (c) 2 84 83 ...
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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