CHAPTER 25. Answer to Checkpoint Questions

Transcription

1 CHAPTER 5 ELECTRIC POTENTIAL 68 CHAPTER 5 Answer to Checkoint Questions. (a) negative; (b) increase. (a) ositive; (b) higher 3. (a) rightward; (b),, 3, 5: ositive; 4: negative; (c) 3, then,, and 5 tie, then 4 4. all tie 5. a, c (zero), b 6. (a), then and 3 tie; (b) 3; (c) accelerate leftward 7. closer (half of 9:3 fm) Answer to Questions. (a) higher; (b) ositive; (c) negative; (d) all tie. (a) left; (b) 50 V; (c) ositive; (d) negative 3. (a) and ; (b) none; (c) no; (d) and yes, 3 and 4 no 4. 4 d 5. (b), then (a), (c) and (d) tie 6. all tie 7. (a) negative; (b) zero 8. (a){(c) Q R; (d) a, b, c 9. (a), then and 3 tie; (b) 3 0. E z, E y, E x. left. (a), 4, and then a tie of, 3, and 5 (where E 0); (b) negative x direction; (c) ositive x direction 3. (a), (b), (c) 4. (a) 3 and 4 tie, then and tie; (b) and, increase; 3 and 4, decrease

2 68 CHAPTER 5 ELECTRIC POTENTIAL 5. (a) c, b, a; (b) zero 6. (a) { (d) zero 7. (a) ositive; (b) ositive; (c) negative; (d) all tie 8. farther 9. (a) no; (b) yes 0. no. no (a article at the intersection would have two dierent otential energies). (a) { (c) C, B, A; (d) all tie 3. (a) { (b) all tie; (c) C, B, A; (d) all tie 4. situation Solutions to Exercises & Problems E The magnitude is U ev : 0 9 ev : GeV: E (a) An amere is a coulomb er second, so 84 Ah (84 Chs) (3600 sh) 3:0 0 5 C : (b) The change in otential energy is U V (3:0 0 5 C)( V) 3:6 0 6 J. 3P (a) When charge moves through a otential dierence V its otential energy changes by U V. In this case, U (30 C)(:0 0 9 V) 3:0 0 0 J. (b) Euate the nal kinetic energy of the automobile to the energy released by the lightning: U mv, where m is the mass of the automobile and v is its nal seed. Thus r s U v m (3:0 0 0 J) 7:7 0 3 m/s : 000 kg (c) Euate the energy reuired to melt mass m of ice to the energy released by the lightning: U ml F, where L F is the heat of fusion for water. Thus m U L F 3:0 00 J 3:3 0 5 J/kg 9:0 04 kg :

3 CHAPTER 5 ELECTRIC POTENTIAL 683 4E +λ -λ z axis 5P -λ - λ +λ +λ +λ +λ some electric field lines some euiotential cross sections 6E (a) V B V A U( e) 3: J( : C) :46 V: (b) V C V A V B V A :46 V: (c) V C V B 0 (Since C and B are on the same euiotential line).

4 684 CHAPTER 5 ELECTRIC POTENTIAL 7E V Es (:9 0 5 N/C)(0:050 m) : V: 8E (a) E Fe 3:9 0 5 N(: C) :4 0 4 N/C: (b) V Es (:4 0 4 N/C)(0: m) :9 0 3 V: 9E The electric eld roduced by an innite sheet of charge has magnitude E 0, 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 arallel to the eld and ositive in the direction of the eld. Then the electric otential is V V s Z x 0 E dx V s Ex ; where V s is the otential at the sheet. The euiotential surfaces are surfaces of constant x; that is, they are lanes that are arallel to the lane of charge. If two surfaces are searated by x then their otentials dier by V Ex ( 0 )x. Thus x 0 V (8:85 0 C /Nm )(50 V) 0:0 0 6 C/m 8:8 0 3 m : 0P (a) W Z f (b) Since V V 0 W 0 z 0, i 0 Eds 0 0 Z z V V 0 0 z 0 : dz 0z 0 : P The otential dierence between the wire and cylinder is given, not the linear charge density on the wire. Use Gauss' law to nd an exression 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 resect to r to nd an exression for the otential dierence between the wire and cylinder in terms of the linear charge density. Use this result to obtain an exression for the linear charge density in terms of the otential dierence and substitute the result into the euation for the electric eld. This will give the electric eld in terms of the otential dierence and will allow you to comute numerical values for the eld at the wire and at the cylinder.

5 CHAPTER 5 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 ortion of the cylinder's surface and is uniform over that surface. This means the electric ux through the Gaussian surface is given by r`e, where E is the magnitude of the electric eld. The charge enclosed by the Gaussian surface is `, where is the linear charge density on the wire. Gauss' law yields 0 r`e `. Thus E 0 r : Since the eld is radial, the dierence in the otential V c of the cylinder and the otential V w of the wire is V V w V c Z r w Z r c E dr rc rw 0 r dr rc ln ; 0 r w where r w is the radius of the wire and r c is the radius of the cylinder. This means that and E 0 V ln(r c r w ) 0 r V r ln(r c r w ) : (a) Substitute r w for r to obtain the eld at the surface of the wire: E V r w ln(r c r w ) : V/m : 850 V (0: m) ln [(:0 0 m)(0: m)] (b) Substitute r c for r to nd the eld at the surface of the cylinder: E V r w ln(r c r w ) 8:8 0 3 V/m : 850 V (:0 0 m) ln [(:0 0 m)(0: m)] P (a) V (r) V (0) Z r 0 E(r)dr 0 Z r 0 r R dr r R : 3 (b) V V (0) V (R) 8 0 R: (c) Since V V (0) V (R) > 0; the otential at the center of the shere is higher. 3P (a) Use Gauss' law to nd exressions for the electric eld inside and outside the sherical charge distribution. Since the eld is radial the electric otential can be written as an integral of the eld along a shere radius, extended to innity. Since dierent exressions for the eld aly in dierent regions the integral must be slit into two arts, one from innity to the surface of the distribution and one from the surface to a oint inside.

6 686 CHAPTER 5 ELECTRIC POTENTIAL Outside the charge distribution the magnitude of the eld is E r and the otential is V r, where r is the distance from the center of the distribution. To nd an exression for the magnitude of the eld inside the charge distribution use a Gaussian surface in the form of a shere 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 4r E. The charge enclosed is r 3 R 3. Gauss' law becomes r E r3 R ; 3 so E r R : 3 If V s is the otential at the surface of the distribution (r R) then the otential at a oint inside, a distance r from the center, is V V s Z r R E dr V s Z r R 3 R r r dr V s 8 0 R R : The otential at the surface can be found by relacing r with R in the exression for the otential at oints outside the distribution. It is V s R. Thus V R r R + 3 R 8 0 R 3 (3R r ) : (b) In P the electric otential was taken to be zero at the center of the shere. In this roblem it is zero at innity. According to the exression derived in art (a) the otential at the center of the shere is V c 38 0 R. Thus V V c r 8 0 R 3. This is the result of P. (c) The otential dierence is V V s V c 8 0 R R 8 0 R : The same value as is given by the exression obtained in P. (d) Only otential dierences have hysical signicance, not the value of the otential at any articular oint. The same value can be added to the otential at every oint without changing the electric eld, for examle. Changing the reference oint from the center of the distribution to innity changes the value of the otential at every oint but it does not change any otential dierences. 4P (a) For r > r the eld is like that of a oint charge and V Q r ; where the zero of otential was taken to be at innity.

7 CHAPTER 5 ELECTRIC POTENTIAL 687 (b) To nd the otential in the region r < r < r, rst use Gauss's law to nd an exression for the electric eld, then integrate along a radial ath from r to r. The Gaussian surface is a shere 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 is 4r E. The volume of the shell is (43)(r 3 r), 3 so the charge density is 3Q 4(r 3 r 3 ) and the charge enclosed by the Gaussian surface is 4 3 Gauss' law yields and the magnitude of the electric eld is (r 3 r 3 ) Q r 3 r 3 r 3 r 3 r E Q r 3 r 3 r 3 r 3 E Q r 3 r 3 r (r 3 r 3 ) : If V s is the electric otential at the outer surface of the shell (r r ) then the otential a distance r from the center is given by Z r Z r r Q V V s E dr V s 4 r 0 r 3 r 3 Q r r V s r 3 r 3 + r3 r r : r 3 : r r 3 dr r The otential at the outer surface is found by lacing r r in the exression found in art (a). It is V s Q r. Make this substitution and collect terms to nd V Q r 3 r 3 3r r r 3 r : Since 3Q4(r 3 r 3 ) this can also be written V 3 0 3r r r 3 : r (c) The electric eld vanishes in the cavity, so the otential is everywhere the same inside and has the same value as at a oint on the inside surface of the shell. Put r r in the result of art (b). After collecting terms the result is V Q 3(r r ) (r 3 r 3 ) ;

8 688 CHAPTER 5 ELECTRIC POTENTIAL or in terms of the charge density V ( 0 )(r r ). (d) The solutions agree at r r and at r r. 5E Let V 0 at r!. Then V (r) r: Thus (a) V A V B r A r B (:0 0 6 C)(8: Nm C ) :0 m 4500 V: :0 m (b) Since V (r) deends only on the magnitude of r, the result is unchanged. 6E (a) From symmetry consideration, the euiotential surface with V shere (of radius R) centered at, where V R. Solve for R: 30 V must be a R V (:5 0 8 C)(8: Nm C ) 30 V 4:5 m : (b) Since R ( )(V ), R is not roortional to V (but rather to V the surfaces are not evenly saced. ), so 7E Imagine moving all the charges on the surface of the shere to the center of the the shere. Using Gauss' law you can see that this would not change the electric eld outside the shere. The magnitude of the electric eld E of the uniformly charged shere as a function of r, the distance from the center of the shere, is thus given by E(r) ( r ) for r > R. Here R is the radius of the shere. Thus the otential V at the surface of the shere (where r R) is given by V (R) V r + Z R E(r) dr Z R r dr (8:99 09 Nm C )(: C) 6:0 cm R 8:43 0 V : 8E RV (0 m)( :0 V) 8: Nm C : 0 9 C :

9 CHAPTER 5 ELECTRIC POTENTIAL 689 9E If the electric otential is zero at innity then at the surface of a uniformly charged shere it is V R, where is the charge on the shere and R is the shere radius. Thus RV and the number of electrons is N jj e RjV j e (:0 0 6 m)(400 V) (8: Nm /C )(: C) :8 05 : 0E (a) + + some electric field lines (b) + + some euiotential cross sections E First, convince yourself that V (x) cannot be eual 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

10 690 CHAPTER 5 ELECTRIC POTENTIAL 0 < x < d. For x < 0 the searation between and a oint on the x axis whose coordinate is x is given by d x; while the corresonding searation for is d d x. Now set V (x) d + d x + 3 d x to obtain x d. Similarly, for 0 < x < d we have d x and d d x. Let and solve for x: x d4. V (x) d + d x + 3 d x 0 0 E (a) Since both charges are ositive, so is the electric otential created by each of them. Thus the net otential cannot ossibly be zero anywhere excet at innity. (b) y E0 0 E x E d x In the gure above, let E(x) E E 0, we obtain E (x) x E (x) (d x) : Combine this euation with + and + and solve for x: x d + :0 m + 0:4 m : 3E Since according to the roblem statement there is a oint in between the two charges on the x axis where the net electric eld is zero, the elds at that oint due to and must be directed oosite to each other. This means that and must have the same sign (i.e., either both are ositive or both negative). Thus the otentials due to either of them must be of the same sign. Therefore the net electric otential cannot ossibly be zero anywhere excet at innity.

11 CHAPTER 5 ELECTRIC POTENTIAL 69 4E (a) V R (4:0 0 6 C)(8: Nm C ) 0:0 m (b) The eld just outside the shere would be E R V R 3:6 05 V 0:0 m 3:6 06 V/m ; which would have exceeded 3:0 MV/m. So this situation cannot occur. 3:6 0 5 V : 5E (a) (b) V R (00 V)(0:5 m) 8: Nm C 3:3 0 9 C : 4R 3:3 0 9 C 4(0:5 m) : 0 8 C/m : 6P (a) The electric otential V at the surface of the dro, the charge on the dro, and the radius R of the dro are related by V R. Thus R V (8:99 09 Nm /C )(30 0 C) 500 V 5:4 0 4 m : (b) After the dros combine the total volume is twice the volume of an original dro, so the radius R 0 of the combined dro is given by (R 0 ) 3 R 3 and R 0 3 R. The charge is twice the charge of original dro: 0. Thus V 0 0 R 0 3 R V (500 V) 790 V : 3 3 7P Assume the charge on the Earth is distributed with sherical symmetry. If the electric otential is zero at innity then at the surface of the Earth it is V R, where is the charge on the Earth and R ( 6:370 6 m) is the radius of the Earth. The magnitude of the electric eld at the surface is E R, so V ER (00 V/m)(6: m) 6:4 0 8 V.

12 69 CHAPTER 5 ELECTRIC POTENTIAL 8P A charge 5 is a distance d from oint P, a charge 5 is a distance d from P, and two charges +5 are each a distance d from P, so the electric otential at P is V 5 d 5 d + 5 d + 5 d The zero of the electric otential was taken to be at innity : 9P Use 37; 000 C from Samle Problem -4 to nd V : V (37; 000 C)(8:99 09 Nm C ) R e 6: m : V : 30P The net electric otential at oint P is the sum of those due to the six charges: 6X 6X i V P V P i r i i i 5:0 d + (d) + :0 d + 3:0 d + (d) + 0:94 d : 3:0 d + (d) + :0 d + 5:0 d + (d) 3P (a) and (b). Consider the two oints A and B, with (A x ; A y ; A z ) (R + x c ; 0; 0) and (B x ; B y ; B z ) (R x c ; 0; 0), where the circle intersects the x axis. We have Solve for R and x c : 8> < >: 8 >< >: V A R + x c + V B R x c + x c x R x x (R + x c ) 0 ; x + (R x c ) 0 : (6:0e) (8:6 nm) 4:8 nm ; (6:0e) ( 0e) (6:0e)( 0e)(8:6 nm) 8: nm : (6:0e) ( 0e)

13 CHAPTER 5 ELECTRIC POTENTIAL 693 (c) No. In fact the V 0 one is the only circular one. 3P (a) The net charge carried by the shere after a time t is aet, where a is the activity of the nickel coating. Thus from V R aet8 0 R we solve for t: t 8 0V R ae (000 V)(0:00 m) (8: Nm C )(3: s)(: C) 38 s : (b) The time t 0 reuired satises at0 ct, where 00 kev, c 4:3 J/K, and T 5:0 C. Thus t 0 ct a (4:3 J/K)(5:0 C ) (3: s)( :6 0 9 J) :4 0 9 s :8 0 da : 33E Use E. 5-0: V r (8:99 09 Nm C )(:47 3: Cm) : V : (5:0 0 9 m) 34E A ositive charge is a distance r d from oint P, another ositive charge is a distance r from P, and a negative charge is a distance r + d from P. Sum the individual electric otentials created at P to nd the total: V r d + : r r + d Use the binomial theorem to aroximate (r d) for r much larger than d: Similarly, r d (r d) (r) (r) ( d) r + d r : r + d r d r : Only the rst two terms of each exansion were retained. Thus V r + d r + r r + d r r + d r r + d r :

14 694 CHAPTER 5 ELECTRIC POTENTIAL 35E (a) From E (b) V 0 due to suerosition. V L + (L 4 + d ) ln d : 36E Z V P d Zrod R R rod d Q R : 37E (a) All the charge is the same distance R from C, so the electric otential at C is V Q R 6Q R 5Q R ; where the zero was taken to be at innity. (b) All the charge is the same distance from P. That distance is R + z, so the electric otential at P is V Q 6Q 5Q R + z R + z R + z : 38E The disk is uniformly charged. This means that when the full disk is resent each uadrant contributes eually to the electric otential at P, so the otential at P due to a single uadrant is one-fourth the otential due to the entire disk. First nd an exression for the otential at P due to the entire disk. Consider a ring of charge with radius r and width dr. Its area is r dr and it contains charge d r dr. All the charge in it is a distance r + z from P, so the otential it roduces at P is The total otential at P is V 0 Z R 0 dv r dr r + z r dr 0 r + z : r dr r + z r + z R R + z The otential V s at P due to a single uadrant is V s V R + z z : z :

15 CHAPTER 5 ELECTRIC POTENTIAL P Consider an innitesimal segment of the ring, located between r 0 and r + dr 0 from the center of the ring. The area of this segment is da r 0 dr 0, and the charge it carries is d da r 0 dr 0. The searation between any oint on this segment of the ring and oint P is a r 0 + z. Thus the contribution to V P from this segment is given by dv P d( a). Integrate over the entire ring to obtain V P : V P Z ring Z dv P d Zring a R r r 0 r 0 + z dr 0 : Uon making a change of variable u r 0 the integral above can be readily evaluated to yield the result V P [ r 0 + z ] 0 R r 0 ( R + z r + z ) : For z :00R and r 0:00R this gives V P 0:3R 0. 40P (a) Denote the surface charge density of the disk as for 0 < r < R, and as for R < r < R. Thus the total charge on the disk is given by Z (b) Use E. 5-36: disk d Z R 0 rdr + Z R R rdr 4 R ( + 3 ) 4 (:0 0 m) [: C/m + 3(8: C/m )] : C : V (z) Z disk r 0 dv Z R z + R 4 0 z Z (R 0 )dr 0 R + z + R z + R r R (R 0 )dr 0 z + R 0 z + R 4 Plug in thenumerical values of,, R and z to obtain V (z) 7:95 0 V. : 4P (a) Consider an innitesimal segment of the rod, located between x and x + dx. It has length dx and contains charge d dx, where QL is the linear charge density of the rod. Its distance from P is d + x and the otential it creates at P is dv d d + x dx d + x :

16 696 CHAPTER 5 ELECTRIC POTENTIAL To nd the total otential at P, integrate over the rod: V Z L 0 dx d + x ln(d + x) L 0 Q L ln + L d : 4P (a) Similar to the last roblem, let us consider an innitesimal segment of the rod, located between x and x + dx. It has length dx and contains charge d dx cxdx. Its distance from P is d + x and the otential it creates at P is dv d d + x cx dx d + x : To nd the total otential at P, integrate over the rod: V Z c L 0 x dx d + x c [x d ln(x + d)] L 0 c L d ln + L : d 43E The magnitede of the electric eld is given by jej V s (5:0 V) 0:05 m 6:7 0 V/m : E oints from the ositively charged late to the negatively charged one. 44E Use E x dvdx, where dvdx is the local sloe of the V vs. x curve deicted in Fig The results are: E x (ab) 6:0 V/m, E x (bc) 0, E x (cd) E x (de) 3:0 V/m, E x (ef) 5 V/m, E x (fg) 0, E x (gh) 3:0 V/m. E x (V/m) 5 e f 6 c d e b f g -5 c 0 5 x (m) a b -6 g h

17 CHAPTER 5 ELECTRIC POTENTIAL E On the diole axis 0 or so j cos j. The magnitude of the electric eld is thus given d dr r 0 r 3 : 46E Use E. 5-4: E x (x; y) E y [(:0 V/m )x (3:0 V/m )y ] (:0 V/m [(:0 V/m )x (3:0 V/m )y ] (3:0 V/m )y : Now lug in x 3:0 m and y :0 m to obtain the magnitude of E: E 7 V/m. E makes an angle with the ositive x axis, where tan Ey E x 35 : E x + E y 47E E dv dx i d dx (500x ) i ( 3000x) i ( 3000 V/m )(0:030 m) i ( 39 V/m) i : 48E (a) The E-eld corresonding to V is E dv dr Ze r d Ze dr r r : R 3 3 R + r R 3 (b) The exression for V (r) is valid inside the atom only and cannot be extended to r! :

18 698 CHAPTER 5 ELECTRIC POTENTIAL 49P (a) The charge on every art of the ring is the same distance from any oint P on the axis. This distance is r z + R, where R is the radius of the ring and z is the distance from the center of the ring to P. The electric otential at P is V Z d r Z Z d z + R z + R (b) The electric eld is along the axis and its comonent is given by d z + R d dz (z + R ) (z + R ) 3 (z) z (z + R ) 3 : This agrees with the result of Section P From 4P the electric otential for a oint on the x axis whose x coordinate is given by x d < 0 is V (x d) Q L ln + L : d Relace x d with a general value x to obtain V (x) Q L ln L x : Thus E x (x d) dv dx x d d Q dx L ln L x x d Q d(l + d) : Note that here E x < 0, indicating that E is in the negative x direction, as exected. (b) From symmetry it is obvious that the electric eld does not have any y comonent, i.e., E y 0. 5P (a) Consider an innitesimal segment of the rod from x to x + dx. Its contribution to the otential at oint P is dv P (x)dx x + y cx x + y dx:

19 CHAPTER 5 ELECTRIC POTENTIAL 699 Thus (b) V P E P;y Z rod dv c Z c L 0 x x + y dx d dy ( L + y y) c c ( L + y y) : y : L + y (c) All we obtained above for the otential is its value at any oint P on the y-axis. In order to obtain E x (x; y) we need to rst calculate V (x; y); i.e., the otential for an arbitrary oint located at (x; y). Then E x (x; y) can be obtained from E x (x; (x; y)@x. 5E (a) Use E with e and r :00 nm: U r e r (8:99 09 Nm C )(: C) : m (b) Since U > 0 and U / r the otential energy U decreases as r increases. :5 0 9 J : 53E (a) The charges are eual and are the same distance from C. Use the Pythagorean theorem to nd the distance r (d) + (d) d. The electric otential at C is the sum of the otential due to the individual charges but since they roduce the same otential, it is twice that of either one: V d d (8:99 09 Nm /C )() (:0 0 6 C) :5 0 6 V : 0:00 m (b) As you move the charge into osition from far away the otential energy changes from zero to V, where V is the electric otential at the nal location of the charge. The change in the otential energy euals the work you must do to bring the charge in: W V (:0 0 6 C)(:5 0 6 V) 5: J. (c) The work calculated in art (b) reresents the otential energy of the interactions between the charge brought in from innity and the other two charges. To nd the total otential energy of the three-charge system you must add the otential energy of the interaction between the xed charges. Their searation is d so this otential energy is d. The total otential energy is U W + d 5: J + (8:99 09 Nm /C )(:0 0 6 C) 0:00 m 6:9 J :

20 700 CHAPTER 5 ELECTRIC POTENTIAL 54E The otential energy of the two-charge system is U (x x ) + (y y ) (8:99 09 Nm C )(3:0 0 6 C)( 4:0 0 6 C) (3:5 + :0) + (0:50 :5) cm :9 J : Thus :9 J of work is needed. 55E (a) Denote the side length of the triangle as a. Then the electric otential energy is U 3(e3) a (8:99 09 Nm C )(: C) 3(:8 0 3 ) :7 0 4 J : (b) Uc :7 0 4 J(3: m/s) 3:0 0 3 kg: This is about a third of the acceted value of the electron mass. 56E Choose the zero of electric otential to be at innity. The initial electric otential energy U i of the system before the articles are brought together is therefore zero. After the system is set u the nal otential energy is U f a a a + a a 0: 0 a : a + a Thus the amount of work reuired to set u the system is given by W U U f U i 0: ( 0 a). 57E Let the uark-uark searation be a. Then (a) U u-u (e3)(e3) a 4e a 4(8:99 09 Nm C )(: C)e 9(:3 0 5 m) 4: ev 0:484 MeV:

21 CHAPTER 5 ELECTRIC POTENTIAL 70 (b) U (e3)(e3) a (e3)(e3) a 0 : 58E U X i6j i j r ij d (8:99 09 Nm C ) :3 m ()( 4) + ()(3) + ( 4)(7) + (3)(7) + ()(7) + ( 4)(3 (0 9 C) : 0 6 J : 59P Let 0: C and a :7 m. The change in electric otential energy of the three-charge system as one of the charges is moved as described in the roblem is U a a a (0: C) (8: Nm C ) :7 m Thus the number of days reuired would be :5 0 8 J: n (:5 0 8 J) (0: W)(86400 s/d) : d : 60P (a) Let ` ( 0:5 m) be the length of the rectangle and w ( 0:050 m) be its width. Charge is a distance ` from oint A and charge is a distance w, so the electric otential at A is V A ` + w 5:0 0 6 (8: Nm /C C ) 0:5 m 6:0 0 4 V : + :0 0 6 C 0:050 m

22 70 CHAPTER 5 ELECTRIC POTENTIAL (b) Charge is a distance w from oint b and charge is a distance `, so the electric otential at B is V B w + ` 5:0 0 6 (8: Nm /C C ) 0:050 m 7:8 0 5 V : + :0 0 6 C 0:5 m (c) Since the kinetic energy is zero at the beginning and end of the tri, the work done by an external agent euals the change in the otential energy of the system. The otential energy is the roduct of the charge 3 and the electric otential. If U A is the otential energy when 3 is at A and U B is the otential energy when 3 is at B, then the work done in moving the charge from B to A is W U A U B 3 (V A V B ) (3:0 0 6 C)(6:0 0 4 V + 7:8 0 5 V) :5 J. (d) The work done by the external agent is ositive, 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 ath. 6P The work reuired is W U (4)(5) + d (5)( ) d 0 : 6P The article with charge has both otential and kinetic energy and both these change when the radius of the orbit is changed. Find an exression for the total energy in terms of the orbit radius. Q rovides the centrietal force reuired for to move in uniform circular motion. The magnitude of the force is F Q r, where r is the orbit radius. The acceleration of is v r, where v is its seed. Newton's second law yields Q r mv r, so mv Q r and the kinetic energy is K mv Q8 0 r. The otential energy is U Q r and the total energy is E K + U Q 8 0 r Q r Q 8 0 r : When the orbit radius is r the energy is E Q8 0 r and when it is r the energy is E Q8 0 r. The dierence E E is the work W done by an external agent to change the radius: W E E Q 8 0 r r Q 8 0 r r :

23 CHAPTER 5 ELECTRIC POTENTIAL P (a) V (r) e r (8:99 09 Nm C )(: C) 5:9 0 m (b) U ev (r) 7: ev: (c) Since m e v r e r ; K mv e r (d) The energy reuired is 7: ev V (r) 7: V: 3:6 ev: E 0 [V (r) + K] 0 ( 7: ev + 3:6 ev) 3:6 ev: 64P Use the conservation of energy rincile. The initial otential energy is U i r, the initial kinetic energy is K i 0, the nal otential energy is U f r, and the nal kinetic energy is K f mv, where v is the nal seed of the article. Conservation of energy yields The solution for v is v s r r r r + mv : m s (8: Nm /C )()(3: 0 6 C) :5 0 3 m/s : kg 0: m :5 0 3 m 65P Let r :5 m, x 3:0 m, 9:0 nc, and 6:0 C. The work done is given by W U r r + x ( 9:0 0 9 C)( 6:0 0 C)(8: Nm C ) " :5 m :8 0 0 J : # (:5 m) + (3:0 m)

24 704 CHAPTER 5 ELECTRIC POTENTIAL 66P (a) The otential energy is U d (8:99 09 Nm /C )(5:0 0 6 C) :00 m relative to the otential energy at innite searation. (b) Each shere reels the other with a force that has magnitude F 0:5 J ; d (8:99 09 Nm /C )(5:0 0 6 C) 0:5 N : (:00 m) According to Newton's second law the acceleration of each shere is the force divided by the mass of the shere. Let m A and m B be the masses of the sheres. The acceleration of shere A is a A F 0:5 N 45:0 m A 5:0 0 3 m/s kg and the acceleration of shere B is a B F m B 0:5 N) kg :5 m/s : (c) Energy is conserved. The initial otential energy is U 0:5 J, as calculated in art (a). The initial kinetic energy is zero since the sheres start from rest. The nal otential energy is zero since the sheres are then far aart. The nal kinetic energy is m Av + A m Bv, where B v A and v B are the nal velocities. Thus Momentum is also conserved, so U m Av A + m Bv B : 0 m A v A + m B v B : Solve these euations simultaneously for v A and v B. Substitute v B (m A m B )v A, from the momentum euation, into the energy euation and collect terms. You should obtain U (m Am B )(m A + m B )v A. Thus v A s s Um B m A (m A + m B ) (0:5 J)(0 0 3 kg) (5:0 0 3 kg)(5:0 0 3 kg kg) 7:75 m/s : Now calculate v B : v B m A m B v A 5:0 0 3 kg (7:75 m/s) 3:87 m/s : kg

25 CHAPTER 5 ELECTRIC POTENTIAL P The initial seed v i of the electron satises K i m ev i v i r ev m e s (: J)(65 V) 9: 0 3 kg ev, which gives : m/s : 68P (a) At the smallest center-to-center searation r min the initial kinetic energy K i of the roton is entirely converted to the electric otential energy between the roton and the neucleus. Thus K i e lead 8e : r min r min Thus r min 8e K i :5 0 4 m 5 fm : Note that ev e V. (b) In this case 8e 8(:6 0 9 C)(8: Nm C ) (4: e) V 4: V K i lead r 0 min 8e r 0 8e ; min r min so the new minimum searation is r 0 min r min 50 fm: 69P The idea for solving this roblem is the same as that for the last one. In this case which gives r min Q K. K Q ; r min 70P The change in electric otential energy of the electron-shell system as the electron starts from its initial osition and just reaches the shell is U ( e)( V ) ev: Thus from U K m ev i we nd the initial electron seed to be r r U ev v i : m e m e

26 706 CHAPTER 5 ELECTRIC POTENTIAL 7P Use the conservation of energy rincile. Take the otential energy to be zero when the moving electron is far away from the xed electrons. The nal otential energy is then U f e d, where d is the half the distance between the xed electrons. The initial kinetic energy is K i m ev, where m e is the mass of an electron and v is the initial seed of the moving electron. The nal kinetic energy is zero. Thus K i U f or m ev e d. Hence v s 4e dm e s (8: Nm /C )(4)(: C) (0:00 m)(9: 0 3 kg) 3: 0 ms : 7P The otential dierence between the surface of the shere of charge Q and radius r and a oint innitely far from it is V Q r. Thus the escae seed v esc of the electron of mass m e satises K m ev esc ev, or v esc r s eq m e r : 0 4 m/s : (: C)(: C)(8: Nm C ) (9: 0 3 kg)(:0 0 m) 73P Let the distance in uestion be r. The initial kinetic energy of the electron is K i m ev i, where v i 3: 0 5 m/s. As the seed doubles, K becomes 4K i. Thus or U e r K (4K i K i ) 3K i 3 m ev i ; e r (:6 0 9 C) (8: Nm C ) 3( )m e v i 3(9: 0 9 kg)(3: 0 5 m/s) :6 0 9 m : 74E Since the electric otential throughout the entire conductor is a constant, the electric otential at its center is also +400 V. 75E (a) and (b) For r < R 0 cm the electric eld is zero since this region is inside the conductor. For r > R we have E(r) ( r ), where +3:0 C. In articular, just outside the shere E ( R ) E 0 6:8 0 5 N/C.

27 CHAPTER 5 ELECTRIC POTENTIAL 707 Use V (r) R 0 E(r)dr to calculate V (r). For r < R 0 cm the electric otential is a constant since this region is inside the conductor, an euiotential body. For r R we have V (r) ( r). In articular, at r R V ( R) V 0 :4 0 5 V. This is also the value of V (r) for r < R. Both E and V as functions of r are lotted below. E V E 0 V 0 /r /r 0 cm r 0 cm r V0.4x05 V. 76E If the electric otential is zero at innity, then the otential at the surface of the shere is given by V r, where is the charge on the shere and r is its radius. Thus rv (0:5 m)(500 V) 8: Nm /C :5 0 8 C : 77E (a) Since the two conductors are connected V and V must be the same. (b) Let V R V R and note that + and R R. Solve for and : 3, 3. (c) 4R 4R R R : 78P In sketching the electric eld lines and the euiotential lines ay attention to the following: () The electric eld lines and the euiotential lines and always erendicular to each other wherever they intersect. () The led lines are erenduluar to the surface of the conductor (which is an euiotential surface); while the euiotential lines just outside

28 708 CHAPTER 5 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 otential is a constant. axis some euiotential lines and electric field lines (denoted with arrows) 79P (a) The otential would be V e Q e 4R e e R e R e 4(6: m)(:0 electron m )( :6 0 9 C electron)(8: Nm C ) (b) 0: V : E e 0 V e R e 0: V 6: m :8 0 8 N/C ; where the minus sign indicates that E is radially inward. 80P (a) The electric otential is the sum of the contributions of the individual sheres. Let be the charge on one and be the charge on the other. The oint halfway between them

29 CHAPTER 5 ELECTRIC POTENTIAL 709 is the same distance d ( :0 m) from the center of each shere, so the otential at the halfway oint is V + d (8:99 09 Nm /C )(:0 0 8 C 3:0 0 8 C) :0 m :80 0 V : (b) The distance from the center of one shere to the surface of the other is d R, where R is the radius of either shere. The otential of either one of the sheres is due to the charge on that shere and the charge on the other shere. The otential at the surface of shere is V R + d R :0 0 8 (8: Nm /C C ) 0:030 m :7 0 3 V : 3:0 0 8 C :0 m 0:030 m The otential at the surface of shere is V d R + R :0 0 8 (8: Nm /C C ) :0 m 0:030 m 8:9 0 3 V : 3:0 0 8 C 0:030 m 8P (a) E 0 R (3:0 0 8 C)(8: Nm C ) : 0 4 N/C : (0:5 m) (b) V RE (0:5 m)(: 0 4 N/C) :8 0 3 V: (c) Let the distance be x. Then V V (x) V R + x R 500 V ; which gives x RV V V (0:5 m)( 500 V) 800 V V 5:8 0 m :

30 70 CHAPTER 5 ELECTRIC POTENTIAL 8P Since the charge distribution is sherically symmetric we may write E(r) encl ; r where R encl is the charge enclosed in a shere of radius r centered at the origin. Also V (r) E(r) dr. The results are as follows: r For r > R > R 8 > < >: for R > r > R 8 > < >: V (r) + r ; E(r) + r ; V (r) r + R ; E(r) r ; and for R > R > r 8 < E 0; : V (r) + : R R E(r) (04 N/C) V(r) (0 4 V) r(m) r(m) 83 (c) 4:4 V 84 (a) for N < conguration is less energetic, for N conguration is less energetic; (b) ; (c)