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Halliday/Resnick/Walker 7e
Chapter 24 – Electric Potential
- (a) An Ampere is a Coulomb per second, so
84 A h 84 3600 3 0 105
C h s
s h
⋅ = C
F ⋅
H
G
I
K
J
F
H
G
I
K
J =^.^ ×.
(b) The change in potential energy is ∆ U = q ∆ V = (3.0 × 105 C)(12 V) = 3.6 × 106 J.
- The magnitude is ∆ U = e ∆ V = 1.2 × 109 eV = 1.2 GeV.
3. The electric field produced by an infinite sheet of charge has magnitude E = σ/2ε 0 , where σ is
the surface charge density. The field 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 field and positive in the direction of the field. Then the electric potential is
V V (^) s E dx V Ex
x
= − z 0 = s − ,
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 differ in magnitude by ∆ V = E ∆ x = ( σ/2ε 0 )∆ x. Thus,
x
V
× ⋅
×
= ×
− −
2 2 8 85^10 50 −
12 2 6
C N m V C m
m
2 2
c hb g
- (a) VB – VA = ∆ U / q = – W /(– 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).
5. (a) E = F e = c 3 9. × 10 − 15 N h c 160. × 10 −^19 Ch = 2 4. × 104 N C.
(b) ∆ V = E ∆ s = c2 4. × 10 4 N C hb 012. m g = 2 9. × 103 V.
- (a) The work done by the electric field is (in SI units)
19 12 0 0 21 (^0 ) 0 0
1.87 10 J.
f d i
q q d W q E ds dz
− − − −
× ×
= ⋅ = = = = ×
∫ ∫ ×
G G
(b) Since V – V 0 = – W / q 0 = – σ z /2ε 0 , with V 0 set to be zero on the sheet, the electric potential at P
is (in SI units)
12 2 12 0
1.17 10 V.
z V
− − −
×
= − = − = − ×
×
- The charge is
9 (^0 9 2 ) m /C
(10m) ( 1.0V) 4 1 8.99 10 N
q πε RV −
⋅
.1 10 C.
= = = − ×
×
- (a) The potential difference is
( )
2 6 9 2 0 0 3
N m 1 1 1.0 10 C 8.99 10 4 4 C 2.0 m 1.0 m 4.5 10 V.
A B A B
q q V V
ε r ε r
− = − = × −^ ⎛^ × ⋅^ ⎞ ⎛^ − ⎞
⎝ ⎠ ⎝^ ⎠
= − ×
p p
(b) Since V ( r ) depends only on the magnitude of
G
r , the result is unchanged.
- First, we observe that V ( x ) cannot be equal to zero for x > d. In fact V ( x ) is always negative for x > d. Now we consider the two remaining regions on the x axis: x < 0 and 0 < x < d.
(a) For 0 < x < d we have d 1 = x and d 2 = d – x. Let
V x k
q d
q d
q x d x
F HG^
I KJ^
F H G
I K
(^1) J = 1
2 2 4 0
p ε
and solve: x = d /4. With d = 24.0 cm, we have x = 6.00 cm.
(b) Similarly, for x < 0 the separation between q 1 and a point on the x axis whose coordinate is x is given by d 1 = – x ; while the corresponding separation for q 2 is d 2 = d – x. We set
V x k q d
q d
q x d x
F H G
I K J =^ −
F H G
I K
(^1) J = 1
2 2 4 0
p ε
to obtain x = –d /2. With d = 24.0 cm, we have x = – 12.0 cm.
- A charge –5 q is a distance 2 d from P , a charge –5 q is a distance d from P , and two charges +5 q are each a distance d from P , so the electric potential at P is (in SI units)
9 15 4 2 0 0
5.62 10 V.
q q V
πε d d d d πε d
− − −
⎡ ⎤ ×^ ×
= ⎢ − − + + ⎥ = = = ×
⎣ ⎦ ×
leftward, from the problem description (indicating deceleration of the rightward moving particle),
so that q > 0 (ensuring that F
→ is parallel to E
→ ); it is a proton.
(b) We use conservation of energy:
K 0 + U 0 = K + U ⇒
2 mpv
2 0 +^ qV 1 =
2 mpv
(^2) + qV
Using q = +1.6 × 10 −^19 C,^ m p = 1.67^ ×^10 −^27 kg,^ v 0 = 90^ ×^103 m/s,^ V 1 =^ −70 V and^ V 2^ =^ −50 V^ ,
we obtain the final speed v = 6.53 × 104 m/s. We note that the value of d is not used in the solution.
- (a) Using U = qV we can “translate” the graph of voltage into a potential energy graph (in eV units). From the information in the problem, we can calculate its kinetic energy (which is its total energy at x = 0) in those units: Ki = 284 eV. This is less than the “height” of the potential energy “barrier” (500 eV high once we’ve translated the graph as indicated above). Thus, it must reach a turning point and then reverse its motion.
(b) Its final velocity, then, is in the negative x direction with a magnitude equal to that of its initial velocity. That is, its speed (upon leaving this region) is 1.0 × 107 m/s.
- If the electric potential is zero at infinity, then the potential at the surface of the sphere is given by V = q /4πε 0 r , where q is the charge on the sphere and r is its radius. Thus
q = rV = × ⋅
4 = × −
p ε 0 9 2 5 108
m 1500 V N m C
2 2 C.
b gb g
- (a) The electric potential is the sum of the contributions of the individual spheres. Let q 1 be the charge on one, q 2 be the charge on the other, and d be their separation. The point halfway between them is the same distance d /2 (= 1.0 m) from the center of each sphere, so the potential at the halfway point is
( )( ) 9 2 2 8 8 1 2 8.99^10 N m^ C^ 1.0^10 C^ 3.0^10 C^ 1.8 10 V.^2 4 2 1.0 m
q q V
ε d
× ⋅ × −^ − × −
= = = − ×
p 0
(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
( )
8 8 1 2 9 2 2 3 1 0
1 1.0 10 C 3.0 10 C
8.99 10 N m C 2.9 10 V. 4 0.030 m 2.0 m 0.030 m
q q V
πε R d R
⎡ ⎤ ⎡^ ×^ −^ × − ⎤
= ⎢ + ⎥= × ⋅ ⎢ − ⎥= ×
⎣ −^ ⎦ ⎣ − ⎦
(c) The potential at the surface of sphere 2 is
( )
8 8 1 2 9 2 2 3 2 0
1 1.0 10 C 3.0 10 C
8.99 10 N m C 8.9 10 V. 4 2.0 m 0.030 m 0.030 m
q q V
πε d R R
⎡ × −^ × − ⎤
= ⎡^ + ⎤= × ⋅ − = − ×
⎢ − ⎥ ⎢^ − ⎥
- The escape speed may be calculated from the requirement that the initial kinetic energy (of launch ) be equal to the absolute value of the initial potential energy (compare with the gravitational case in chapter 14). Thus,
1 2 m^ v
(^2) = e q 4 πεo r
where m = 9.11 × 10 −^31 kg, e = 1.60 × 10 −^19 C, q = 10000 e , and r = 0.010 m. This yields the answer v = 22490 m/s ≈ 2.2 × 104 m/s.
- We use Ex = –dV / dx , where dV / dx is the local slope of the V vs. x curve depicted in Fig. 24-
- The results are:
(a) Ex ( ab ) = –6.0 V/m,
(b) Ex ( bc ) = 0,
(c) Ex ( cd ) 3.0 V/m,
(d) Ex ( de ) = 3.0 V/m,
(e) Ex ( ef ) = 15 V/m,
(f) Ex ( fg ) = 0,
