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Unformatted text preview: UTION Solving V = kq
(Equation 19.6) for q gives
r
Vr
q=
k Since the surface area of a sphere is A = 4π r 2 , it follows that r = A
. Substituting this
4π result into the expression for q gives
q= Vr V
=
k
k A
490 V
1.1 m 2
=
= 1.6 × 10−8 C
4π 8.99 × 109 N ⋅ m 2 / C 2
4π 1032 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 33. SSM REASONING The magnitude E of the electric field is given by Equation 19.7a
∆V
(without the minus sign) as E =
, where ∆V is the potential difference between the two
∆s
metal conductors of the spark plug, and ∆s is the distance between the two conductors. We
can use this relation to find ∆V. SOLUTION The potential difference between the conductors is ( )( ) ∆V = E ∆s = 4.7 × 107 V/m 0.75 × 10−3 m = 3.5 × 104 V 34. REASONING Since the electric force does negative work, the electric force must point
opposite to the displacement of the test charge. The point charge is positive, so it exerts an
outwarddirected force on the positive test charge. Since the force and the displacement of
the test charge have opposite directions, the displacement must be directed inward toward
the point charge. Therefore, the radius rB is less than the radius rA.
SOLUTION According to Equation 19.4, we have VB − VA = −WAB
q0 where VB and VA are the potentials of the equipotential surfaces, and WAB is the work done
in moving the charge q0 from surface A to surface B. The potential of a point charge q is
given by V = kq/r (Equation 19.6). With this substitution for VB and VA, we have kq kq −WAB
−
=
rB rA
q0 or kq kq WAB
=
−
rB rA
q0 or 1
1W
= − AB
rB rA kqq0 Solving for rB gives: 1
1W
= − AB
rB rA kqq0
= rB = 1
−8.1 × 10−9 J
−
= 0.83 m −1
1.8 m 8.99 × 109 N ⋅ m 2 / C2 +7.2 × 10−8 C +4.5 × 10−11 C ( 1
= 1.2 m
0.83 m −1 )( )( ) Chapter 19 Problems 1033 35. SSM WWW REASONING AND SOLUTION From Equation 19.7a we know that
∆V
E=−
, where ∆V is the potential difference between the two surfaces of the membrane,
∆s
and ∆s is the distance between them. If A is a point on the positive surface and B is a point
on the negative surface, then ∆V = VA − VB = 0.070 V. The electric field between the
surfaces is E= − V − VA
V − VB
∆V
0.070 V
=− B
=A
=
= 8.8 × 106 V/m
−9
∆s
∆s
∆s
8.0 × 10 m 36. REASONING The net work WAB done by the electric force on the point charge q0 as it
moves from A to B is proportional to the potential difference VB − VA between those
positions, according to VB − V A = −W AB
(Equation 19.4). Positions A and B are on different
q0 equipotential surfaces, so we will read the potentials VB and VA from the drawing. We will
employ the same procedure to solve part (b).
SOLUTION a. Solving VB − V A = −W AB
q0 (Equation 19.4) for WAB, we obtain WAB = −q0 (VB − VA ) (1) From the drawing, we see that VA = +350.0 V and VB = +550.0 V. Therefore, from
Equation (1), ( ) WAB = − +2.8 ×10−7 C ( 550.0 V − 350.0 V ) = −5.6 ×10−5 J b. Positions A and C are both on the +350.0V equipotential surface. Adapting Equation (1),
then, we obtain ( ) WAC = −q0 (VC − VA ) = − +2.8 ×10−7 C ( 350.0 V − 350.0 V ) = 0 J 1034 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 37. REASONING The drawing shows the electric field E
and the three points, A, B, and C, in the vicinity of point
P, which we take as the origin. We choose the upward
3
direction as being positive. Thus, E = −4.0 × 10 V/m,
since the electric field points straight down. The electric
potential at points A and B can be determined from
Equation 19.7a as ∆V = −E ∆s, since E and ∆s are known.
Since the path from P to C is perpendicular to the electric
field, no work is done in moving a charge along such a
path. Thus, the potential difference between these two
points is zero. A
6.0 × 10 −3 m
−3 8.0 × 10 P 3.0 × 10 −3 m C m B
E SOLUTION
a. The potential difference between points P and A is VA − VP = − E ∆s . The potential at A is ( )( ) VA = VP − E ∆s = 155 V − −4.0 ×103 V/m 6.0 × 10−3 m = 179 V
b. The potential difference between points P and B is VB − VP = − E ∆s . The potential at B is ( )( ) VB = VP − E ∆s = 155 V − −4.0 ×103 V/m −3.0 ×10−3 m = 143 V
c. Since the path from P to C is perpendicular to the electric field and no work is done in
moving a charge along such a path, it follows that ∆V = 0 V. Therefore, VC = VP = 155 V . 38. REASONING The electric force F is a conservative force, so the total energy (kinetic
energy plus electric potential energy) remains constant as the electron moves across the
capacitor. Thus, as the electron accelerates and its kinetic energy increases, its electric
potential energy decreases. According to Equation 19.4, the change in the electron’s electric
potential energy is equal to the charge on the electron (−e) times the potential difference
between the plates, or
EPEpositive − EPEnegative = (−e)(Vpositive − Vnegative) (19.4) The electric field E is related to the potential difference between the plates and the
displacement ∆s by E...
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 Spring '13
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 Physics, The Lottery

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