Homework for the week of October 13. 3rd week of classes.
Ch. 23: 3, 7, 15, 22, 26, 35, 51, 54, 61
3.
The kinetic energy gained by the electron is the work done by the electric force. Use Eq. 232b to calculate the potential difference.
5.25 × 10−16 J
W
= 3280 V
Vba = − ba = −
q
( −1.60 × 10−19 C)
The electron moves from low potential to high potential, so plate B is at the higher
potential.
7.
The maximum charge will produce an electric field that causes breakdown in the air. We
use the same approach as in Examples 23-4 and 23-5.
1 Q
→
Vsurface = r0 Ebreakdown and Vsurface =
4πε 0 r0
1


2
Q = 4πε 0 r02 Ebreakdown = 
( 0.065m) 3 × 106 V m = 1.4 × 10−6 C
9
2
2 
 8.99 × 10 N m C 
(
)
After the connection, the two spheres are at the same potential. If they were at
15.
( a)
different
potentials, then there would be a flow of charge in the wire until the potentials were
equalized.
(b) We assume the spheres are so far apart that the charge on one sphere does not influence
the charge on the other sphere. Another way to express this would be to say that the
potential due to either of the spheres is zero at the location of the other sphere. The
charge splits between the spheres so that their potentials (due to the charge on them
only) are equal. The initial charge on sphere 1 is Q, and the final charge on sphere 1 is
Q1.
Q1
Q − Q1
Q1
Q − Q1
r1
=
→ Q1 = Q
V1 =
; V2 =
; V1 = V2 →
4πε 0 r1
4πε 0 r2
4πε 0 r1 4πε 0 r2
( r1 + r2 )
Charge transferred Q − Q1 = Q − Q
r1
( r1 + r2 )
= Q
r2
( r1 + r2 )
22. Because of the spherical symmetry of the problem, the electric field in each region is the
same as that of a point charge equal to the net enclosed charge.
(a) For r > r2 : E =
1 Qencl
4πε 0 r
2
=
1
3
2
Q
4πε 0 r
2
=
3 Q
8πε 0 r 2
For r1 < r < r2 : E = 0 , because the electric field is 0 inside of conducting material.
For 0 < r < r1 : E =
(b)
1 Qencl
4πε 0 r
2
=
1
1
2
Q
4πε 0 r
2
=
1 Q
8πε 0 r 2
For r > r2 , the potential is that of a point charge at the center of the sphere.
V=
1 Q
4πε 0 r
=
1
3
2
Q
=
4πε 0 r
3 Q
8πε 0 r
, r > r2
(c) For r1 < r < r2 , the potential is constant and equal to its value on the outer shell, because
there is no electric field inside the conducting material.
V = V ( r = r2 ) =
3 Q
8πε 0 r2
, r1 < r < r2
(d) For 0 < r < r1 , we use Eq. 23-4a. The field is radial, so we integrate along a radial line
r r
so that E d l = Edr.
r
r
r
r r
1 Q
Q 1 1
Vr − Vr = − ∫ E d l = − ∫ E dr = − ∫
dr =
 − 
2
8πε 0 r
8πε 0  r r1 
r
r
r
1
1
Vr = Vr +
1
1
1
Q 1
Q  1 1
Q  1 1
1
− =
+ =


 +  , 0 < r < r1
8πε 0  r r1  8πε 0  2r1 r  8πε 0  r2 r 
To plot, we first calculate V0 = V ( r = r2 ) =
(e)
3Q
8πε 0 r2
and E0 = E ( r = r2 ) =
Then we plot
V V0 and E E0 as functions of r r2 .
For 0 < r < r1 :
Q  1 1
1 Q
+ 

V 8πε 0  r2 r  1
E 8πε 0 r 2 1 r22 1
−1
−2
=
= 3 1 + ( r r2 )  ;
=
= 3 2 = 3 ( r r2 )
Q
Q
3
3
V0
E0
r
8πε 0r2
8πε 0r22
3 Q
0
V 8πε 0 r2
E
=
=1 ;
=
=0
3Q
3Q
V0
E0
8πε 0 r2
8πε 0 r22
For r1 < r < r2 :
For r > r2 :
3 Q
3 Q
V 8πε 0 r r2
E 8πε 0 r 2 r22
−1
−2
=
= = ( r r2 ) ;
=
= 2 = ( r r2 )
Q
Q
3
3
V0
r
E0
r
8πε 0 r2
8πε 0 r22
5.0
4.0
V /V 0
3.0
2.0
1.0
0.0
0.00
0.25
0.50
0.75
1.00
r /r 2
1.25
1.50
1.75
2.00
3Q
8πε 0 r22
.
5.0
4.0
E /E 0
3.0
2.0
1.0
0.0
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
r /r 2
26.
(a)
Because of the inverse square nature of the
x
d
electric
q2 < 0
field, any location where the field is zero must be
q1 > 0
closer to the weaker charge ( q2 ) . Also, in between the
two charges, the fields due to the two charges are parallel to each other (both to the left)
and cannot cancel. Thus the only places where the field can be zero are closer to the
weaker charge, but not between them. In the diagram, this is the point to the left of q2 .
Take rightward as the positive direction.
q1
1 q2
1
2
−
= 0 → q2 ( d + x ) = q1 x 2 →
E=
2
2
4πε 0 x
4πε 0 ( d + x )
q2
2.0 × 10−6 C
( 5.0cm) = 16cm left of q2
3.4 × 10−6 C − 2.0 × 10−6 C
(b) The potential due to the positive charge is positive
d
x1
x2
everywhere, and the potential due to the negative
charge is negative everywhere. Since the negative
q2 < 0
q1 > 0
charge is smaller in magnitude than the positive charge,
any point where the potential is zero must be closer to the negative charge. So consider
locations between the charges (position x1 ) and to the left of the negative charge
x=
q1 −
d=
q2
(position x2 ) as shown in the diagram.
Vlocation 1 =
Vlocation 2 =
x2
1 
q1

4πε 0  ( d − x1 )
+
q2 
 = 0 → x1 =
x1 
( −2.0 × 10 C) (5.0cm) = 1.852 cm
(q − q )
( −5.4 × 10 C)
−6
q2d
2
=
−6
1
1 
q1
q 
+ 2=0 →

4πε 0  ( d + x2 ) x2 
( −2.0 × 10 C) ( 5.0cm) = 7.143cm
=−
=−
(q + q )
(1.4 × 10 C)
−6
q2 d
−6
1
2
So the two locations where the potential is zero are 1.9 cm from the negative charge
towards the positive charge, and 7.1 cm from the negative charge away from the
positive charge.
35. We follow the development of Example 23-9, with Figure 23-15. The charge on a thin ring
of radius R and thickness dR is dq = σ dA = σ ( 2π RdR ) . Use Eq. 23-6b to find the potential
of a continuous charge distribution.
R
R
R
σ ( 2π RdR ) σ
1
dq
1
R
σ
2
2 1/ 2
V =
dR
x
R
=
=
=
+
R
4πε 0 ∫ r
4πε 0 R∫
2ε 0 R∫ x 2 + R 2
2ε 0
x2 + R2
2
2
1
1
(
)
2
1
=
51.
σ
2ε 0
(
x 2 + R22 −
x 2 + R12
)
We use Eq. 23-9 to find the components of the electric field.
Ex = −
∂V
∂x
= −2.5 y + 3.5 yz ; E y = −
∂V
∂y
= −2 y − 2.5 x + 3.5 xz ; E z = −
∂V
∂z
= 3.5 xy
r
E = ( −2.5 y + 3.5 yz ) ˆi + ( −2 y − 2.5 x + 3.5 xz ) ˆj + ( 3.5 xy ) kˆ
54. Let the side length of the equilateral triangle be L. Imagine bringing the
−e
l
electrons in from infinity one at a time. It takes no work to bring the first
electron to its final location, because there are no other charges present.
l
l
Thus W1 = 0 . The work done in bringing in the second electron to its
final location is equal to the charge on the electron times the potential
−e
(due to the first electron) at the final location of the second electron.
 1 e
1 e2
=
Thus W2 = ( −e )  −
. The work done in bringing the third electron to its

 4πε 0 l  4πε 0 L
final
location is equal to the charge on the electron times the potential (due to the first two
 1 e
1 e
1 2e 2
−
=
electrons). Thus W3 = ( −e )  −
. The total work done is the

 4πε 0 l 4πε 0 l  4πε 0 l
sum W1 + W2 + W3 .
W = W1 + W2 + W3 = 0 +
1 e2
4πε 0 l
+
1 2e 2
4πε 0 l
=
1 3e2
=
4πε 0 l
(
)(
3 8.99 × 109 N m2 C2 1.60 × 10−19 C
(1.0 × 10
−10
m
)
)
2
1eV

 = 43eV
−19 
 1.60 × 10 J 
= 6.9 × 10−18 J = 6.9 × 10−18 J 
61. (a) The electron was accelerated through a potential difference of 1.33 kV (moving from low
potential to high potential) in gaining 1.33 keV of kinetic energy. The proton is
accelerated through the opposite potential difference as the electron, and has the exact
opposite charge. Thus the proton gains the same kinetic energy, 1.33 keV .
(b) Both the proton and the electron have the same KE. Use that to find the ratio of the
speeds.
1
2
mp vp2 = 12 me ve2 →
ve
vp
=
mp
me
=
1.67 × 10 −27 kg
9.11 × 10−31 kg
= 42.8
The lighter electron is moving about 43 times faster than the heavier proton.
−e