Unformatted text preview: sides of this result gives 1
4
=
2
2
2
L
L + ( 2.00 m ) L2 + ( 2.00 m ) = 4 L2
2 or Solving for L, we obtain
3L2 = ( 2.00 m ) 2 or L= 2.00 m
= 1.15 m
3 2 1022 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 21. REASONING The potential V created by a point charge q at a spot that is located at a
kq
distance r is given by Equation 19.6 as V = , where q can be either a positive or negative
r
quantity, depending on the nature of the charge. We will apply this expression to obtain the
potential created at the empty corner by each of the three charges fixed to the square. The
total potential at the empty corner is the sum of these three contributions. Setting this sum
equal to zero will allow us to obtain the unknown charge.
SOLUTION The drawing at the right shows the three
charges fixed to the corners of the square. The length of each
side of the square is denoted by L. Note that the distance
between the unknown charge q and the empty corner is L.
Note also that the distance between one of the 1.8µC charges
and the empty corner is r = L , but that the distance between
the other 1.8µC charge and the empty corner is +1.8 µC L L L +1.8 µC q
L
r = L + L = 2 L , according to the Pythagorean theorem.
Using Equation 19.6 to express the potential created by the unknown charge q and by each
of the 1.8µC charges, we find that the total potential at the empty corner is
2 2 Vtotal ( )( ) −6
−6
kq k +1.8 ×10 C k +1.8 ×10 C
=
+
+
=0
L
L
2L In this result the constant k and the length L can be eliminated algebraically, leading to the
following result for q:
q + 1.8 ×10−6 C + 1.8 ×10−6 C
=0
2 or ( ) 1 −6
q = −1.8 ×10−6 C 1 + = −3.1×10 C
2 22. REASONING The radius R of the circle is the distance between each charge and the center
of the circle, where we know the total electric potential V. The total electric potential is the
kq
sum of the contributions V1, V2, and V3 made by the three charges. We will use V =
r
9
22
(Equation 19.6), where k = 8.99×10 N·m /C and r = R to determine each of the three
contributions to the total electric potential. Summing the contributions and equating the
result to V = −2100 V, we will obtain the radius R of the circle.
kq
(Equation 19.6), the total potential V at the center of the circle is
r
k ( q1 + q2 + q3 )
kq
kq kq
(1)
V = V1 + V2 + V3 = 1 + 2 + 3 =
R
R
R
R SOLUTION From V = Solving Equation (1) for R, we find that Chapter 19 Problems R= 1023 k ( q1 + q2 + q3 )
V (8.99 ×109 N ⋅ m 2 /C2 )( −5.8 ×10−9 C − 9.0 ×10−9 C+7.3 ×10−9 C ) = 0.032 m
=
−2100 V 23. SSM REASONING The only force acting on the moving charge is the conservative
electric force. Therefore, the sum of the kinetic energy KE and the electric potential energy
EPE is the same at points A and B:
1
2 2
2
mvA + EPE A = 1 mvB + EPE B
2 Since the particle comes to rest at B, v B = 0 . Combining Equations 19.3 and 19.6, we have kq EPE A = qVA = q 1 d
and kq EPE B = qVB = q 1 r
where d is the initial distance between the fixed charge and the moving charged particle, and
r is the distance between the charged particles after the moving charge has stopped.
Therefore, the expression for the conservation of energy becomes
1
2 2
mvA + kqq1
d = kqq1
r This expression can be solved for r. Once r is known, the distance that the charged particle
moves can be determined.
SOLUTION Solving the expression above for r gives 1024 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL r= = kqq1
1 mv 2 + kqq1
A
2
d
(8.99 ×109 N ⋅ m 2 / C2 )(–8.00 ×10− 6 C)(−3.00 ×10− 6 C)
9
2
2
−6
−6
1 (7.20 × 10 –3 kg)(65.0 m/s) 2 + (8.99 × 10 N ⋅ m / C )( – 8.00 × 10 C)( −3.00 × 10 C)
2
0.0450 m = 0.0108 m
Therefore, the charge moves a distance of 0 .0450 m – 0.0108 m = 0.0342 m . 24. REASONING The speed vB of the test charge
at position B, the center of the square, is related q 2
to its kinetic energy by KE B = 1 mvB
2
(Equation 6.2). No forces other than the
conservative electric force act on the test
charge, so the total mechanical energy of the
test charge is conserved as it accelerates from
its initial position A at one corner of the square
to its final position B. There is no rotational
motion and no elastic potential energy, and the
test charge is at rest at position A, so the
conservation of energy takes the following form:
1 mv 2 + EPE
B
B
2
{{
Electric
Kinetic
potential
energy
energy at B
at B = 1 m ( 0 m/s ) + EPE A
24
14 244
3{
2 Kinetic
energy at A q0 rA A rB q0
vB B rA = 0.480 m rB
q= +7.20 µC or 1
2 2
mvB + EPE B = EPE A (1) Electric
potential
energy at A In Equation (1), m is the mass of the test charge. At both position A and position B, the
electric potential energy EPE of the test charge is determined by EPE = q0V
(Equation 19.3), where V is the total electric potential due to both of the charges q. We will
kq
use V =
(Equation 19.6), where k = 8.99×109 N·m2/C2 to obtain the total electric
r
potential at positions A and B. SOLUTION At positions A and B, the total electric potential is the sum of the electric
potentials due to the charges q. Since the charges are identical, and both are equidistant from
positions A and B, the total potentials VA and VB are twice the potentials due to either charge
kq
alone. Therefore, from V =
(Equation 19.6), we h...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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