Unformatted text preview: ential at B, indicating that the charge q of the particle is positive. The work WAB
done on the particle by the electric force is proportional to q, according to
WAB = −q (VB − VA ) (Equation 19.4), where VB − V A is the electric potential difference
between positions B and A. According to the workenergy theorem, the work done on the
particle by the electric force is also equal to the difference between the final and initial
kinetic energies of the particle: WAB = KE B − KE A (Equation 6.3). The kinetic energies are
given in electronvolts (eV), so we will use the equivalence 1 eV = 1.60×10−19 J to convert
from electronvolts to joules.
SOLUTION Solving WAB = −q (VB − VA ) (Equation 19.4) for q, we obtain
q=− W AB
VB − V A (1) Substituting WAB = KE B − KE A (Equation 6.3) into Equation (1) yields
q=− KE B − KE A
VB − V A Since 1 eV = 1.60×10−19 J, Equation (2) gives the particle’s charge as (2) Chapter 19 Problems q=− 1015 KE B − KE A
( 7060 eV − 9520 eV ) 1.60 × 10−19 J = +4.80 × 10−18 C
=− VB − V A
1 eV +27.0 V − ( −55.0 V ) 11. SSM WWW REASONING The only force acting on the moving charge is the
conservative electric force. Therefore, the total energy of the charge remains constant.
Applying the principle of conservation of energy between locations A and B, we obtain
1
2 2
2
mvA + EPE A = 1 mvB + EPE B
2 Since the charged particle starts from rest, vA = 0 . The difference in potential energies is
related to the difference in potentials by Equation 19.4, EPE B − EPE A = q (VB − VA ) . Thus,
we have
2
(1)
q (VA − VB ) = 1 mvB
2
Similarly, applying the conservation of energy between locations C and B gives
q (VC − VB ) = 1 m (2vB ) 2
2 (2) Dividing Equation (1) by Equation (2) yields VA – VB 1
=
VC – VB 4
This expression can be solved for VB .
SOLUTION Solving for VB , we find that
VB = 4VA – VC
3 = 4(452 V)–791 V
= 339 V
3 12. REASONING The gravitational and electric forces are conservative forces, so the total
energy of the particle remains constant as it moves from point A to point B. Recall from
Equation 6.5 that the gravitational potential energy (GPE) of a particle of mass m is
GPE = mgh, where h is the height of the particle above the earth’s surface. The conservation
of energy is written as
1
2 mv 2 + mghA + EPE A =
A 1
2 2
mvB + mghB + EPE B (1) 1016 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL We will use this equation several times to determine the initial speed vA of the negatively
charged particle.
SOLUTION
When the negatively charged particle is thrown upward, it attains a maximum height of h.
For this particle we have:
vA = ? EPEA = (−q) VA
hA = 0 (ground level) vB = 0 (at maximum height) EPEB = (−q)VB
hB = h (the maximum height) Solving the conservation of energy equation, Equation (1), for vA and substituting in the data
above gives
2 mgh + ( −q ) (VB − VA ) (2)
vA = m Equation (2) cannot be solved as it stands because the height h and the potential difference
(VB − VA) are not known. We now make use of the fact that a positively charged particle,
when thrown straight upward with an initial speed of 30.0 m/s, also reaches the maximum
height h. For this particle we have:
vA = 30.0 m/s vB = 0 (at maximum height) EPEA = (+q) VA
hA = 0 (ground level) EPEB= (+q)VB
hB = h Solving the conservation of energy equation, Equation (1), for the potential difference
(VB − VA) and substituting in the data above gives
VB − V A = 1 1
2 2 m ( 30.0 m/s ) − mgh +q (3) Substituting Equation (3) into Equation (2) gives, after some algebraic simplifications,
v A = 4 gh − ( 30.0 m/s ) 2 (4) Equation (4) cannot be solved because the height h is still unknown. We now make use of
the fact that the uncharged particle, when thrown straight upward with an initial speed of
25.0 m/s, also reaches the maximum height h. For this particle we have: Chapter 19 Problems vA = 25.0 m/s
EPEA = qVA = 0 (since q = 0)
hA = 0 (ground level) 1017 vB = 0 (at maximum height)
EPEB = qVB = 0 (since q = 0)
hA = h Solving Equation (1) with this data for the maximum height h yields ( 25.0 m/s )2
h=
2g = ( 25.0 m/s )2 ( 2 9.80 m/s 2 ) = 31.9 m Substituting h = 31.9 m into Equation (4) gives vA = 18.7 m / s . 13. REASONING The electric potential at a distance r from a point charge q is given by
V = kq/r (Equation 19.6). The total electric potential due to the two charges is the algebraic
sum of the individual potentials.
SOLUTION Using Equation 19.6, we find that the total electric potential due to the two
charges is
kq kq
k
V = 1 + 2 = ( q1 + q2 )
r
r
r The distance r is onehalf the distance between the charges, so r =
charges. Thus, the total electric potential midway between the charges is V= ( 1
2 (1.20 m ) for both ) k
8.99 ×109 N ⋅ m 2 / C2
q1 + q2 ) =
+3.40 ×10−6 C − 6.10 ×10−6 C = −4.05 ×104 V
(
1 1.20 m
r
)
2( 14. REASONING
The electric potential difference between the two points is
kq kq
VB − VA = −
(Equation 19.5). We can use this expression directly to calculate the
rB rA
electric potential difference. SOLUTION According to Equation 19.5, the electric potential difference is VB − VA = 1 1
kq kq
−
= kq − r rB rA B rA ( )( ) 1
1
= 8.99 ×109 N ⋅ m 2 / C2 −2.1×10−9 C − = 38 V 0....
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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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