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Unformatted text preview: 50 m 0.25 m 1018 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 15. SSM REASONING The potential of each charge q at a distance r away is given by
Equation 19.6 as V = kq/r. By applying this expression to each charge, we will be able to
find the desired ratio, because the distances are given for each charge.
SOLUTION According to Equation 19.6, the potentials of each charge are VA = kqA
rA and VB = kqB
rB Since we know that VA = VB, it follows that
kqA kqB
=
rA
rB or qB rB 0.43 m
=
=
= 2.4
qA rA 0.18 m 16. REASONING The electric potential energy EPE of the third charge q3 is given by
EPE = q3VTotal (see Equation 19.3), where VTotal is the total potential at the point where the
third charge is placed. The total potential at either of the empty corners is the sum of the
individual potentials created by each of the charges. The individual potential created by a
point charge q is V = kq/r (Equation 19.6), where r is the distance between the charge and an
empty corner.
SOLUTION We begin by noting that each side of the square has a length L, so that the
diagonal has a length of 2 L , according to the Pythagorean theorem. Using Equation 19.6,
we can express the total potential at corners A and B as follows:
VTotal, A = kq2 kq1
+
L
2L and VTotal, B = kq1 kq2
+
L
2L Using Equation 19.3, we find that the electric potential energy of the third charge at each
corner is: Chapter 19 Problems 1019 kq q k q kq
EPE A = q3VTotal, A = q3 2 + 1 = 3 q2 + 1 L
2L 2
L ( −6.0 × 10−9 C )(8.99 × 109 N ⋅ m2 /C2 ) +4.0 × 10−9 C + +1.5 × 10−9 C =
0.25 m 2 = −1.1 × 10−6 J
kq q k q kq
EPE B = q3VTotal, B = q3 1 + 2 = 3 q1 + 2 L
2L 2
L
= ( −6.0 × 10−9 C ) (8.99 × 109 N ⋅ m2 /C2 ) +1.5 × 10−9 C + +4.0 × 10−9 C 0.25 m 2 = −0.93 × 10−6 J 17. SSM WWW REASONING Initially, suppose that one
charge is at C and the other charge is held fixed at B. The charge
at C is then moved to position A. According to Equation 19.4,
the work WCA done by the electric force as the charge moves
from C to A is WCA = q (VC − VA ) , where, from Equation 19.6, VC = kq / d and VA = kq / r . From the figure at the right we see
that d = r 2 + r 2 = 2r . Therefore, we find that d r B kq kq kq 2 1 WCA = q – =
– 1 r 2 2r r SOLUTION Substituting values, we obtain
WCA = C A (8.99 ×109 N ⋅ m 2 / C2 )(3.0 ×10−6 C) 2 1 – 1 = –4.7 ×10−2 J 0.500 m
2 r 1020 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 18. REASONING The electric potential at a distance r from
a point charge q is given by Equation 19.6 as V = kq / r .
The total electric potential at location P due to the four
point charges is the algebraic sum of the individual
potentials. q
d SOLUTION The total electric potential at P is (see the
drawing)
V= k ( −q )
d + k ( +q )
2d + k ( +q )
d + k ( −q )
d = −q q
d d
P − kq
2d d
−q Substituting in the numbers gives N ⋅ m2 − 8.99 ×109
2.0 × 10−6 C
2
C
− kq
=
= −9.4 ×103 V
V=
2d
2 ( 0.96 m ) ( ) 19. REASONING The electric potential at a distance r from a point charge q is given by
Equation 19.6 as V = kq / r . The total electric potential at location P due to the six point
charges is the algebraic sum of the individual potentials. +7.0q +5.0q +3.0q
d d d d
P
d d
−5.0q −3.0q +7.0q SOLUTION Starting at the upper left corner of the rectangle, we proceed clockwise and
add up the six contributions to the total electric potential at P (see the drawing): V= k ( +7.0q )
d d + 2 2 2 = k ( +14.0q )
d d + 2
2 2 + k ( +3.0q )
k ( +5.0q )
k ( +7.0q )
k ( −3.0q )
k ( −5.0q )
+
+
+
+
d
d
2
2
2
2 d 2 d 2 d d + d + d + 2
2
2
2
2 Chapter 19 Problems Substituting q = 9.0 × 10 −6 1021 C and d = 0.13 m gives 2 9 N⋅m +14.0 ) 9.0 × 10−6 C 8.99 ×10
2 (
k ( +14.0q )
C
=
= +7.8 × 106 V
V=
2
2
d 0.13 m d2 + ( 0.13 m )2 + 2
2 ( ) 20. REASONING The potential from the positive charge is positive, while the potential from
the negative charge is negative at the spot in question. The magnitude of the positive
potential must be equal to the magnitude of the negative potential. Only then will the
algebraic sum of the two potentials give zero for the total potential.
The drawing shows the arrangement of the charges.
the two charges and the spot in question form a right
triangle. Therefore, the distance between the positive
charge and the zeropotential spot is the hypotenuse
of the right triangle and, according to the Pythagorean
theorem, is greater than L. Zero potential
L
–q +2q 2.00 m There are two spots on the dashed line where the total potential is zero. One spot is shown in
the drawing. The other spot is on the dashed line at a distance L below the negative charge.
SOLUTION We begin by noting that the distance between the positive charge +2q and the
zeropotential spot is given by the Pythagorean theorem as L2 + ( 2.00 m ) . With this in
2 mind and using V = kq/r (Equation 19.6), we write the total potential at the spot in question
as follows: VTotal = k ( −q )
k ( +2 q )
+
=0
2
L
L2 + ( 2.00 m ) or +2 1
=
L L2 + ( 2.00 m ) Squaring both...
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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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