CH25 - CHAPTER 25 ELECTRIC POTENTIAL ActivPhysics can help...

CHAPTER 25 ELECTRIC POTENTIAL
ActivPhysics can help with these problems: Activities 11.9, 11.10
Section 25 - 2: Potential Difference
Problem
1. How much work does it take to move a 50- C m charge against a 12-V potential difference?
Solution
The potential difference and the work per unit charge, done by an external agent, are equal in magnitude, so
W q V = = = C V J. ∆ ( )( ) 50 12 600 m m (Note: Since only magnitudes are needed in this problem, we omitted the subscripts
A and B .)
Problem
2. The potential difference between the two sides of an ordinary electrical outlet is 120 V. How much energy does an
electron gain when it moves from one side to the other?
Solution
Moving from the negative to the positive side (i.e., opposite to the electric field), an electron gains e V eV ∆ = = 120
( . )( ) . 16 10 120 1 92 10
19 17
× = ×
- -
C V J
of energy.
Problem
3. It takes 45 J to move a 15-mC charge from point A to point B . What is the potential difference ∆ V
AB
?
Solution
The work done by an external agent equals the potential energy change, ∆ ∆ U q V
AB AB
= = 45 J , hence ∆ V
AB
=
45 15 3 J mC kV.
=
= (Since the work required to move the charge from A to B is positive, V V V
B A AB
and ∆ is positive.)
Problem
4. Show that 1 V/m is the same as 1 N/C.
Solution
Since a volt joule/coulomb newton-meter/coulomb, = = it follows that a V/m N/C = .
Problem
5. Find the magnitude of the potential difference between two points located 1.4 m apart in a uniform 650 N/C electric
field, if a line between the points is parallel to the field.
Solution
For l in the direction of a uniform electric field, Equation 25-2b gives ∆ V E = = = l ( )( .4 ) 650 1 910 N/C m V. (See note
in solution to Problem 1. Since dV = ⋅ - E d l , the potential always decreases in the direction of the electric field.)
Problem
6. A charge of 31 . C moves from the positive to the negative terminal of a 9.0-V battery. How much energy does the
battery impart to the charge?

CHAPTER 25 595
Solution
∆ ∆ U q V
AB AB
= = = C V J. ( . )( . ) . 31 9 0 27 9
Problem
7. Two points A and B lie 15 cm apart in a uniform electric field, with the path AB parallel to the field. If the potential
difference ∆ V
AB
is 840 V, what is the field strength?
Solution
Equation 25-2b for a uniform field gives E V = = = ∆
= =
l 840 5 60 V 0.15 m kV/m. . (See notes in solutions to Problems 1
and 5.)
Problem
8. Figure 25-37 shows a uniform electric field of magnitude E . Find expressions for (a) the potential difference ∆ V
AB
and
(b) ∆ V
BC
. (c) Use your result to determine ∆ V
AC
.
FIGURE
25-37 Problem 8.
Solution
(a) On the line A to B , d l is antiparallel to E , so Equation 25-2 gives V V E d Ed
B A A
B
A
B
- = -
z
⋅ =
z
= E d l l . (b) The line
B to C makes an angle of 45 ° with E , so V V E d Ed
C B B
C
- = -
z
° = - cos . 45 2 l
=
(c) Addition yields V V
C A
- =
V V V V Ed Ed
C B B A
- + - = - = ( ) . . 1 1 2 0 293
=
Problem
9. A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a
potential difference of 100 V. Find the energy each gains.