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Chapter 23
Electrical Potential
2
••
Picture the Problem A charged particle placed in an electric field experiences an
accelerating force that does work on the particle. From the work-kinetic energy theorem
we know that the work done on the particle by the net force changes its kinetic energy
and that the kinetic energy K acquired by such a particle whose charge is q that is
accelerated through a potential difference V is given by K = qV. Let the numeral 1 refer
to the alpha particle and the numeral 2 to the lithium nucleus and equate their kinetic
energies after being accelerated through potential differences V1 and V2.
*11 •
Picture the Problem We can use Coulomb’s law and the superposition of fields to find E
at the origin and the definition of the electric potential due to a point charge to find V at
the origin.
Apply Coulomb’s law and the
superposition of fields to find the
electric field E at the origin:
Express the potential V at the origin:
 

E  E Q at a  E Q at a

kQ ˆ kQ ˆ
i  2 i 0
a2
a
V  VQ at a  VQ at a
kQ kQ 2kQ


a
a
a
and (b) is correct.

21 •
Picture the Problem We can use the definition of finite potential difference to find the
potential difference V(4 m)  V(0) and conservation of energy to find the kinetic energy
of the charge when it is at x = 4 m. We can also find V(x) if V(x) is assigned various
values at various positions from the definition of finite potential difference.
(a) Apply the definition of finite
potential difference to obtain:
4m
 
V 4 m   V 0     E  d     Ed
b
a
0
 2 kN/C 4 m 
  8.00 kV
(b) By definition, U is given by:
U  qV  3 C  8 kV 
  24.0 mJ
(c) Use conservation of energy to
relate U and K:
K  U  0
or
K 4 m  K 0  U  0
Because K0 = 0:
Use the definition of finite potential
difference to obtain:
(d) For V(0) = 0:
K 4 m  U  24.0 mJ
V x   V x0    E x x  x0 
 2 kV/m x  x0 
V x  0  2 kV/m x  0
or
V x    2 kV/m x
(e) For V(0) = 4 kV:
V x  4 kV  2 kV/m x  0
or
V x   4 kV  2 kV/m x
(f) For V(1m) = 0:
V x  0  2 kV/m x  1
or
V x   2 kV  2 kV/m x
27 ••
Picture the Problem We know that energy is conserved in the interaction between the 
particle and the massive nucleus. Under the assumption that the recoil of the massive
nucleus is negligible, we know that the initial kinetic energy of the  particle will be
transformed into potential energy of the two-body system when the particles are at their
distance of closest approach.
K  U  0
(a) Apply conservation of energy to
the system consisting of the  particle
and the massive nucleus:
or
Because Kf = Ui = 0 and Ki = E:
 E Uf  0
K f  Ki  U f  U i  0
Letting r be the separation of the
particles at closest approach, express
Uf:
Uf 
kqnucleusq k Ze2e 2kZe2


r
r
r
Substitute to obtain:
E
2kZe2
0
r
Solve for r to obtain:
r
(b) For a 5.0-MeV  particle and a gold nucleus:



2kZe2
E

2 8.99  109 N  m 2 /C 2 79 1.6  1019 C
r5 
 4.55  1014 m  45.4 fm
5 MeV  1.6 1019 J/eV

2

For a 9.0-MeV  particle and a gold nucleus:




2 8.99 109 N  m2 /C 2 79 1.6 1019 C
r9 
 2.53 1014 m  25.3 fm
19
9 MeV  1.6 10 J/eV

2

33 ••
Picture the Problem For the two charges, r  x  a and x  a respectively and the
electric potential at x is the algebraic sum of the potentials at that point due to the charges
at x = +a and x = a.
(a) Express V(x) as the sum of the
potentials due to the charges at
x = +a and x = a:
 1
1
V  kq

 xa xa




(b) The following graph of V(x) versus x for kq = 1 and a = 1 was plotted using a
spreadsheet program:
10
V (V)
8
6
4
2
0
-3
-2
-1
0
1
2
3
x (m)
(c) At x = 0:
dV
dV
 0 and E x  
 0
dx
dx
*47 ••
Picture the Problem Let the charge per
unit length be  = Q/L and dy be a line
element with charge dy. We can express
the potential dV at any point on the x axis
due to dy and integrate of find V(x, 0).
(a) Express the element of potential
dV due to the line element dy:
dV 
k
dy
r
where r 
Integrate dV from y = L/2 to
y = L/2:
L2
kQ
dy
V x,0  

2
L L 2 x  y 2

(b) Factor x from the numerator and
denominator within the parentheses to
obtain:
x2  y2
2
2
kQ  x  L 4  L 2 
ln
L  x 2  L2 4  L 2 


2


 1 L  L 
kQ 
4x 2 2x 
V x,0 
ln 

2
L
 1 L  L 


4x 2 2x 

Use ln
a
 ln a  ln b to obtain:
b

kQ 
L2
L 
L2
L 
 


ln
1



ln
1


 

2
2



L  
4x
2x 
4x
2 x 



L2
L2
12
1
1 2
Let  
and
use
to
expand
:
1



1



1





...
2
8
4x 2
4x 2
V x,0 
12

L2 
1  2 
 4x 
2
1 L2 1  L2 
  ...  1 for x >> L.
1
 
2 4 x 2 8  4 x 2 
Substitute to obtain:
V x,0 
Let  
kQ  
L
L 

ln 1    ln 1  
L   2x 
 2 x 
L
L 

and use ln 1       12  2  ... to expand ln 1 
:
2x
 2x 
L L
L2
L
L
L2


ln 1   
 2 and ln 1      2 for x >> L.
2x 4x
 2x  2x 4x
 2x 
Substitute and simplify to obtain:
V x,0 
kQ  L
L2  L
L2 
kQ

 





2
2 

L  2 x 4 x  2 x 4 x 
x
77 ••
Picture the Problem We can use Wqfinal position  qVi f to find the work required to
move these charges between the given points.
(a) Express the required work in
terms of the charge being moved and
the potential due to the charge at x =
+a:
WQ a  QV a
 QV a   V  
kQ2
 kQ 
 QV a   Q

2a
 2a 
(b) Express the required work in
terms of the charge being moved and
the potentials due to the charges at
x = +a and x = a:
WQ0  QV0
 QV 0   V  
 QV 0 

 Q Vcharge at -a  Vcharge at  a

 2kQ2
 kQ kQ 
 Q


a 
a
 a
(c) Express the required work in
terms of the charge being moved and
the potentials due to the charges at
x = +a and x = a:
WQ2 a  QV02 a
 QV 2a   V 0

 kQ kQ 2kQ 
 Q



a
a 
 3a

2kQ2
3a
Chapter 24
Electrostatic Energy and Capacitance
21 ••
Picture the Problem The diagram shows
the four charges fixed at the corners of the
square and the fifth charge that is released
from rest at the origin. We can use
conservation of energy to relate the initial
potential energy of the fifth particle to its
kinetic energy when it is at a great distance
from the origin and the electrostatic
potential at the origin to express Ui.
Use conservation of energy to relate
the initial potential energy of the
particle to its kinetic energy when it
is at a great distance from the origin:
Express the initial potential energy
of the particle to its charge and the
electrostatic potential at the origin:

 Q Vcharge at -a  Vcharge at a  V 0
K  U  0
or, because Ki = Uf = 0,
Kf Ui  0
U i  qV 0
Substitute for Kf and Ui to obtain:
1
2
mv2  qV 0  0
2qV 0
m
Solve for v:
v
Express the electrostatic potential at
the origin:
V 0 
Substitute and simplify to obtain:
kq
2kq  3kq 6kq



2a
2a
2a
2a
6kq

2a
v
2q  6kq 
6 2k

 q
m  2a 
ma