Chapter 25
Electric Potential. Solutions of Home
Work Problems
25.1
Problem 25.14 (In the text book)
A particle having charge q = +2.00 C and mass m = 0.010 0 kg is connected to a string
that is L = 1.50 m long and is tied to the pivot point P in Figure (25.14). The particle,
string and pivot point all lie on a frictionless horizontal table. The particle is released from
rest when the string makes an angle θ = 60.0◦ with a uniform electric field of magnitude E
= 300 V /m. Determine the speed of the particle when the string is parallel to the electric
field (point a in Figure (25.14)).
Figure 25.14:
2
CHAPTER 25. ELECTRIC POTENTIAL. SOLUTIONS OF HOME WORK
PROBLEMS
Solution
Taking point P as the reference point and setting the electric potential there at zero, then,
the electric potential at the original position of the particle is:
Vi = −E · s = −EL cos θ
at the final point a the electric potential is Vf = −EL. Since the table is frictionless, that is
there is not energy loss, then:
(K + U )i = (K + U )f
1 2
0 − qEL cos θ =
mv − qEL
2
which give the velocity as:
r
r
2qEL(1 − cos θ)
2 × 2.00 × 10−6 × 300 × 1.5 × (1 − cos 60)
=
= 0.300 m/s
v=
m
0.010
Physics 111:Introductory Physics, Chapter 25
Winter 2005
Ahmed H. Hussein
25.2. PROBLEM 25.20 (IN THE TEXT BOOK)
25.2
3
Problem 25.20 (In the text book)
Two point charges, Q1 = + 5.00 nC and Q2 = −3.00 nC, are separated by 35.0 cm.
(a) What is the potential energy of the pair? What is the significance of the algebraic sign
of your answer?
(b) What is the electric potential at a point midway between the charges?
Solution
(a) The potential energy of the pair of charges is:
U=
ke Q 1 Q 2
8.99 × 109 × 5.00 × 10−9 × −3.00 × 10−9
=
= −3.86 × 10−7 J
r
0.35
The negative sign in the answer means that one needs to spend 3.86 × 10−7 J of energy
to take the charges from a 35 cm to a much larger separation.
(b) The electric potential at a point due to two charges is the sum of the potentials due
to the individual charges. The electric potential at a point midway between the two
charges is:
ke Q 1 ke Q 2
5.00 × 10−9 −3.00 × 10−9
9
V = V1 + V2 =
+
= 8.99 × 10
+
= 103 V
r
r
0.175
0.175
Physics 111:Introductory Physics, Chapter 25
Winter 2005
Ahmed H. Hussein
4
25.3
CHAPTER 25. ELECTRIC POTENTIAL. SOLUTIONS OF HOME WORK
PROBLEMS
Problem 25.46 (In the text book)
Calculate the electric potential at point P on the axis of the annulus shown in Figure (25.46),
which has a uniform charge density σ.
Figure 25.46:
Solution
We can find the potential at point be as the potential due to a solid disk with radius b minus
the potential due a disk with radius a, i.e.
i
i
h√
h√
h√
i
√
2
2
2
2
2
2
2
2
x + b − x − 2πσke
x + a − x = 2πke
x +b − x +a
V = 2πσke
OR
We can choose a ring on the annulus with radius r and thickness dr and find the potential
dV due to this ring, which is:
ke dq
dV =
s
where, dq is the charge on the √
ring which is dq = σA = σ2πrdr and s is the distance from
the ring to point P where s = r2 + x2 , so the potential of the annulus is then:
Z b
Z b
h√
h√
ib
i
√
rdr
2
2
2
2
2
2
√
V =
dv = 2πke σ
= 2πke σ
r +x
= 2πke σ
x +b − x +a
a
r 2 + x2
a
a
Physics 111:Introductory Physics, Chapter 25
Winter 2005
Ahmed H. Hussein
25.4. PROBLEM 25.56 (IN THE TEXT BOOK)
25.4
5
Problem 25.56 (In the text book)
An electron is released from rest on the axis of a uniform positively charged ring, 0.100 m
from the rings center. If the linear charge density of the ring is +0.100 C/m and the radius
of the ring is 0.200 m, how fast will the electron be moving when it reaches the center of the
ring?
Solution
The potential Vi created by the ring at the electrons starting point is:
ke (2πλa)
=p 2
xi + a2
+a
ke Q
Vi = p
x2i
2
where Q is the total charge of the ring, xi is the initial position of the electron from the
center of the ring before it was released and λ is the linear charge density of the ring. The
final position of the electron is at the center of the ring xf = 0 so the final electric potential
is Vf = 2πke λ. The electron’s energy conservation gives:
1
1 2
mvi + Ui = mvf2 + Uf
2
2
where Ui = eVi is the initial potential energy of the electron and Uf = eVf is the potential
energy of the electron at the final location. Noting that vi = 0 we get;
0 + eVi =
1 2
mv + eVf
2 f
1 2
mv = e(Vi − Vf )
2 f
2e
(Vi − Vf )
vf2 =
m
!
2e
2πke λa
p
=
− 2πke λ
m
x2i + a2
!
4eπke λ
a
p
=
−1
m
x2i + a2
4π(−1.602 × 10−19 ) × 8.99 × 109 × 1.00 × 10−7
=
9.11 × 10−31
vf
!
0.200
p
(0.100)2 + (0.200)2
−1
= 2.097 × 1014
= 1.45 × 107 m/s
Physics 111:Introductory Physics, Chapter 25
Winter 2005
Ahmed H. Hussein
6
25.5
CHAPTER 25. ELECTRIC POTENTIAL. SOLUTIONS OF HOME WORK
PROBLEMS
Problem 25.66 (In the text book)
Consider two thin, conducting, spherical shells as shown in Figure (25.66). The inner shell
has a radius r1 = 15.0 cm and a charge of 10.0 nC. The outer shell has a radius r2 = 30.0
cm and a charge of −15.0 nC. Find
(a) the electric field E and
(b) the electric potential V in regions A, B, and C, with V = 0 at r = ∞.
Figure 25.66:
Solution
(a) The electric field at a point a distance r from the center can be found using Gauss law
by drawing a Gauss surface at the point and finding the charge qin within the surface:
qin
E = ke 2
r
applying this at each of the specified points we get:
EA = 0
EB
EC
no charge within
qA
1.00 × 10−8
89.9
= 2 V /m
= ke 2 = 8.99 × 109 ×
2
r
r
r
qA + qB
(10 − 15) × 10−9
45.0
9
= ke
=
8.99
×
10
×
= − 2 V /m
2
2
r
r
r
Physics 111:Introductory Physics, Chapter 25
Winter 2005
Ahmed H. Hussein
25.5. PROBLEM 25.66 (IN THE TEXT BOOK)
7
(b) It is easier to do this part in reverse order. The potential at C is:
VC = EC r = −
45.0
r
The potential at r2 is:
45
= −150 V
0.3
So, the potential inside r2 at say B is then:
Z r
Z r
dr
1
1
89.9
VB = V (r2 )+
−
= −450+
V
EB dr = −150+
89.9 2 = −150+89.9
r
r 0.300
r
r2
r2
V (r2 ) = −
Similarly:
V (r1 ) = −450 +
89.9
= +150 V
0.150
So the potential inside r1 is:
VA = +150 V
Physics 111:Introductory Physics, Chapter 25
Winter 2005
Ahmed H. Hussein