<Chap. 20 Electric Charge, Force, and Field: Suggested Problems and
Solutions>
20.19, 20.28, 20.51, 20.56, 20.69, 20.70
20.19
A Charge q is at the point x=5m, y=0m. Write expressions for the unit vectors you would use in
Coulomb’s law if you were finding the force that q exerts on other charges located at (a) x=5m, y=2.5m;
(b) the origin; (c) x=7m, y=3.5m. You are not given the sign of q. Why doesn’t it matter?
Solution
A unit vector from charge located at x=5m, y=0m to other point r=(x,y) can be described as,
𝑛=
𝑟 − 𝑟𝑞
|𝑟 − 𝑟𝑞 |
(𝑥 − 5, 𝑦)
=
√(𝑥 − 5)2 + 𝑦 2
a) For r=(5,2.5)
𝑛=
𝑟 − 𝑟𝑞
|𝑟 − 𝑟𝑞 |
=
(0, 2.5)
√2.52
= (0, 1)
b) For r=(0,0)
𝑛=
𝑟 − 𝑟𝑞
|𝑟 − 𝑟𝑞 |
=
(−5,0)
√52
= (−1, 0)
c) For r=(7,3.5)
𝑛=
𝑟 − 𝑟𝑞
|𝑟 − 𝑟𝑞 |
=
(2,3.5)
√4 + 12.25
≅ (0.496, 0.868)
Sign of q is not given since it does not affect the unit vector as shown in the expression
𝑛=
𝑟 − 𝑟𝑞
|𝑟 − 𝑟𝑞 |
20.28
The water molecule’s dipole moment is 6.17E-30C*m. What would be the separation distance if the
molecule consisted of charges ±𝑒? (The effective charge is actually less because H and O atoms share
the electrons.)
Solution
From equation 20.5, the electric dipole moment p can be expressed as the product of charge q
and separation distance d.
𝑝 = 𝑞𝑑
Substituting p=6.17E-30C*m and elementary charge e=1.60E-19C, one finds
6.17 ∗ 10−30 = 1.60 ∗ 10−19 𝑑
𝑑 ≅ 3.86 ∗ 10−11 𝑚 = 38.6𝑝𝑚
20.51
A proton is at the origin and an ion is at x = 5.0 nm. If the electric field is zero at x =-5.0 nm, what’s the
ion’s charge?
Answer)
The field at the point x = 5 nm due to the proton is ⃗⃗⃗⃗⃗⃗⃗⃗
𝐸𝑝𝑟𝑜 = (5
𝑘𝑒
𝑛𝑚)2
𝑘𝑞
And, the field at the same point due to the ion is ⃗⃗⃗⃗⃗⃗⃗
𝐸𝑖𝑜𝑛 = (10
𝑛𝑚)2
𝑘𝑒
𝑘𝑞
Solving ⃗⃗⃗⃗⃗⃗⃗⃗
𝐸𝑝𝑟𝑜 + ⃗⃗⃗⃗⃗⃗⃗
𝐸𝑖𝑜𝑛 = [(5
2 + (10
𝑛𝑚)
𝑛𝑚)2
] (−𝑥̂) = 0, ∴ 𝑞 = −4𝑒
(−𝑥̂).
(−𝑥̂).
20.56
Three identical charges q form an equilateral triangle of side a, with two charges on the x-axis and one
on the positive y-axis.
(a) Find an expression for the electric field at points on the y-axis above the uppermost charge.
(b) Show that your result reduces to the field of a point charge 3q for y ≫ a.
Answer)
(a) The electric field on the y axis (above the uppermost charge) due to the charges on the x axis
𝑦
2𝑘𝑞𝑦
⃗⃗⃗⃗1 = 𝑘𝑞 2 ×
is 𝐸
̂) =
𝑦̂.
1 × 2 (𝑦
𝑎
2 3⁄2
𝑦2 +
4
(𝑦 2 +
𝑎
)
4
𝑎2 2
)
4
(𝑦 2 +
And the electric field due to the charge on the y axis is ⃗⃗⃗⃗
𝐸2 =
∴ ⃗⃗⃗⃗
𝐸1 + ⃗⃗⃗⃗
𝐸2 = 𝑘𝑞 [
2𝑦
3⁄2
𝑎2
(𝑦 2 + )
4
+
1
2
(𝑦−√3𝑎 ⁄2)
] 𝑦̂.
(b) For y ≫ a, 𝑦 − 𝑎 ~ 𝑦 and 𝑦 2 + 𝑎2 ~ 𝑦 2 .
2𝑦
1
2
1
∴ ⃗⃗⃗⃗
𝐸1 + ⃗⃗⃗⃗
𝐸2 ~ 𝑘𝑞 [(𝑦2)3⁄2 + (𝑦)2] 𝑦̂ = 𝑘𝑞 [ 2 + (𝑦)2] 𝑦̂ =
𝑦
𝑘(3𝑞)
𝑦2
𝑦̂.
𝑘𝑞
2
(𝑦−√3𝑎 ⁄2)
𝑦̂.
20.69
An electric quadrupole consists of two oppositely directed dipoles in close proximity. (a) Calculate the
field of the quadrupole shown in Fig. 20.33 for points to the right of x = a and (b) Show that for x >> a
the quadrupole field falls off as 1/x4.
Fig. 20.33
(a) For x > a, electric field can be calculated by Coulomb’s law as below:
𝐸⃗ (𝑥) = [𝑘
(+𝑞)
(𝑥 − (−𝑎))
2+𝑘
(−2𝑞)
(+𝑞)
+𝑘
] 𝑥̂
(𝑥 − 𝑎)2
𝑥2
If you are confused to determine the sign of the electric field, remind that electric field is simply
“force per unit (+) charge”. So, you can determine the direction of the field by putting unit (+)
charge on the position where you want to know it.
(b) Let us consider the binomial expansion (one kind of Taylor expansion):
(𝟏 + 𝒙)𝒏 = 𝟏 + 𝒏𝒙 +
𝒏(𝒏 − 𝟏) 𝟐 𝒏(𝒏 − 𝟏)(𝒏 − 𝟐) 𝟑
𝒙 +
𝒙 +⋯
𝟐!
𝟑!
(n can be negative integer and -1 < x < 1)
Then, for x >> a, some expressions in (a) can be changed:
1
1
𝑎
2 = 𝑥 2 (1 + (𝑥 ))
(𝑥 − (−𝑎))
1
1
𝑎
= 2 (1 + (− ))
2
(𝑥 − 𝑎)
𝑥
𝑥
−2
−2
=
=
(−2)(−2 − 1) 𝑎 2
1
𝑎
(−2)
(1
+
(
)
+
( ) + ⋯)
𝑥2
𝑥
2
𝑥
(−2)(−2 − 1)
1
𝑎
𝑎 2
(−2)
(1
+
(−
)
+
(−
) + ⋯)
𝑥2
𝑥
2
𝑥
So, the electric field can be approximated as follows:
𝐸⃗ (𝑥) = [𝑘
(+𝑞)
2+𝑘
(𝑥 − (−𝑎))
(−2𝑞)
(+𝑞)
+𝑘
] 𝑥̂
(𝑥 − 𝑎)2
𝑥2
=
(−2)(−2 − 1) 𝑎 2
(−2)(−2 − 1)
𝑘𝑞
𝑎
𝑎
𝑎 2
[1 + (−2) ( ) +
( ) − 2 + 1 + (−2) (− ) +
(− ) + ⋯ ] 𝑥̂
2
𝑥
𝑥
2
𝑥
𝑥
2
𝑥
=
𝑘𝑞
𝑎2
6𝑘𝑞𝑎2
1
[0
+
0
+
6
]
𝑥
̂
=
𝑥̂ ∝ 4
2
2
4
𝑥
𝑥
𝑥
𝑥
Therefore, for x>>a, the quadrupole field falls off as 1/x4.
Note 1: If you want to know a kind of poles in the given system, then you should find the xdependence of electric field or potential for far distance. Also, charge configuration or dipole
configuration can be evidence for determining poles. For example, the quadrupole can be
expressed like below:
Note 2: Different types of poles can be mixed simultaneously.
20.70
Four charges lie at the corners of a square of side a, with the center of the square at the origin. The two
charges with y=a/2 have magnitude Q and are positive. The two charges with y=-a/2 also have
magnitude Q but are negative. (a) Find an expression for he magnitude of the electric field for points on
the y-axis with y>a/2. (b) Show that, for y>>a, your result exhibits the 1/y3 fall off you would expect for an
electric dipole. (c) Compare the result of (b) with Equation 20.6b and write an expression for the
magnitude of the dipole moment of this four-charge distribution. Hint: Be careful with your approximation
in (b)! If you get 0 for your answer, then you’ve gone too far. You can neglect a2 when compared with y2,
but you can’t neglect a when compared with y or you’ll be throwing out the charge separation that makes
this distribution resemble dipole at large distances.
(a) For y>a/2, electric field by the four charges can be drown like below:
By symmetry, what you only have to do first is to calculate E1 and E2. From the Coulomb’s law,
the magnitude of two electric fields are given by
E1 = k
𝑄
−𝑄
,E = k
𝑎 2
𝑎 2 2
𝑎 2
𝑎 2
( ) + (𝑦 − )
( ) + (𝑦 + )
2
2
2
2
From the vector sum and symmetry, total electric field can be obtained:
𝐸⃗ (0, 𝑦) = (2𝐸1 cos 𝜃1 + 2𝐸2 cos 𝜃2 )𝑦̂
,where cos 𝜃1 =
𝑎
2
(𝑦− )
, cos 𝜃2 =
2
2
√(𝑎) +(𝑦−𝑎)
2
2
𝑎
2
(𝑦+ )
2
√(𝑎) +(𝑦+𝑎)
2
2
2
By substituting two expressions into the electric field, then the solution is
𝑎
𝑄 (𝑦 − )
2
𝐸⃗ (0, 𝑦) = 2𝑘
3+
𝑎 2
𝑎 2 2
[( ) + (𝑦 − ) ]
2
2
(
𝑎
(−𝑄) (𝑦 + )
2
3
𝑦̂
𝑎 2
𝑎 2 2
[( ) + (𝑦 + ) ]
2
2
)
(b) When y >> a/2, electric field can be approximated by the following procedure:
𝐸⃗ (0, 𝑦) = 2𝑘 (
𝐸⃗ (0, 𝑦) ≈ 2k (
1
𝑄
3
𝑎 2
2 2
(𝑦− )
𝑎/2
2
[1+((𝑦−𝑎/2)) ]
𝑄
𝑎 2
(𝑦− )
2
−
𝑄
𝑎 2
(𝑦+ )
2
)=
+
(−𝑄)
1
3
𝑎 2
2 2
(𝑦+ )
𝑎/2
2
[1+((𝑦+𝑎/2)) ]
2𝑘𝑄
((1 −
𝑦2
𝑎
2𝑦
)
−2
) 𝑦̂
− (1 +
−2
𝑎
2𝑦
) ) 𝑦̂
Also, by using binomial expansion in the previous problem,
(1 −
(1 +
𝑎
2𝑦
𝑎
2𝑦
)
)
−2
−2
= 1 + (−2) (−
𝑎
2𝑦
)+
(−2)(−3)
2
𝑎
(−2)(−3)
2𝑦
2
= 1 + (−2) ( ) +
(−
𝑎
2
𝑎
2𝑦
2
) +
( ) +
(−2)(−3)(−4)
6
(−2)(−3)(−4)
2𝑦
6
𝑎
(−
𝑎
2𝑦
3
) +⋯
3
( ) +⋯
2𝑦
Therefore, electric field can be approximated by following term:
2𝑘𝑄
𝑎
𝑎
4𝑘𝑄𝑎
1
𝐸⃗ (0, 𝑦) ≈ 2 ((−2) (− ) − (−2) ( )) 𝑦̂ = 3 𝑦̂ ∝ 3
𝑦
2𝑦
2𝑦
𝑦
𝑦
2𝑘𝑝
(c) Comparing with the equation 20.6(b), 𝐸⃗ = |𝑥|3 𝑖̂, dipole moment is 𝑝 = 2Q𝑎𝑦̂.
So, the magnitude is 2Q𝑎. Also, the dipole can be expressed like below: