Homework Problem Set 1 – With Solutions
Homework due by 8:00 am on Tuesday January 20.
Partial credit may be given even if the final answer is incorrect so please show all work!
Question 1 (1 point)
List three basic properties of electric charge.
Possible Correct Answers:
Opposite charges attract and like charges repel, charge is conserved, there are two types of
charge (+ & -), and charge is quantized.
Question 2 (3 points)
Four charged objects are sitting on a horizontal surface, each at the corners of a square. The first
object is located at the origin with a charge of +0.25 mC. The second object is located at (x, y) =
(0, 0.5 m) with a charge of -0.25 mC. The third object is located at (x, y) = (0.5 m, 0) with a
charge of -0.25 mC. The fourth object is located at (x, y) = (0.5 m, 0.5 m) and a charge of +0.25
mC. What are the forces on each of the objects? What is the strength of the electric field at the
center of the square?
Solution: Call the four corners of the square A, B, C and D. Then:
𝐹⃑ BA =-keqAqB = (8.99x109 N m2/kg)(0.25x10-3C)2 = 2250N in the plus y direction
r2
0.52
𝐹⃑ DA =-keqAqC = (8.99x109 N m2/kg)(0.25x10-3C)2 = 2250N in the plus x direction
r2
0.52
𝐹⃑ DA =-keqAqC = (8.99x109 N m2/kg)(0.25x10-3C)2 = 1125N in a direction 45º between -x and -y
r2
(2)(0.5)2
∑ 𝐹⃑ = 𝐹⃑ AB + 𝐹⃑ AC + 𝐹⃑ AD
∑ 𝐹⃑ x= 𝐹⃑ ABx + 𝐹⃑ ACx + 𝐹⃑ ADx = 0 - 1125Nsin 45 + 2250N = 1455N x
∑ 𝐹⃑ y= 𝐹⃑ ABy + 𝐹⃑ ACy + 𝐹⃑ ADy = 2250N – 1125Ncos 45 + 0 = 1455 y
𝐹⃑ A = (1455 N , 1455 N)
𝐹⃑ B = (1455 N , -1455 N)
𝐹⃑ C = (-1455 N , -1455 N)
𝐹⃑ D = (-1455 N , 1455 N)
What is the strength of the electric field at the center of the square?
𝐸�⃑ CENTER = 0 (by symmetry of the charges on the corners of the square)
Question 3 (3 points)
Two charged objects are sitting on a horizontal surface. The first object is located at the origin
with a charge of +0.25 mC. The second object is located at (x, y) = (0, 0.5 m) with a charge of 0.25 mC. What is the energy required to move both charges from an infinite distance away to the
positions they are in now? If the electric potential an infinite distance away is set to be 0V, what
is the electric potential at the point halfway between the two charges?
Work =
ds where s is some path opposite to r (which points outward):
⋅dr
Work= Work = -
and since F = kq1q2/r2
kq1q2/r2⋅dr in the direction of r
*Since the charges and coulombs constant do not change, they can be pulled out of the
integral
Work = -kq1q2 dr/r2
Work = kq1q2/r from a to ∞
Work = k q1q2/a
Work = (9 x 109Nm2/C2)(0.25mC)(-0.25mC) = -1125J (no work done-gives energy)
0.5m
Electric Potential :
kq1/r + kq2/r = 0
*Since this is halfway between 2 equal and opposite charges there is no potential
Question 4 (3 points)
a) An infinitely long rod sits horizontally along the x-axis with a continuous uniform charge
distribution such that the linear charge density λ is 0.015 C/m. At a point 0.35 m directly
below the rod what is the strength of the electric field? Solve WITHOUT Gauss’ Law.
b) Same as part (a), but now DO use Gauss’ Law to solve. Which method is easier?
Solution to part (a): Horizontal component of the force sums to zero.
𝐷
𝑘 𝑑𝑑
𝐹⃑ y = 𝐹⃑ sin (θ) = 𝐸 𝑟 = 𝑟2
∞
𝐸�⃑ y = ∫−∞
∞
𝐸�⃑ y = ∫−∞
𝐸�⃑ y=
𝑘𝑘𝑘𝐷 2
3
�√𝑥 2 +𝐷 2 �
𝑘𝑘𝑘𝑘𝑘
3
�√𝑥 2 +𝐷 2 �
and 𝜆 =
For part (b):
=
𝑞
𝑥
∞
= 𝑘𝑘𝑘 ∫−∞
𝐶
𝑁𝑚2
(−2)(9𝑥109 2 )(0.015 )
𝑚
𝐶
0.35𝑚
𝐷
𝑟
𝑘𝑘𝑘𝐷2
3
�√𝑥 2 +𝐷2 �
𝑑𝑑
𝑑𝑑
so
𝑑𝑑
= λ so 𝑑𝑑 = 𝜆𝜆𝜆
2
��𝑥2 +𝐷 �
= −7.7 ×
108 𝑁
𝐶
=
3
𝑦
−∞
𝑘𝑘
𝑥
�
𝐷 √𝑥 2 +𝐷2 𝐷2 ∞
=
𝑘𝑘
𝐷
(1 + 1) =
2𝑘𝑘
𝐷
= 0.01λ5 C/m
∫∫E⋅dA
(for curved side only- since the angle between the field and normal force is 0 the dot
product becomes a scalar
E∫∫dA = EAcurved = E2πrl = qin / εo = λl / εo → E = λ/(2πrεo) = 2kλ/r
E= (2)(9x109 Nm2/C2)(0.015C/m)
= 7.7 x 108 N/C or 7.7 x 108 V/m
0.35 m
This method is easier (it doesn’t require solving a complicated integral!)