PHYS 222
Worksheet 1 - Coulomb’s law
Supplemental Instruction
Iowa State University
Leader:
Course:
Instructor:
Date:
Alek Jerauld
PHYS 222
Dr. Paula Herrera-Siklódy
1/17/12
Useful Equations

The force exerted on two charged point particles due to their
charges

The electrical field created by a charged point particle
𝐸𝑛𝑒𝑡 =

2𝑘𝑒 𝑄
𝑟𝐿 The electrical field at some point r distance above an infinite
line of charge
Constants
Related Problems
1) Two uncharged metal spheres, L and M, are in contact. A negatively charged plastic
rod is brought close to L, but not touching it, as shown below. The two spheres are then
slightly separated and, after this separation, the rod is removed. As a result of these
steps:
A) both spheres are neutral
B) both spheres are positively charged
C) both spheres are negatively charged
D) L is negatively and M is positively charged
E) L is positively and M is negatively charged
++
++
-
++
++
-
y
2) Find the magnitude and direction of the electrical field
at the origin (0,0,0) in terms of q. (Drawing is NOT to
scale)
-q
120 mm
𝐸
𝐸
𝑘
𝑘
16 cm
𝑘
4 cm
3 cm
z
𝐸
𝐸
𝐸
𝐸
+q
𝑘
𝑘
𝑘
𝑘 𝑘
(magnitude)
𝑘 (Unit vector shows direction)
3) The diagram below shows the electric field lines in a region of space containing two
small charged spheres, Y and Z. We can deduce that:
A) Y is positive and Z is negative
B) the magnitude of the electric field is the
same everywhere
C) the electric field is strongest midway
between Y and Z
D) a small negative charge placed at point X
would experience a force to the left
E) Y and Z must have the same sign
A) Incorrect because particle Y is NEGATIVE and particle Z is POSSITIVE.
B) Incorrect because electrical field depends on the distance from the charged
particle, thus cannot be constant throughout space
C) Incorrect because there is a smaller E field line density halfway between the
particles than it is right next to the particles. And because E is INVERSELY
related to the square of the distance, thus increasing the E magnitude the closer
it is to the charged particle
D) CORRECT because the negative X particle will be “attracted” to the positive Z
particle since opposite charges attract one another. This “attraction” is a Force
pointing to the left
E) Incorrect because the field lines are going away from Z and coming into Y, thus
they cannot be the same.
-q
x
4) A particle with a mass of 10-4 kg and a charge of +2 µC is moving at a constant
horizontal velocity of 3 m/s into an area with an electrical field of 50 N/C. Which point
most accurately describes the final resting position of the charged particle?
0.4 m
0.2 m
(a)
(b)
0.2 cm
E
0.2 cm
(c)
(d)
V0 x  3m / s
V0 y  0
F  Eq
Eq
m
2
ayt
Eqt 2
y  y0  v0 y t 
 y  0  0t 
t 
2
2m
F  ma y  a y 
x  x0  v0 xt 
2(0.0002)(104 )
 0.02s
50(2)(10 6 )
2 ym

Eq
ax t 2
 x  0  v0 xt  0  3(.02)  0.6m
2
The particle is positive, so it will move in the direction of the electric field. B is the
correct answer.
5) Three point charges are arranged on a line. Charge q3 = +5 nC and is at the origin.
Charge q2 = -3 nC and is at x= +4 cm. Charge q1 is at x= +2 cm. What is q1 (magnitude
and sign) if the net force on q3 is zero?
F3/1
q3
+
0
F3/2
q1
?
0.02 m
q2
-
x
0.04 m
kq3q2 kq3q1
kq3q2
kq3q1
0.0022
 F3x 0  0.0042  0.022  0.0042  0.022  q1  0.0042 q2  0.75nC
CHALLENGE!!
6) Positive charge Q is uniformly distributed around a semicircle of radius A.
dθ
θ
Ey

Q


Enet  E y  
0
k
2kQ
cos(

/
2


)
d


a2
 a2
Ex cancels out, so the net electric field is the summation of Ey around the semicircle