Chater 23: Electric Fields
Michelle Chen
Conductor: material in which electrons move freely
Insulator: material in which electrons do not move freely
Induction: process of charging in which a charged object is brought near a neutrally charged objects,
causing the electrons in the neutral object to move
Equations
!!!⃗
𝐹!
𝑞
!!!⃗
𝐹! = 𝑞𝐸!⃗
𝑞! 𝑞!
𝐹 = 𝑘! ! 𝑟̂
𝑟
𝑞
𝐸!⃗ = 𝑘 ! 𝑟̂
𝑟
𝑑𝑞
𝐸!⃗ = 𝑘! ! ! 𝑟̂
𝑟
𝐸!⃗ =
Constants/Values
1
= 8.99×10! 𝑁 ∙ 𝑚! /𝐶 !
4𝜋𝜀!
𝑞!"#$#% = +1.602𝑒 − 19 𝐶
𝑞!"!#$%&' = −1.602𝑒 − 19 𝐶
𝑘=
Field lines
•
•
•
Always away from positive and
towards negative
Number of lines drawn correspond
to the magnitude of the charge
Use 8 lines per magnitude of
charge
Field Diagram Example
Charging by induction
Charging by conduction
Problem 1
An electron accelerates at rest at 780 N/C. Later, the electron reaches a speed of 5.30
Mm/s. (a) What is the electron’s acceleration? (b) How long does it take the electron to
reach this speed? (c) How far does the electron move?
Solutions:
(a):
𝐹 = 𝑚𝑎
𝑞𝐸 = 𝑚𝑎
−1.602×10!!" 780 = 9.11×10!!" 𝑎
𝑎 = 1.37×10!" 𝑚/𝑠 !
(b):
𝑣! = 𝑣! + 𝑎𝑡
5.30×10! = 0 + 1.37×10!" 𝑡
𝑡 = 3.87×10!! 𝑠
(c):
𝑣! ! = 𝑣! ! + 2𝑎∆𝑥
(5.30×10! )! = 0 + 2 1.37×10!" 𝑥
𝑥 = .103 𝑚
Problem 2
Two 1-μC charges lie at (1,0) [A] and at (0,1) [B]. A third negative charge, -Q, lies
along the y-axis at (0,10) [C]. (a) Calculate the magnitude of the force that charge A
exerts on charge B. (b) Calculate the components of this force in the x- and y-directions
and express the force as a vector. (c) Given that the y-component of the net force
(due to charges B and C) on charge A is zero, calculate the unknown charge –Q.
Solutions:
(a):
C
𝐹 = 𝑘!
𝐹=
B
𝑞! 𝑞!
𝑟!
(8.99×10! )
ϕ
The angle is 45˚ because it’s a
45-45-90 triangle
𝐹 = 3.18×10!! 𝑁 𝚤 − (3.18×10!! 𝑁)𝚥
𝐹 = 𝑘!
𝑞! 𝑞!
𝑟!
1.0×10!! 𝑄
𝐹 = 8.99×10!!
1.0! + 10.0!
!
= 89.0 𝑄 𝑁/𝐶
𝐹! = 𝑠𝑖𝑛ϕ 89𝑄 =
88.6𝑄 = 3.18×10!!
𝑄 = 3.59×10!! 𝐶
! !√1.0! + 1.0! !
𝐹 = 4.5×10!! N A
(b):
𝐹! = 4.5×10!! cos 45 = 3.18×10!! 𝑁
𝐹! = − 4.5×10!! sin 45 = −3.18×10!! 𝑁
(c):
(1.0×10!! )!
!"
!"! !!!
89𝑄 = 88.6𝑄
Problem 3
An insulating rod (L=20.0 cm) of uniform charge is being into a semicircle. The total charge
of the rod is 15.5 μC. What is the magnitude and direction of the rod at the center of
the circle?
Solutions:
𝐸=
!∙!"
𝑑𝐸𝑐𝑜𝑠𝜃 =
𝑘
𝐸= !
𝑟
𝑘
𝐸= !
𝑟
𝑘𝜆
𝐸=
𝑟
!!
𝑑𝑞 = 𝜆𝑑𝐿
𝜆 𝑑𝐿 𝑐𝑜𝑠𝜃
𝜆𝑑𝐿 = 𝜆𝑟 ∙ 𝑑𝜃
𝑘𝜆
𝜆𝑟 ∙ 𝑑𝜃 𝑐𝑜𝑠𝜃 = ! 𝑟 ∙ 𝑑𝜃 𝑐𝑜𝑠𝜃
𝑟
!!
!
!
𝑐𝑜𝑠𝜃 𝑑𝜃
!
3𝜋
𝑘𝜆
𝑠𝑖𝑛𝜃 𝜋 2 =
𝑟
2
𝐸 = −2.19×10!! 𝑁/𝐶
𝐸=
We use cos because
vertical components of
force would cancel
𝑐𝑜𝑠𝜃
1.55×10!!
)
3𝜋
𝜋
. 20
sin
− sin ( )
6.37
2
2
(8.99×10! )(