Chapter 23: Electric Fields By Courtney Yang Background/Summary Electromagnetic forces exist everywhere in nature, and are governed by the existence of positive and negative charges in the atmosphere. The relationship between positive and negative charges (a lack or abundance of electrons, respectively) and the way charges move relative to each other are both illustrated by electric fields. Major Topics • Test charges—opposites attract ! !
• Coulomb’s Law, ! = ! !! ! ! Electric fields and electric field lines Motion of charged particles in an electric field Important Vocabulary • Proton = +1.6e-­‐19 C, mass = 1.67e-­‐27 • Electron= -­‐1.6e-­‐19 C, mass = 9.11e-­‐31 • Conductor = material in which electrons can move freely • Insulator = material in which electrons cannot move freely • Coulomb = SI unit of charge • Coulomb constant ! = 8.99!9 ! ∙ !! /! ! • !! = 8.85! − 12 ! ! /! ∙ !! • Electric field = exists in space around a charge or charged object (units N/C) Diagrams Here are 4 basic scenarios illustrating the behavior of positive and negative charges (F=force on test charge, E=direction of electric field) q
q0
q
q0
+
-­‐ r
F
F
P
P
q
q
q0
-­‐ +
r
E
r
E
P
P
•
•
Formulas !! !!
!"
!
!
!"
!! !"
! = ! ! !! = !" !" = ! ! ! = ! ! = !
=!
! = = !
!
!
!
!
!
!
! !
!
!
!
! = ! = ! = !
!
!
Free Response Problems (See solutions for diagrams) 1. [Easy] The electron and proton of a hydrogen atom are separated by a distance of approximately 5.3e-­‐11 m. Find the magnitudes of the electric force and the gravitational force between the two particles. 2. [Medium] A ring of radius a carries a uniformly distributed positive total charge Q. Calculate the electric field due to the ring at a point P lying a distance x from its center along the central axis perpendicular to the plane of the ring. 3. [Medium] Point P is at the origin, as shown below, and is the center of curvature of the charge distribution. a. On the dot representing point P below, indicate the direction of the electric field at point P due to the charge Q. b. Derive an expression for the magnitude of the electric field at point P. Solutions: 1. ! = !
!
!
!^!
=
!.!!!! !.!"!!!" !
!.!!!!! !
= 8.2! − 8 ! We use the magnitude of the charges for Coulomb’s law (otherwise we would have a negative force, because although the charges of a proton and electron are of the same magnitude, an electron has a negative charge), and find the force of attraction between the proton and electron. The force is an attractive force because of the opposite charges of the proton and electron. p
+ e
-­‐ 5.3e-­‐11
Hydrogen atom !
!"
!
2. !! = ! ! ! , ! = ! ! + !! ! , cos ! = ! !" !
!"
!" = !"#$%& = ! !
=
! !" ! !
! ! + !! !
!"
!"
!"#
!=
!"
=
!"
=
!
!
! ! ! + !! !
! ! + !! !
! ! + !! !
dq
+
+
+
r +
+
P
+
+
x
θ
+
+
+
+
+
!
Using the Pythagorean Theorem, we find that ! = ! ! + !! ! , and we use this value for r in the differential equation for an electric field. Since we are only using the cosine (or x) component of each dq section around the ring (the y-­‐values cancel each other out), we can substitute x/r for cosine in the differential equation. Finally, after the integration, we find that the electric field at point P relative to a ring of equally !"#
distributed charge can be expressed as ! . ! ! !! ! !
3. ! =!=
!" =
2!
!" !
!" =
!"#$%& =
! !"
!"#$ !!
!
2!"
2!"
! =
!"#$ !" =
!"#$ 4! !
!
!!
!!
−4
!!/!
2 2
! =
!" !!!
!/!
Like Problem #2, we only need to consider the cosine (x) components of the electric field because the vertical components cancel each other out. Therefore, since we are integrating with respect to !"#$, we need to alter dq so it is with respect to the angle theta. Plugging everything in and integrating from π/4 to –π/4 (equivalent part of circle over which charge Q is distributed), we solve for the magnitude of the electric field at point P.