Transcript Electric Forces and Fields

Electric Forces and Fields
AP Physics C
Electrostatic Forces (F)
(measured in Newtons)
Given 2 charges of magnitude q1 & q2 separated by a distance d, the
magnitude of the force that each of the charges exerts on the other is
calculated by the relationship given by Coulomb’s Law. The
direction of the force is determined by the fundamental law of
charges (i.e. attraction or repulsion). The forces must be equal and
opposite … Newton’s 3rd Law!
q1
F
F
+
Coulomb’s Law:
q2
-
d
F
F
+
+
q1
q2
k = 9 x 109 N*m2/C2
kq1q2
F
2
d
This is known as “Coulomb’s constant”
Electric Fields (E)
measured in Newtons per Coulomb
• Electric fields are the “energy field” that surrounds
any charged particle”
• Any other charge in this field will experience a force.
• Electric fields are vectors (mag. & dir.)
• The direction of an Electric field is defined by the
direction of the force on a “+” test charge placed in
that field, thus Electric fields are always in a direction
that is away from “+” and toward “-”.
• Electric Field strength is measured in units of N/C
(newtons per coulomb)
Force on a charge due
to an Electric Field
If a charge, q, is placed at point “x” in the field
where the Electric Field strength is E, it will
experience a force F.
F  qE
F
E
+
q
F
E
q
d
Q
+
Electric Potential Energy (Uelec)
measured in Joules
When two charges are placed near each other, they have
the potential to start moving under the influence of the
force between them. If they have the “potential” to have
“kinetic energy” then according to the Law of Conservation
of Energy, they must have Potential Energy when they are in
that arrangement. To calculate the potential energy of this
arrangement of charges…
q1
+
q2
d
U elec
kq1q2

d
Remember: energy of any kind is always a scalar quantity…magnitude only!
Charge Distributions
Line charge density (or charge per unit of length) is represented by the greek
letter lambda, λ, and calculated as   Q
, where Q is the total charge
L
and L is the length, measured in C/m.
++++++++++++++++++++++++++++++++++++
L
+++++
Surface charge density (or
charge per unit of area) is
represented by the greek
letter sigma, σ and Q
calculated as
A

and measured in C/m2
++++++++
++++++++++
++++++++
+++++
Q is the total charge and A
is the surface area of the
charged object. You may
need to know how to
calculate the surface area
of circle (disk), sphere,
cube, or cylinder.
Volume charge density, ρ is coming soon (Gauss’s Law…next)
E Field due a Line Charge Distribution
y
Basic Formulas:
 kq
E 2
r
dEy
y
r
θ
dEx
r  x2  y2
cos  
dE



E   dE   dE y
θ
r
Assume a line charge of length L and
charge density of λ=Q/L. An infinitesimal
piece of charge, dq, creates an
infinitesimal Electric Field, dE, at a point in
space. dE has components dEx and dEy.
y

 dEx  0
and

kdq
kdx
y
dE y  2 cos   2

r
x  y2 x2  y2

L/2
L/2
L/2
E   dE y  2 dE y  2
L / 2
dq = λdx
0
x
++++++++++++++++++++++++++++++++++++
- L/2
0
0
x
kydx
2
y
x
+ L/2
2k
Ey 

y
1
2
L
 L
1
2
2
 y2
2

3
2
E Field on the Axis of a Ring of Charge
Assume a ring of uniform charge distribution. Let dq be
an infinitesimal piece of charge on the ring and dE be the
part of the total electric field at a point on the ring’s axis
that is created by dq.
Basic Formulas:
dq
+ + +
+
+
+
+
+
a
+ + +
+
+
r
 kq
E 2
r
+
+
+
x
θ
θ
dEy



E   dE   dEx and
r  x2  a2
dEx
cos  
dE

 dE y  0

kdq
kdq
x
dE y  2 cos   2

r
x  y2 x2  y2

Q
Q
E   dEx  
0
0
x
kxdq
2
a
2

3

2
x
kQx
2
a
2

3
2
x
r
E field along the Axis of a Disk of
Uniform Surface Charge Density
da
Assume a disk of total charge Q, radius R, and surface
charge density σ=Q/A. Find the electric field at a point along
the axis of the disk at a distance, x, away.
We will divide the disk into rings of radius a,
infinitesimal width, da, and charge dq. Lets call the
area of the disk A and the area of the ring dA. If you
imagine the ring straightened out into a rectangle
you see that the area of the ring dA = 2πa da .
Since Q = σA, then dq = σda = 2πaσ da
a
R
Since we already know

kQx
the equation for the field E 
x2  a2
due to a ring of charge is:


3
2
We can start with:

kx2a
dE 
3 da
2
2 2
x a

Then we substitute for dq:


R
E   dE  kx 
0
x

2a
2
a
2

3
2

E
x
kxdq
2
a
2

3
2
We can now sum up (integrate)
the rings from radius 0 to R…

x
da  2k 1 
2
2
x

R




