Electrostatics
Electric Fields
Electric Field Strength
Earth
mass
Gravity
Gravitational field found
using a test mass
FG = GmEm/r²
mg = GmEm/r²
g = GmE/r²
Measured in N/kg
+
proton
Electrostatics
Electric field found
using a test proton
FE = KQq/r²
Eq = KQq/r²
E = KQ/r²
Measured in N/C
Electric Field Strength
Electric field is always
found using a test proton
kQ
E 2
r
+
FE
E
q
Vector
Measured in Newtons per Coulombs


proton
Electric Field Strength Example
What is the magnitude and direction of the
electric field at the proton?
E = KQ/r²
E = (9x109Nm²/C²)(0.003C)/(1.5m)²
E = +1.2x107N/C
Positive means that
test charge will
experience repulsion
3mC
+
1.5m
Electric Field Strength Example 2
What is the magnitude and direction of the electric
field that the proton experiences?
Strategy to Solve:
1. Find Distances (radii) & Angles
2. Find Electric Fields (E1 & E2)
3. Vector Addition
+
1m
4C
-3C
1m
2m
Drawing Electric Fields
A field is a collection of vectors showing an
overall force at any point in space
If you were to test the electric field with a
proton in many places using the previous
examples you would get a collection of vectors.
-
+
proton
proton
Test particle is
always a proton
proton
Drawing Electric Fields
Let’s test some harder situations:
-
+
-
+
-
+
Electric Field Strength
2 small point charges separated by a large
distance
Now, move the particles further apart:
What happens to the electric force?
Decreases
FE = kQ1Q2/r²
What happens to the electric field?
Non-Uniform
E = kQ/r²
Decreases
Electric Field
+
+
+
rfi
+
Electric Field Strength
2 large charged plates separated by a
small distance
Let’s use a test particle
+
+
+ +
+
+
-
ri
Electric Field Between Parallel Plates
Let’s look at a test particle:
At Position 1:
 Whole lot of repulsion force from positive plate
 Little bit of attraction force from negative plate
At Position 2:
 Average repulsion force from positive plate
 Average attraction force from negative plate
At Position 3:
 Little bit of repulsion force from positive plate
 Whole lot of attraction force from negative plate
Electric
Force Is
Uniform
Everywhere
FNET1 = FNET2 = FNET3
FEP1
+
+
+
+
FEN3
-
1 FEP2 2 FEP3 3
+
+
+
FEN1
FEN2
rP1rP2rP3
rN1
rN2
rN3
Electric Field Strength
2 large charged plates separated by a
small distance
Stays
Now, move the plates further apart:
The
Same
What happens to the electric force?
Stays
What happens to the electric field?
The
DO NOT USE FE and E EQUATIONS! Same
Uniform
Electric Field
+
+
+
+
+
+
+
+
rf ri
-
Capacitor Electric Field Example
Two parallel capacitor plates are 0.02m apart with
an electric field of 600N/C between them. A
proton is released from rest at point A, what
force does the proton experience? What is its
kinetic energy when it gets to point B?
QP =
+1.6x10-19C
mP = 1.67x10-27kg
+
A
+
+ +
+
E = 600N/C
0.02m
B -
Capacitor Electric Field Example
What force does the proton experience?
F = E * QP
F = (600N/C)(1.6E-19C)
F = 9.6x10-17N
Find acceleration:
a=F/m
a = (9.6x10-17N) / (1.67x10-27kg)
a = 5.75x1010m/s²
+
A
+
+ +
+
E = 600N/C
QP = +1.6x10-19C B mP = 1.67x10-27kg
-
0.02m
Capacitor Electric Field Example
What is its kinetic energy when it gets to point B?
Find final velocity:
vf² = vi² + 2ad
vf² = (0m/s)² + 2(5.75x1010m/s²)(0.02m)
vf = 48000m/s
Find kinetic energy:
EK = ½mv²
EK = ½(1.67x10-27kg)(48000m/s)²
EK = 1.92x10-18J
+
E = 600N/C
B
+
-19C
+1.6x10
Q
=
+ P
+ A
mP = 1.67x10-27kg
+
-
0.02m