1. Electrostatics
Electric Fields

2. Electric Field Strength
Earth
mass +
proton
Gravity
Gravitational field found using a test mass
FG = GmEm/r²
mg = GmEm/r²
g = GmE/r²
Measured in N/kg
Electrostatics
Electric field found using a test proton
FE = KQq/r²
Eq = KQq/r²
E = KQ/r²
Measured in N/C

3. Electric Field Strength+
Electric field is always found using a test proton
Vector
Measured in Newtons per Coulombs
proton

4. 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
3mC
Positive means that test charge will experience repulsion +
1.5m

5. Electric Field Strength Example 2
What is the magnitude and direction of the electric field that the proton experiences?
Strategy to Solve:
Find Distances (radii) & Angles
Find Electric Fields (E1 & E2)
Vector Addition + 1m 4C
-3C 1m 2m

6. 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
proton
Test particle is always a proton

7. Drawing Electric Fields
Let’s test some harder situations: - + + + - -

8. Electric Field Strength
2 small point charges separated by a large distance
Now, move the particles further apart:
What happens to the electric force?
FE = kQ1Q2/r²
What happens to the electric field?
E = kQ/r²
Decreases
Non-Uniform Electric Field
Decreases + rf + ri

9. Electric Field Strength
2 large charged plates separated by a small distance
Let’s use a test particle + + ri

10. 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
FNET1 = FNET2 = FNET3
FNET1
FEP2
FEP3
rP3
rN3 + 3
rP1
rN1 + 1
rP2
rN2 + 2
FEP1
FEN1
FEN2
FEN3

11. Electric Field Strength
2 large charged plates separated by a small distance
Now, move the plates further apart:
What happens to the electric force?
What happens to the electric field?
DO NOT USE FE and E EQUATIONS!
Stays The Same
Stays The Same
Uniform Electric Field rf ri

12. 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 + A
E = 600N/C B
mP =
1.67x10-27kg
0.02m

13. 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
0.02m + A B
QP =
+1.6x10-19C
mP =
1.67x10-27kg
E = 600N/C

14. 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
0.02m + B A
QP =
+1.6x10-19C
mP =
1.67x10-27kg
E = 600N/C