1. Chapter 21: Electric Fields
Honors Physics
Bloom High School

2. 21.1 Creating and Measuring Electric Fields
Electric Field- comparison to a gravitational field
Exists around any charged object (objects with mass)
Similar- acts at a distance
Dissimilar- can be positive OR negative
Test Charge- used to determine the strength of a field and/or the direction of the field
Test charge is always positive
Field Strength- equal to the force on the (+) test charge divided by the strength of the test charge
E=F/q (N/C)

3. Relative Field Strength
Value (N/C)
Near a charged, hard-rubber rod
1x103
In a TV picture tube
1x105
Needed to create a spark in air
3x106
At an electron’s orbit in H
5x1011

4. Practice Problem 1: Solving for Field Strength
1. Known/Unknown
q=5.0x10-6C, F=2.0x10-4N, E=?
2. Formula
E=F/q
3. Solve
E=(2.0x10-4N)/(5.0x10-6C)
4.0x101N/C or 40N/C

5. Practice Problem 4: Gravity vs. Electricity
1. Known/Unknown
Fg=2.1x10-3N, E=6.5x104N/C (down), q=?
2. Formulae
E=F/q (q=F/E)
3. Solve
q=(2.1x10-3N)/(6.5x104N/C)=3.2x10-8C
Should the charge be (+) or (-)?

6. Example Problem 2 1. Known/Unknown 2. Formulae 3. Solve
d=0.3m, q=-4.0x10-6C, E=?
2. Formulae
E=F/q1, F=kq1q2/d2
Solve both for q1 and set equal to each other
Solve new equation for E
E=kq2/d2
3. Solve
E=(9.0x109Nm2/C2)(-4.0x10-6C)/(0.3m)2=4.0x105N/C

7. Picturing the Electric Field
Electric field is a vector quantity
Magnitude and direction matter
Arrows extend from positive charges and toward negative charges
Lines closer together represent a stronger field
Physics Physlets I.23.2, I.23.3, P.23.2

8. Section 21.1 Quiz
An electric charge, q, produces an electric field. A test charge, q, is used to measure the strength of the field generated by q. Why must q be relatively small?
Define each variable in the formula E=F/q.
Describe how electric field lines are drawn around a freestanding positive charge and a freestanding negative charge.
A charge of +1.5x108C experiences a force of 0.025N to the left in an electric field. What are the magnitude and direction of the field?
A charge of +3.4x106C is in an electric field with a strength of 5.1x105N/C. What is the force it experiences?

9. 21.2 Applications of Electric Fields
Just as g=F/m describes the field strength per mass of gravity, E=F/q describes the field strength per charge
Changing the distance of either is work! (W=Fd)
Performing work on an object gives it DPE
Electric potential energy- DV=W/q (V=J/C)
See Figure 21-5 (page 569)
Physics Physlets I.25.2
Equipotential- when DV is zero
Moving a (+) charge around a (-) charge, keeping d constant

10. Grounding Charges will move until the electric PE is zero
No DV between the conductors
Grounding- makes the electric PE between an object and the Earth 0V
Can prevent sparks resulting form a neutral object making contact with a charged object

11. DV in a Uniform Field
By moving a charge between parallel plates, only the distance change in the field matters
Because W=Fd and DV=Fd/q
DV=Ed
The potential difference is equal to the field strength multiplied by the distance the charge is moved

12. Practice Problem 16 1. Known/Unknown 2. Formula 3. Solve
E=6000N/C, d=0.05m, DV=?
2. Formula
DV=Ed
3. Solve
DV=(6000N/C)(0.05m)=300V

13. Milikan’s Oil Drop Experiment

14. Oil Drop Rationale
If a known electric field is applied to the plates (F=E/q) and the mass is found of each droplet (F=mg), the charge can be found for a single droplet!
mg=Eq  q=mg/E
The charges were found to always be multiples of 1.60x10-19C, which we now know is the charge of a e-

15. How many electrons? 1. Known/Unknown 2. Formulae 3. Solve
Fg=2.4x10-14N, DV=450V, d=1.2cm, q=?, ne-=?
2. Formulae
Fe=Fq (qDV/d=Fg, solve for q)
q=Fg/DV
3. Solve
q=(2.4x10-14N)/(450V)=6.4x10-19C
q/1.60x10-19C=4e-

16. Sharing Charges All systems desire equilibrium
Charges we distribute themselves evenly across any available surface
When 2 spheres touch, they act as a single object
Charge to area ratio is what counts
Charge density the greatest near points

17. The Van de Graff Generator
Van de Graff Generator- high voltages are built up on a surface
Charges are distributed evenly on surface
Very large charge is possible (MV range!)
Ours builds to 750,000V
MythBusters: Van de Graff Generator

18. Storing Charges: The Capacitor
Capacitor- stores electrical charge
Used in all circuitry
Storage based on voltage, size of plates and gap between plates
Capacitance (C)- ratio of charge stored to electric potential difference
C=q/DV
Measured in Farads (F=C/V)
Typically to 10-6 F
Physics Physlets I.26.1

19. Practice Problem 31 1. Known/Unknown 2. Formula 3. Solve
C1=3.3mF, C2=6.8mF, DV=24V, q1=?, q2=?
2. Formula
C=q/DV  q=DVC
3. Solve
q1=(24V)(3.3x10-6F)=7.92x10-5C
q2=(24V)(6.8x10-6F)=1.63x10-4C