Chapter 21: Electric Fields

Honors Physics

Bloom High School

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)

Relative Field Strength

Field 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

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

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 (-)?

Example Problem 2

• 1. Known/Unknown– 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

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

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?

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

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

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

Practice Problem 16

• 1. Known/Unknown– E=6000N/C, d=0.05m, DV=?

• 2. Formula– DV=Ed

• 3. Solve– DV=(6000N/C)(0.05m)=300V

Milikan’s Oil Drop Experiment

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-

How many electrons?

• 1. Known/Unknown– 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-

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

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– http://www.youtube.com/watch?v=7qgM1A3pgkQ

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 10-12 to 10-6 F

• Physics Physlets I.26.1

Practice Problem 31

• 1. Known/Unknown– 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