### Chapter 32: Electrostatics

```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
– q=Fg/DV
(qDV/d=Fg, solve for q)
• 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
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
```