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The Electric Field
26.1
I)
Electric Field Models
Review
A) Fields point away from positive, toward negative source charges
B) Field strength is proportional to source charge, inversely proportional to distance from
source charge
C) Net electric field is the vector sum of the fields due to each charge (superposition)
1
𝑞
II) 𝐸⃗ = 4𝜋𝜖 𝑟2 𝑟̂
0
26.2
I)
The Electric Field of Multiple Point Charges
Superposition
A) (𝐸𝑛𝑒𝑡 )𝑥 = (𝐸1 )𝑥 + (𝐸2 )𝑥 + ⋯ = ∑(𝐸𝑖 )𝑥
B) (𝐸𝑛𝑒𝑡 )𝑦 = (𝐸1 )𝑦 + (𝐸2 )𝑦 + ⋯ = ∑(𝐸𝑖 )𝑦
C) (𝐸𝑛𝑒𝑡 )𝑧 = (𝐸1 )𝑧 + (𝐸2 )𝑧 + ⋯ = ∑(𝐸𝑖 )𝑧
D) Problem Solving Strategy for Electric Field due to multiple point charge
II) The electric Field of a Dipole
A) Derivation of Equation
2𝑝
B) 𝐸⃗𝑑𝑖𝑝𝑜𝑙𝑒 = 𝑘 𝑟3 , where 𝑝 = (𝑞𝑠, from the negative to the positive charge) along the dipole
axis
𝑝
C) 𝐸⃗𝑑𝑖𝑝𝑜𝑙𝑒 = −𝑘 𝑟 3 along the bisecting plane
III) Picturing the Electric Field
A) Drawing the Field
26.3
I)
The Electric Field of a Continuous Charge Distribution
Charge Density
A) Linear
𝜆=
𝑄
𝐿
B) Surface
𝜂=
𝑄
𝐴
C) Example 26.3
II) Infinite Line of Charge
A) 𝐸rod = 𝑘
2|𝜆|
1
𝑟 √1+4𝑟 2 ⁄𝐿2
B) But if 𝐿 → ∞ then 𝐸line = 𝑘
26.4
I)
2|𝜆|
𝑟
The Electric Field of Rings, Disks, Planes and Spheres
Example 26.4 derives (𝐸𝑟𝑖𝑛𝑔 )
𝑧
A) Different than the field generated by a line of charge or rod, all the segments of charge are
at the same position relative to the point in question along the z-axis
B) Consequently, no summation of distances is needed, nor summation of segments of charge
II) A Disk of Charge
A) 𝜂 =
𝑄
𝐴
=
𝑄
𝜋𝑅2
B) Method includes using segments of charge in the shape of rings, since the 𝐸𝑟𝑖𝑛𝑔 is already
derived
C) A bit of calculus…
D) (𝐸disk )𝑧 =
𝜂
2𝜖0
[1 −
𝑧
√𝑧 2 +𝑅2
]
III) A Plane of Charge
A) Plane could be a disk of infinite radius
B) if 𝑅 → ∞ in the equation (𝐸disk )𝑧 =
𝜂
2𝜖0
[1 −
𝑧
√𝑧 2 +𝑅2
] then 𝐸plane =
𝜂
2𝜖0
IV) A Sphere of Charge
A) If 𝑟 is the radius of the sphere and 𝑅 is the distance from the center of the sphere
𝑄
B) 𝐸⃗sphere =
2 𝑟̂
4𝜋𝜖0 𝑟
26.5
I)
The Parallel-Plate Capacitor
Two parallel plates, essential two planes of charge, comprise a parallel plate capacitor
𝜂
𝐸⃗capacitor = 𝐸⃗+ + 𝐸⃗− = ( , from positive to negative)
𝜖0
𝑄
𝐸⃗capacitor = 𝐸⃗+ + 𝐸⃗− = (
, from positive to negative)
𝜖0 𝐴
II) Uniform electric fields
26.6 Motion of a Charged Particle in an Electric Field
I) Recall that electric fields exert forces onto charges within the electric field
A) 𝐹on 𝑞 = 𝑞𝐸⃗
B) 𝑎 =
𝐹on 𝑞
𝑚
=
𝑞
𝐸⃗
𝑚
II) Motion in a Uniform Field
A) A charged particle in a uniform electric field will move with constant acceleration
B) 𝑎 =
26.7
I)
𝐹on 𝑞
𝑚
=
𝑞
𝐸⃗
𝑚
Motion of a Dipole in an Electric Field
Factors influencing polarization force
A) The charge polarizes a neutral object, creating the dipole
B) The charge then exerts an attractive force on one end, that is larger than the repulsive force
on the other end
II) Dipoles in a Uniform Field
A) 𝐹net = 𝐹+ + 𝐹− = 0, but the forces do create unbalanced torques
1
2
B) 𝜏 = 2 × 𝑑𝐹+ = 2( 𝑠 sin 𝜃)(𝑞𝐸)
C) 𝜏 = 𝑝 × 𝐸⃗
III) Dipoles in a Nonuniform Field
A) Net force on a dipole is toward the strongest field
B) A dipole will experience a net force with any charged object