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Electric Fields
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1
Fields
How does one charged particle attract or repel another particle at a distance? ⇒ Action at a distance
This is more prevalent than you think. Consider how you touch something and cause it to move by applying a force. In fact, you never actually “touch” the object. The electrons in the atoms of your hand start repelling the electrons in the atoms of the object. The electrons never really touch! (they are point particles anyway…) So touching something is Coulomb repulsion at a distance.
In physics, we say that the force exerted by one object onto another a distance away is conveyed through a field.
Electric Field
The electric field is defined as the force acting on a positive test charge, per unit charge.
0 0
q 0 points in direction of q
F
E E F
Units are thus N/C for the electric field.
It is similar to the gravitational field on the surface of the Earth for a test mass m 0 :
m 0
F
g
The electric field is a vector field. It has a magnitude and a direction. Another example vector field is the wind distribution. Temperature distribution is an example of a scalar field.
Field Lines
Consider a positive test charge immersed in the electric field of a positive charge.
q 0
- Lines indicate the direction of F if a test charge is placed there
- Field lines point away from positive charge, and toward negative charge
- Density of lines indicates the strength of the field
- Field lines cannot cross. If they did, there would be more than one possible direction for a particle to travel at a given point.
- The lines themselves have no meaning. The field is present everywhere.
The magnitude of the electric field a distance r away from a point charge q :
2 0
q K q r
F
E i.e. dropped the q 0 from Coulomb’s Law
Electric Field of Several Point Charges
Apply the superposition principle. This principle states that the resulting electric field is the sum of all fields, without any interference of one field upon another. It is generally true for electromagnetism at least for fields that are not enormously strong.
For example, the total electric field at some point A from N charges is:
1 1 2 1
N N (^) i A (^) i iA i A
q K = = r E = (^) ∑ E =∑ r iA
In other words, the fields from each point particle i a distance riA away from point A all add together, without any field affecting another.
TOT (^2 )
TOT (^22) 3/ 2 3 2 3/ 2
q d K x d (^) x d
Kqd Kqd
x d x d x
⎡ ⎤^ ⎡^ ⎤
= − ⎢ ⎥ ⎢^ ⎥
⎢ + ⎥ ⎢^ + ⎥
E y
E
By the binomial expansion, ( 1 ) 1 " for small x.
n
2 3/ 2
TOT (^3)
1 / 2 1 1 for 2 2
d d x x d x qd K x
− ⇒ ⎡^ + ⎤ − ⎣ ⎦
⇒ E
Let’s define the electric dipole moment p ≡ qd
TOT 3
p K r
⇒ E
where we have generalized from anywhere along the x axis to anywhere in the x-z plane, a distance r away from the dipole.
Note that this field falls off faster than a single point charge, r −^2 , because the opposite- sign charges contribute to the canceling of the electric field. This is a general behavior for the dipole field strength in any direction. For example, along the y axis, one gets a dipole field 2 times larger than the expression above (but all other terms the same).
Continuous Charge Distributions
Consider the electric field arising from an infinitesimal charge dq at a point a distance r away:
2 ˆ
dq d K r
E = r y r d E dq x
Projecting along each axis yields the following components of the infinitesimal electric field: ˆ ˆ
x y
dE d dE d
E x E y
Field from a Continuous Line Charge
Now consider electric charge distributed uniformly along a 1-dimensional line from − L to L along the z -axis.
z
+L
dq
r
y
dE
dE’
dq'
x
The charge per unit length is λ (units: C/m)
In terms of the total charge q , 2
q L
So an infinitesimal length of the line, dz , has a charge dq = λ dz.
Let’s calculate the electric field along the y -axis (any axis perpendicular to the line will do).
By symmetry, only the y ˆ component of the field (that is, a field pointing perpendicular to
the line) can survive when we integrate the field contributions from each infinitesimal charge dq along the length.
2
2 2 2
2 2
cos
where
cos
y y
dq dE K r
r y z y y z
E E ∫ ∫
Important to remember to project the field along the appropriate axis!
An infinitesimal length along the ring, ds , has a charge dq = λ ds.
It generates an electric field of:
2 ˆ^2 ˆ
dq ds d K K r r
E = r = r
Only the z-component matters, so:
( )
( )
2
2 2 2 2
(^2 2 2 )
2 2
cos
where and cos
z
z
ds dE K r z r R z R z
ds z dE K R z (^) R z K z ds R z
∫ ∫
∫
E
E
The integration is a line integral of the function around the ring. ds represents an infinitesimal length around this path.
since R and z are held constant when integrating around the ring.
But (^) ∫ ds = 2 π R , the circumference of the circle. We can see this explicitly by switching
to polar coordinates:
2 0 ds Rd 2 R
π ∫ =^ ∫ θ^ = π
So: ( ) 2 2 3/ 2
2 RK z R z
E
Now, 2 π R λ = q , the total charge of the ring. So we can re-write the field arising from a
ring of charge as:
( ) 2 2 3/ 2
qz K R z
E
If (^2)
q z R K z
⇒ E → , the field of a point charge as expected when the ring
becomes very small when observed far away.
Field of a Charged Disk
Fig. from HRW 7/e
Let’s find the field along the z -axis again. Disk has radius R , and a charge per unit area of σ, such that an infinitesimal area has a charge of dq = σ dA =σ dxdy
We can solve for the electric field by first determining the electric field arising from a ring of infinitesimal thickness dr and radius r. (Note that this is a different definition of “ r ” from the previous example).
In fact, we solved for the field of a charged ring already in the previous example! Just make the following substitutions:
R r q dq σ dA σ π rdr
So the electric field of this thin ring points in the z -direction and has a magnitude of:
( ) ( ) 2 2 3/ 2^2 2 3/ 2
z dq z 2 rdr d K r z r z
E
The total field arising from the entire disk is then:
( ) 0 2 2 3/ 2
R (^) 2 rdr dE K z r z
E ∫ ∫
Let’s change variables:
