Download Chapter 23: Electric Fields

Chapter 23: Electric Fields
Previously (Ch. 22) we have seen that two electric charges will either attract or repel each other.
How is it that one electric charge even knows about the existence of the other charge??
Based on an idea of Michael Faraday’s we say that one of the charges sets up an electric field
throughout space and the other charge reacts to this field.
The electric field (E) due to a charged object at a location in space is defined to be the
force (F) per unit charge (q0) as felt by a test charge placed at that location in space.
F ( x, y , z )
E ( x, y , z ) =
q0
Important Points about the above definition of the electric field:
1) The electric field is a vector (just like force). It has magnitude (units=N/C) and direction.
2) The test charge, q0, in the definition is assumed to be positive. It is also assumed to be so
small that it does not change the distribution of charge that is causing the electric field.
3) Since the test charge, q0, is positive, by DEFINITION the electric field
is in the same in the same direction as the force.
positive charge: electric field points away from the charge.
negative charge: electric field points towards the charge.
Direction of E for
a negative charge.
HRW Fig 23-2
R. Kass
P132 Sp04 Ch23
1
Electric Field of a point Charge
Recall Coulomb’s Law for a point charge and our test charge, q0.
The magnitude of force (F) on q0 due to q is:
qq0
F =k 2
r
k≡
1
4πε 0
= 8.99 × 109 Nm 2 / C 2
Therefore by definition the magnitude of the electric field (E) due to q is:
q
F
E=
=k 2
r
q0
What about the direction of E? There are two cases, positive (+) and negative (-) q.
+
example: what is the magnitude of E
r
q0 (+)
1 m away from a charge of 1C?
E
q (8.99 × 109 Nm 2 / C 2 )(1C )
9
-
E=k
r
E
R. Kass
q0 (+)
r2
=
(1m) 2
= 8.99 × 10 N / C
This is a very large electric field!
~103 times large than a small spark in air
P132 Sp04 Ch23
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Visualizing the Electric Field
One way to visualize the electric field is the use the concept of electric field lines.
1) The direction of the electric field is given by the direction of a straight field line or the
E
tangent to a curved field line.
Field line
E is tangent to the line
E
2) The magnitude of the electric field is related to the number of field lines per unit area in a
plane perpendicular to the lines. In other words, the electric field is large when the field
lines are close together and weaker when the lines are further apart.
Region of low electric field.
Lines are far apart.
Electric field lines for
a positive (+) charge
Region of high electric field
Lines are close together
Both boxes have the same area.
The box with more lines in it
has the larger electric field.
+q
For more examples of electric field
lines and charges see Figs.23-3,4,5
of your text.
Question: If in the above diagram we had a negative charge what would the field lines look like?
Answer: The arrows on the field lines would all point inward, otherwise the same.
R. Kass
P132 Sp04 Ch23
3
Electric field due to a dipole
So far we have discussed the electric field due to a single charge. What about 2 charges?
A very important charge configuration is two opposite charges separated by a distance d.
“ELECTRIC DIPOLE”
For example many molecules (e.g. water) have this charge configuration.
We can calculate the electric field using the principle of superposition:
E = E+ + E−
If we consider the electric field at a point along the z axis (x=y=0, +z) from
two equal but opposite charges (q) we find for the magnitude of the electric field, E:
z
E=k
+
x
d
-
E=
We are interested in the case where we are far from the charges (z>>d).
So, we re-write E in the following form:
kq
kq
−
=
2
2
( z − d / 2)
( z + d / 2)
E=
R. Kass
q
q
−
k
( z − d / 2) 2
( z + d / 2) 2
kq
z 2 (1 −
d 2
)
2z
−
kq
z 2 (1 +
d 2
)
2z
kq
d −2
d −2
[(
1
−
)
−
(
1
+
) ]
2
2z
2z
z
P132 Sp04 Ch23
4
Electric field due to a dipole
If z>>d then we can use the approximation: (1+α)-2≈ 1+2α with α= -d/(2z) or +d/(2z)
Actually, this is just the binomial theorem (page A10 in your book) where we only keep the first term (other than “1”)
in the expansion (1+x)n≈1+nx.
How well does this approximation work? Assume z=1m and d=1mm so d/(2z) = 5x10-4.
exact:
approximate:
1
1
=
= 1.001001
d 2 (1 − 5 × 10 − 4 ) 2
(1 − )
2z
1
≈ 1 + 2 × (5 × 10 − 4 ) = 1.001000
d
(1 − ) 2
2z
Pretty good appoximation!
difference is 1 part in 106.
Continuing on, we find the electric field of a dipole to be:
kq
d −2
d − 2 kq
d
d
2kqd
−
−
+
≈
+
−
−
=
[(
1
)
(
1
)
]
[(
1
2
)
(
1
2
)]
z2
2z
2z
z2
2z
2z
z3
p
E=
with: p≡qd=“electric dipole moment”
2πε 0 z 3
+q
E=
The electric dipole moment is a vector, its direction
is from the negative to the positive charge.
Note: The dipole moment of water is 6.2x10-30C-m
p = qd ,
d
| p |= qd
+q
-q
There lots of other “moments” possible. e.g. 4 charges give a quadrupole moment!
R. Kass
P132 Sp04 Ch23
-q
-q
+q
5
How do we calculate the electric field for many charges?
In principle we can always add the contributions together. For example, 10,000 charges:
E = E1 + E 2 + + + E100 + + + + + + E1000 + + + + + + + + E10000
Clearly, this will become very tedious. Especially since the number of charges can be
extremely large (e.g. 1010 or more!).
Mathematically, we move from summation to integration:
n
E = ∑ E i ⇒ E = ∫ dE
i =1
Note: this is actually 3 equations,
one for Ex, Ey, and Ez!
Since we are interested in situations where there are huge numbers of charges it is convenient
to use charge densities rather than individual charges.
There are three common types of charge densities:
I) Linear Charge Density (λ)
Imagine we take a rod of length Land spread a total charge Q all over the rod λ=Q/L
The amount of charge (dQ) in a small length (dL) of the rod is: dQ= λdL
II) Surface Charge Density (σ)
Here we distribute the charge (Q) over the area (A) of a surface, σ=Q/A. For example if we take a sphere
and uniformly spread the charge (Q) over the surface of the sphere (A=4πR2) then the
surface charge density is: σ=Q/4πR2.
III) Volume Charge Density (ρ)
In this case charge is distributed throughout the volume (V) of some object, ρ=Q/V. So for example, if we
spread the charge (Q) uniformly throughout the volume of a sphere (V=(4/3)πR3) and ρ is:
ρ=
R. Kass
Q
4 / 3πR 3
P132 Sp04 Ch23
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Calculation of the Electric Field from a line of charge
As a 1st example of the calculation of the electric field using integration consider a line of charge.
Assume that we have a total charge Q uniformly distributed along the x-axis.
The charge starts at x=0 and ends at x=d. We are assuming that the line is very thin,
it only extends along the x-axis and has negligible width along the y-axis (and z-axis).
We want to calculate the electric field at a point (x0) on the x-axis.
First, remember that the electric field is a vector so we need to worry about Ex, Ey, Ez!
For this problem, since the charge is along the x-axis and the point we are interested in is on the
x-axis too, the electric field at x0 will only have a component along the x-axis (Ey=Ez=0).
line of charge with linear charge density λ
y
dx
x
x
d
E
x0-x
x0
The amount of charge (dq) in an infinitesimal (dx) piece of this line centered at x is: dq=λdx.
The electric field (dEx) due to this charge (dq=λdx)is:
dE x =
kdq
kλdx
=
( x0 − x) 2 ( x0 − x) 2
To get the total contribution for the entire line we must integrate from x=0 to x=d!
R. Kass
P132 Sp04 Ch23
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Calculation of the Electric Field from a line of charge
line of charge with linear charge density λ
y
E
dx
x
x
d
x0
x0-x
To get the total contribution for the entire line we must integrate from x=0 to x=d:
Ex = ∫
d
0
Ex =
d
kλdx
kλ
=
( x0 − x) 2 ( x0 − x) 0
A wire with charge on
it is an example of a “line
of charge”.
kλ
kλ
kλd
−
=
( x0 − d ) x0 x0 ( x0 − d )
We can simplify this problem by remembering that λ=Q/d or Q=λd.
Ex =
kλd
kQ
=
x0 ( x0 − d ) x0 ( x0 − d )
Recall: Ey=Ez=0, so E=Ex
for this example
What happens if we are very far away (x0>>d) from the line of charge?
kQ
kQ
kQ
≈
= 2
Ex =
Far from line charge get same E field as a point charge!
x0 ( x0 − d ) x0 ( x0 ) x0
R. Kass
P132 Sp04 Ch23
8
Electric Field from a Circular Line of Charge
Suppose we deposit electric charge, +Q, uniformly along the arc of a semi-circle with radius r.
(x,y)
If we move an angle θ along the circle then the arc length (s) is: s=θr
The linear charge density is: λ=Q/(arc length of semi-circle)
λ=Q/(πr)
r
What is the electric field at the center (x=0,y=0) of the semi-circle?
θ
Let’s write down the electric field (Ex, Ey) at (0,0) due to
(0,0)
a small piece of the arc, ds, located at (x,y) which contains
E=?
y
a small amount of charge, dq (=λds):
Let’s just calculate the magnitude of Ex and Ey and figure
out their directions later.
kdq
kλds
x
dE x = dE cos θ = 2 cos θ = 2 cos θ
r
r
kdq
kλds
dE y = dE sin θ = 2 sin θ = 2 sin θ
r
r
To calculate Ex and Ey we must add up all the charge contributions by integrating!
recall: ds=rdθ.
π /2
π /2
π /2
π /2
kλds
kλ
kλ
kλ
2kλ
λ
Ex = ∫
rd
cos
θ
θ
cos
θ
d
θ
cos
θ
sin
θ
=
=
=
=
=
2
2
∫
∫
r
r
r
2πε 0 r
r
r
−π / 2
−π / 2
−π / 2
−π / 2
Ex is in the negative x direction since we have a positive charge density. Finally we use λ=Q/(πr):
Ex =
R. Kass
−Q
2π 2 ε 0 r 2
To see that Ex is in the negative x direction draw some electric field lines
from the arc to (0,0). The x component of all these field lines point in the
negative x direction.
P132 Sp04 Ch23
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Electric Field from a Circular Line of Charge continued
Now let’s calculate Ey.
kdq
kλds
sin
θ
=
sin θ
2
2
r
r
π /2
π /2
π /2
kλds
kλ
kλ
− kλ
Ey = ∫
rd
d
sin
θ
=
θ
sin
θ
=
θ
sin
θ
=
cos θ
2
2
∫
∫
r
r
r
r
−π / 2
−π / 2
−π / 2
dE y = dE sin θ =
(x,y)
π /2
=0
−π / 2
E(x,y)
θ
Ey = 0
−θ
y
In fact, Ey=0 for any symmetric angle interval as long as
we are on the x axis. For this problem the angle interval
was ±π/2. However if the interval was ±θ then the integral
would still give zero since: [cosθ-cos(-θ)]=0 since cosθ=cos(-θ).
E(x,-y)
x
(x,-y)
We can also see that Ey is zero by drawing pairs of electric field lines (see above drawing)
and seeing that y components always cancel. This is an example where symmetry tells us the
answer without resorting to integrations.
Finally, we can solve one more problem using the above as a guideline:
What is the electric field at the center of a circle of linear charge density λ?
Answer: Ex=Ey=0!
You can get this result by integration (over ±π) or using symmetry arguments.
R. Kass
P132 Sp04 Ch23
E=?
10