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Chapter 2
Phys 2180
Electric Field
Two field forces have been introduced into our study so far—the gravitational force
Fg and the electric force Fe .
The gravitational field g at a point in space equal to the gravitational force Fg acting
on a test particle of mass m divided by that mass:
g =
Fg
m
.
The electric field vector E at a point in space is defined as the electric force Fe
acting on a positive test charge q0 placed at that point divided by the test charge:
E =
Fe
qo
The vector E has the SI units of newtons per coulomb (N/C).
Note that E is the field produced by some charge or charge distribution separate from
the test charge—it is not the field produced by the test charge itself. Also, note that the
existence of an electric field is a property of its source—the presence of the test charge is
not necessary for the field to exist. The test charge serves as a detector of the electric
field. The Equation electric field can be rearranged as
Fe = E q o
where we have used the general symbol q for a charge. This equation gives us the force
on a charged particle placed in an electric field. If q is positive, the force is in the same
direction as the field. If q is negative, the force and the field are in opposite directions ( as
shown in the following figure).
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Chapter 2
Phys 2180
Figure (1)
To determine the direction of an electric field, consider a point charge q as a source
charge. This charge creates an electric field at all points in space surrounding it. A test
charge q0 is placed at point P, a distance r from the source charge, as in Figure 1-a.
According to Coulomb’s law, the force exerted by q on the test charge is
q qo ∧
r
r2
Fe = K e
Q E=
Fe
qo
∴ E = Ke
q ∧
r
r2
where rˆ is a unit vector directed from q toward q0.
If the source charge q is positive, Figure 1-b shows the situation with the test
charge removed—the source charge sets up an electric field at point P, directed away
from q. If q is negative, as in Figure 1-c, the force on the test charge is toward the source
charge, so the electric field at P is directed toward the source charge, as in Figure 1-d.
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Note that at any point P, the total electric field due to a group of source charges equals
the vector sum of the electric fields of all the charges. Thus, the electric field at point P
due to a group of source charges can be expressed as the vector sum
∧
Where ri is the distance from the i th source charge qi to the point P and ri is a unit
vector directed from qi toward P.
Example 1 ( Electric Field Due to Two Charges)
A charge q1= 7.0 µC is located at the origin, and a second charge q2 = -5.0 µC is located
on the x axis, 0.30 m from the origin (Figure 2). Find the electric field at the point P,
which has coordinates (0, 0.40) m.
Figure (2)
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Phys 2180
Solution:
First, let us find the magnitude of the electric field at P due to each charge. The fields E1
due to the 7.0 µC charge and E2 due to the - 5.0 µC charge are shown in Figure 2. Their
magnitudes are
The vector E1 has only a y component. The vector E2 has an x component given by E2
Cos θ =
3
4
E2 and a negative y component given by E2 Sin θ =
E2. Hence, we can
5
5
express the vectors as
The resultant field E at P is the superposition of E1 and E2:
From this result, we find that E makes an angle φ of 66° with the positive x axis and has
a magnitude of 2.7 x 105 N/C
Example 2 ( Electric Field of a Dipole)
An electric dipole is defined as a positive charge + q and a negative charge - q separated
by a distance 2a. For the dipole shown in Figure 3, find the electric field E at P due to the
dipole, where P is a distance y >> a from the origin.
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Phys 2180
Solution:
At P, the fields E1 and E2 due to the two charges are equal in magnitude because P is
equidistant from the charges. The total field is E = E1 + E2 , where
Figure (3)
The y components of E1 and E2 cancel each other, and the x components are both in the
positive x direction and have the same magnitude. Therefore, E is parallel to the x axis
and has a magnitude equal to 2 E1 Cos θ. From Figure 3 we see that
Cos θ =
a
=
r
a
(y
Therefore
5
2
+ a2
)
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Chapter 2
Phys 2180
Because y >>a , we can neglect a2 compared to y2 and write
Electric Field Lines
Representative electric field lines for the field due to a single positive point charge are
shown in Figure 8 a. The electric field lines representing the field due to a single negative
point charge are directed toward the charge (Fig.8 b). In either case, the lines are along
the radial direction and extend all the way to infinity. Note that the lines become closer
together as they approach the charge; this indicates that the strength of the field increases
as we move toward the source charge.
Figure (8)
The rules for drawing electric field lines are as follows:
•
The lines must begin on a positive charge and terminate on a negative charge. In
the case of an excess of one type of charge, some lines will begin or end infinitely
far away.
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Chapter 2
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Phys 2180
The number of lines drawn leaving a positive charge or approaching a negative
charge is proportional to the magnitude of the charge.
•
No two field lines can cross or intersect with one another.
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Problems
(1) In the figure bellow, determine the point (other than infinity) at which the electric
field is zero.
Solution:
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Phys 2180
(2) Three point charges are arranged as shown in the figure bellow:
(a) Find the vector electric field that the 6.00-nC and "3.00-nC charges together
create at the origin.
(b) Find the vector force on the 5.00-nC charge.
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Solution:
(3) Four identical point charges (q = +10.0 µC) are located on the corners of a
rectangle as shown in the figure bellow. The dimensions of the rectangle are L=
60.0 cm and W = 15.0 cm. Calculate the magnitude and direction of the resultant
electric force exerted on the charge at the lower left corner by the other three
charges.
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Solution:
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