Download Example 16-7 Field of an Electric Dipole

Example 16-7 Field of an Electric Dipole
A combination of two point charges of the same magnitude but opposite signs is called
an electric dipole. Figure 16-13 shows an electric dipole made up of a point charge +q
and a point charge 2q separated by a distance 2d. Derive expressions for the magnitude and direction of the net electric field due to these two charges at a point P a
distance y along the midline of the dipole.
E+
P
E–
r
Figure 16-13 ​An electric dipole The field produced by an electric dipole at any point is the
–q
+q
vector sum of the fields Es + and Es - caused by the positive charge +q and the negative charge 2q,
respectively.
Set Up
The net field is the vector sum of the field
s + due to the charge +q (which points
E
s - due to the
away from +q) and the field E
charge 2q (which points toward 2q). Note
that this problem is very similar to Example
16-4 in Section 16-4, in which we used vector
addition to find the net electric force exerted
by two charges (one positive and one negative)
on a third charge; here we use vector addition
to find the net electric field due to the positive
and negative charge. As in Example 16-4,
we’ll choose the positive x direction to be to
the right and the positive y axis to be upward
and add the two vectors using components.
Solve
Use Equation 16-4 to find the magnitudes of
s + and E
s - at P.
the fields E
2d
Magnitude of the electric field due
to a point charge Q:
E =
k Q 
r
2
y
y
E+
(16-4)
E–
Total electric field:
s = E
s1 + E
s 2
E
(16-5)
+q
O
O
–q
O
x
The distance from +q to point P is the same as the distance from 2q
to P. Call this distance r:
r = 2y 2 + d 2
Since +q and 2q have the same magnitude (q, which is positive) and are
the same distance from P, Equation 16-4 tells us that the fields that the
two charges produce at P have the same magnitude:
E + = E- =
kq
r
2
=
kq
2
y + d2
s + and E
s-,
Find the x and y components of E
and use these to calculate the components of
the net field at P.
s + and E
s - are
The components of E
E+, x = E+ cos u
E+, y = E+ sin u
E2, x = E2 cos u
E2, y = 2E2 sin u
Since the magnitudes E+ and E2 are equal, the components of the net
field at P are
E x = E + , x + E -, x = E + cos u + E - cos u
2kq
cos u
= 2
y + d2
E y = E + , y + E -, y = E + sin u + 1 -E - sin u2
kq
kq
= 2
sin u - 2
sin u = 0
2
y + d
y + d2
From the figure,
cos u =
d
d
=
2
r
2y + d 2
So the components of the net electric field are
2kq
Ex =
2
d
2
1y + d 2 2y + d
Ey = 0
Reflect
We can check our result by substituting
y = 0, so that the point P is directly between
the two charges and a distance d from each
s - both point to the
s + and E
charge. Then E
right, and the magnitude of the net electric
field should be equal to the sum of the mags-.
s + and E
nitudes of E
Note that at points very far from the dipole,
so that y is much greater than d, the magnitude of the field is inversely proportional
to the cube of y: At double the distance,
the field of a dipole is 11>22 3 = 1>8 as
great. This is a much more rapid decrease
with distance than the field of a single point
charge, for which E is inversely proportional
to the square of the distance: At double
the distance, the field of a point charge is
11>22 2 = 1>4 as great. The dipole field decreases much more rapidly because the fields
of +q and 2q partially cancel each other.
2
2
=
2kqd
2
1y + d 2 2 3>2
The net electric field at point P is to the right and has magnitude
E = 2kqd> 1y 2 + d 2 2 3>2.
At y = 0, the net electric field has magnitude
E =
2kqd
10 + d 2 2 3>2
=
2kqd
d3
=
2kq
d2
= 2a
kq
d2
b
This is just twice the magnitude of the field due to each individual charge:
E + = E- =
kq
d2
If y is much greater than d, y2 + d2 is approximately equal to y2.
Then the magnitude of the net electric field due to the dipole is
­approximately
E net =
2kqd
2 3>2
1y 2
=
2kqd
y3