642
CHAPTE R 22 ELECTRIC FIELDS
duces and resolve it into components. One of the components produced by dq is a canceling component; it is
canceled by the corresponding component produced by
dqЈ and needs no further attention. The other component produced by dq is an adding component; it adds to
the corresponding component produced by dqЈ.Add the
adding components of all the elements via integration.
Step 5. Here are four general types of uniform charge
distributions, with strategies for the integral of step 4.
Ring, with point P on (central) axis of symmetry, as
in Fig. 22-11. In the expression for dE, replace r
2
with
z
2 ϩ R
2
, as in Eq. 22-12. Express the adding component
of
in terms of u. That introduces cos u, but u is identical for all elements and thus is not a variable. Replace
cos u as in Eq. 22-13. Integrate over s, around the circumference of the ring.
Circular arc, with point P at the center of curvature, as in Fig. 22-13. Express the adding component of
in terms of u. That introduces either sin u or cos u.
Reduce the resulting two variables s and u to one, u, by
replacing ds with r du. Integrate over u from one end
of the arc to the other end.
Straight line, with point P on an extension of the
line, as in Fig. 22-14a. In the expression for dE, replace
r with x. Integrate over x, from end to end of the line of
charge.
dE
:
dE
:
Problem-Solving Tactics A Field Guide for Lines of Charge
Here is a generic guide for finding the electric field produced at a point P by a line of uniform charge, either circular or straight. The general strategy is to pick out an element
dq of the charge, find
due to that element, and integrate
over the entire line of charge.
Step 1. If the line of charge is circular, let ds be the arc
length of an element of the distribution. If the line is
straight, run an x axis along it and let dx be the length of
an element. Mark the element on a sketch.
Step 2. Relate the charge dq of the element to the length of
the element with either dq ϭ l ds or dq ϭ l dx. Consider
dq and l to be positive, even if the charge is actually negative. (The sign of the charge is used in the next step.)
Step 3. Express the field
produced at P by dq with
Eq. 22-3, replacing q in that equation with either l ds or
l dx. If the charge on the line is positive, then at P draw a
vector
that points directly away from dq. If the charge
is negative, draw the vector pointing directly toward dq.
Step 4. Always look for any symmetry in the situation. If
P is on an axis of symmetry of the charge distribution,
resolve the field
produced by dq into components
that are perpendicular and parallel to the axis of symmetry. Then consider a second element dqЈ that is located
symmetrically to dq about the line of symmetry. At P
draw the vector
that this symmetrical element prodE
: Ј
dE
:
dE
:
dE
:
dE
:
dE
:
E
:
the component dE x set up by ds as
(22-20)
Equation 22-20 has two variables, u and s. Before we can
integrate it, we must eliminate one variable. We do so by
replacing ds, using the relation
ds ϭ r du,
in which du is the angle at P that includes arc length ds
(Fig. 22-13g). With this replacement, we can integrate
Eq. 22-20 over the angle made by the rod at P, from u ϭ Ϫ60Њ
to u ϭ 60Њ; that will give us the field magnitude at P:
(22-21)
ϭ
1.73l
4p´ 0 r
.
ϭ
l
4p´ 0 r
[sin 60Њ Ϫ sin(Ϫ60Њ)]
ϭ
l
4p´ 0 r
͵
60Њ
Ϫ60Њ
cos u du ϭ
l
4p´ 0 r ΄ sin u ΅
60Њ
Ϫ60Њ
E ϭ ͵dEx ϭ ͵
60Њ
Ϫ60Њ
1
4p´ 0
l
r
2
cos u r du
dE x ϭ dE cos u ϭ
1
4p´ 0
l
r
2
cos u ds.
(If we had reversed the limits on the integration, we would
have gotten the same result but with a minus sign. Since the
integration gives only the magnitude of , we would then
have discarded the minus sign.)
Charge density: To evaluate l, we note that the full rod
subtends an angle of 120Њ and so is one-third of a full circle.
Its arc length is then 2pr/3, and its linear charge density
must be
Substituting this into Eq. 22-21 and simplifying give us
(Answer)
The direction of is toward the rod, along the axis of symmetry
of the charge distribution. We can write in unit-vector notation as
.
E
: ϭ
0.83Q
4p´ 0 r
2
i ˆ
E
:
E
:
ϭ
0.83Q
4p´ 0 r
2
.
E ϭ
(1.73)(0.477Q)
4p´ 0 r
2
l ϭ
charge
length
ϭ
Q
2pr/3
ϭ
0.477Q
r
.
E
:
CHAPTE R 22 ELECTRIC FIELDS
duces and resolve it into components. One of the components produced by dq is a canceling component; it is
canceled by the corresponding component produced by
dqЈ and needs no further attention. The other component produced by dq is an adding component; it adds to
the corresponding component produced by dqЈ.Add the
adding components of all the elements via integration.
Step 5. Here are four general types of uniform charge
distributions, with strategies for the integral of step 4.
Ring, with point P on (central) axis of symmetry, as
in Fig. 22-11. In the expression for dE, replace r
2
with
z
2 ϩ R
2
, as in Eq. 22-12. Express the adding component
of
in terms of u. That introduces cos u, but u is identical for all elements and thus is not a variable. Replace
cos u as in Eq. 22-13. Integrate over s, around the circumference of the ring.
Circular arc, with point P at the center of curvature, as in Fig. 22-13. Express the adding component of
in terms of u. That introduces either sin u or cos u.
Reduce the resulting two variables s and u to one, u, by
replacing ds with r du. Integrate over u from one end
of the arc to the other end.
Straight line, with point P on an extension of the
line, as in Fig. 22-14a. In the expression for dE, replace
r with x. Integrate over x, from end to end of the line of
charge.
dE
:
dE
:
Problem-Solving Tactics A Field Guide for Lines of Charge
Here is a generic guide for finding the electric field produced at a point P by a line of uniform charge, either circular or straight. The general strategy is to pick out an element
dq of the charge, find
due to that element, and integrate
over the entire line of charge.
Step 1. If the line of charge is circular, let ds be the arc
length of an element of the distribution. If the line is
straight, run an x axis along it and let dx be the length of
an element. Mark the element on a sketch.
Step 2. Relate the charge dq of the element to the length of
the element with either dq ϭ l ds or dq ϭ l dx. Consider
dq and l to be positive, even if the charge is actually negative. (The sign of the charge is used in the next step.)
Step 3. Express the field
produced at P by dq with
Eq. 22-3, replacing q in that equation with either l ds or
l dx. If the charge on the line is positive, then at P draw a
vector
that points directly away from dq. If the charge
is negative, draw the vector pointing directly toward dq.
Step 4. Always look for any symmetry in the situation. If
P is on an axis of symmetry of the charge distribution,
resolve the field
produced by dq into components
that are perpendicular and parallel to the axis of symmetry. Then consider a second element dqЈ that is located
symmetrically to dq about the line of symmetry. At P
draw the vector
that this symmetrical element prodE
: Ј
dE
:
dE
:
dE
:
dE
:
dE
:
E
:
the component dE x set up by ds as
(22-20)
Equation 22-20 has two variables, u and s. Before we can
integrate it, we must eliminate one variable. We do so by
replacing ds, using the relation
ds ϭ r du,
in which du is the angle at P that includes arc length ds
(Fig. 22-13g). With this replacement, we can integrate
Eq. 22-20 over the angle made by the rod at P, from u ϭ Ϫ60Њ
to u ϭ 60Њ; that will give us the field magnitude at P:
(22-21)
ϭ
1.73l
4p´ 0 r
.
ϭ
l
4p´ 0 r
[sin 60Њ Ϫ sin(Ϫ60Њ)]
ϭ
l
4p´ 0 r
͵
60Њ
Ϫ60Њ
cos u du ϭ
l
4p´ 0 r ΄ sin u ΅
60Њ
Ϫ60Њ
E ϭ ͵dEx ϭ ͵
60Њ
Ϫ60Њ
1
4p´ 0
l
r
2
cos u r du
dE x ϭ dE cos u ϭ
1
4p´ 0
l
r
2
cos u ds.
(If we had reversed the limits on the integration, we would
have gotten the same result but with a minus sign. Since the
integration gives only the magnitude of , we would then
have discarded the minus sign.)
Charge density: To evaluate l, we note that the full rod
subtends an angle of 120Њ and so is one-third of a full circle.
Its arc length is then 2pr/3, and its linear charge density
must be
Substituting this into Eq. 22-21 and simplifying give us
(Answer)
The direction of is toward the rod, along the axis of symmetry
of the charge distribution. We can write in unit-vector notation as
.
E
: ϭ
0.83Q
4p´ 0 r
2
i ˆ
E
:
E
:
ϭ
0.83Q
4p´ 0 r
2
.
E ϭ
(1.73)(0.477Q)
4p´ 0 r
2
l ϭ
charge
length
ϭ
Q
2pr/3
ϭ
0.477Q
r
.
E
: