1. We note that the symbol q2 is used in the problem statement to mean the absolute value
of the negative charge which resides on the larger shell. The following sketch is for
q1 = q2 .
The following two sketches are for the cases q1 > q2 (left figure) and q1 < q2 (right figure).
2. (a) We note that the electric field points leftward at both points. Using F = q0 E , and
orienting our x axis rightward (so ˆi points right in the figure), we find
. × 10−19 C −40 i = −6.4 × 10−18 N i
F = +16
which means the magnitude of the force on the proton is 6.4 × 10–18 N and its direction
(−ˆi) is leftward.
(b) As the discussion in §22-2 makes clear, the field strength is proportional to the
“crowdedness” of the field lines. It is seen that the lines are twice as crowded at A than at
B, so we conclude that EA = 2EB. Thus, EB = 20 N/C.
3. The following diagram is an edge view of the disk and shows the field lines above it.
Near the disk, the lines are perpendicular to the surface and since the disk is uniformly
charged, the lines are uniformly distributed over the surface. Far away from the disk, the
lines are like those of a single point charge (the charge on the disk). Extended back to the
disk (along the dotted lines of the diagram) they intersect at the center of the disk.
If the disk is positively charged, the lines are directed outward from the disk. If the disk is
negatively charged, they are directed inward toward the disk. A similar set of lines is
associated with the region below the disk.
4. We find the charge magnitude |q| from E = |q|/4πε0r2:
q = 4πε 0 Er
(1.00 N C )(1.00 m )
8.99 × 10 N ⋅ m C
= 1.11× 10−10 C.
5. Since the magnitude of the electric field produced by a point charge q is given by
E =| q | / 4πε 0 r 2 , where r is the distance from the charge to the point where the field has
magnitude E, the magnitude of the charge is
( 0.50 m ) ( 2.0 N C ) = 5.6 ×10−11 C.
q = 4πε 0 r 2 E =
8.99 ×109 N ⋅ m 2 C2
6. With x1 = 6.00 cm and x2 = 21.00 cm, the point midway between the two charges is
located at x = 13.5 cm. The values of the charge are q1 = –q2 = – 2.00 × 10–7 C, and the
magnitudes and directions of the individual fields are given by:
E1 = −
| q1 |
ˆi = −(3.196 ×105 N C)iˆ
4πε 0 ( x − x1 )
E2 = −
ˆi = −(3.196 ×105 N C)iˆ
4πε 0 ( x − x1 ) 2
Thus, the net electric field is
Enet = E1 + E2 = −(6.39 ×105 N C)iˆ
7. Since the charge is uniformly distributed throughout a sphere, the electric field at the
surface is exactly the same as it would be if the charge were all at the center. That is, the
magnitude of the field is
4 πε 0 R 2
where q is the magnitude of the total charge and R is the sphere radius.
(a) The magnitude of the total charge is Ze, so
. × 10−19 C
8.99 × 109 N ⋅ m2 C 2 94 160
= 3.07 × 1021 N C .
4 πε 0 R 2
6.64 × 10 m
(b) The field is normal to the surface and since the charge is positive, it points outward
from the surface.
8. (a) The individual magnitudes E1 and E2 are figured from Eq. 22-3, where the
absolute value signs for q2 are unnecessary since
G this charge is positive. Whether we add
the magnitudes or subtract them depends on if E1 is in the same, or opposite, direction as
E 2 . At points left of q1 (on the –x axis) the fields point in opposite directions, but there is
no possibility of cancellation (zero net field) since E1 is everywhere bigger than E2 in
this region. In the region between the charges (0 < x < L) both fields point leftward
there is no possibility of cancellation. At points to the right of q2 (where x > L), E1 points
leftward and E2 points rightward so the net field in this range is
Enet = | E2 | − | E1 | ˆi .
Although |q1| > q2 there is the possibility of E net = 0 since these points are closer to q2
than to q1. Thus, we look for the zero net field point in the x > L region:
| E1 |=| E2 |
1 | q1 |
4πε 0 x
4πε 0 ( x − L )2
which leads to
| q1 |
Thus, we obtain x =
≈ 2.72 L .
1− 2 5
(b) A sketch of the field lines is shown in the figure below:
9. At points between the charges, the individual electric fields are in the same direction
and do not cancel. Since charge q2= − 4.00 q1 located at x2 = 70 cm has a greater
magnitude than q1 = 2.1 ×10−8 C located at x1 = 20 cm, a point of zero field must be closer
to q1 than to q2. It must be to the left of q1.
Let x be the coordinate of P, the point where the field vanishes. Then, the total electric
field at P is given by
| q1 | ·
1 § | q2 |
4πε 0 ¨© ( x − x2 ) ( x − x1 )2 ¸¹
If the field is to vanish, then
| q2 |
| q1 |
( x − x2 ) ( x − x1 )2
| q2 | ( x − x2 ) 2
| q1 | ( x − x1 )2
Taking the square root of both sides, noting that |q2|/|q1| = 4, we obtain
x − 70
= ±2.0 .
x − 20
Choosing –2.0 for consistency, the value of x is found to be x = −30 cm.
10. We place the origin of our coordinate system at point P and orient our y axis in the
direction of the q4 = –12q charge (passing through the q3 = +3q charge). The x axis is
perpendicular to the y axis, and
G the identical q1 = q2 = +5q charges.
The individual magnitudes | E1 |, | E2 |, | E3 |, and | E4 | are figured from Eq. 22-3, where the
absolute value signs for q1, q2, and q3 are unnecessary since those charges are positive
G q > 0). We note that the contribution from q1 cancels that of q2 (that is,
| E1 | = | E2 | ), and the net field (if there is any) should be along the y axis, with magnitude
E net =
4 πε 0
GH b2d g
1 12q 3q
4 πε 0 4d 2 d 2
which is seen to be zero. A rough sketch of the field lines is shown below:
11. The x component of the electric field at the center of the square is given by
ª | q1 |
| q3 |
| q2 |
| q4 | º
¬ (a / 2) (a / 2) (a / 2) (a / 2) ¼
(| q1 | + | q2 | − | q3 | − | q4 |)
4πε 0 a / 2
Similarly, the y component of the electric field is
| q3 |
| q1 |
| q2 |
| q4 | º
» cos 45°
4πε 0 ¬ (a / 2) 2 (a / 2) 2 (a / 2) 2 (a / 2) 2 ¼
( − | q1 | + | q2 | + | q3 | − | q4 |)
4πε 0 a / 2
N ⋅ m 2 / C2 ) (2.0 ×10−8 C) 1
= 1.02 ×105 N/C.
(0.050 m) / 2
Thus, the electric field at the center of the square is E = E y ˆj = (1.02 ×105 N/C)j.
12. By symmetry we see the contributions from the two charges q1 = q2 = +e cancel each
other, and we simply use Eq. 22-3 to compute magnitude of the field due to q3 = +2e.
(a) The magnitude of the net electric field is
9 4(1.60 ×10
| Enet |=
= 160 N/C.
4πε 0 r
4πε 0 (a / 2) 4πε 0 a
(6.00 ×10 )
(b) This field points at 45.0°, counterclockwise from the x axis.
13. (a) The vertical components of the individual fields (due to the two charges) cancel,
by symmetry. Using d = 3.00 m, the horizontal components (both pointing to the –x
direction) add to give a magnitude of
Ex, net =
= 1.38 × 10−10 N/C .
4πεo (d2 + y2)3/2
(b) The net electric field points in the –x direction, or 180° counterclockwise from the +x
14. For it to be possible for the net field to vanish at some x > 0, the two individual fields
(caused by q1 and q2) must point in opposite directions for x > 0. Given their locations in
the figure, we conclude they are therefore oppositely charged. Further, since the net field
points more strongly leftward for the small positive x (where it is very close to q2) then
we conclude that q2 is the negative-valued charge. Thus, q1 is a positive-valued charge.
We write each charge as a multiple of some positive number ξ (not determined at this
point). Since the problem states the absolute value of their ratio, and we have already
inferred their signs, we have q1 = 4 ξ and q2 = −ξ. Using Eq. 22-3 for the individual fields,
Enet = E1 + E2 =
4πεo (L + x)
for points along the positive x axis. Setting Enet = 0 at x = 20 cm (see graph) immediately
leads to L = 20 cm.
(a) If we differentiate Enet with respect to x and set equal to zero (in order to find where it
is maximum), we obtain (after some simplification) that location:
x = ¨3 2 + 3 4 + 3¸L = 34 cm.
We note that the result for part (a) does not depend on the particular value of ξ.
(b) Now we are asked to set ξ = 3e, where e = 1.60 × 10−19 C, and evaluate Enet at the
value of x (converted to meters) found in part (a). The result is 2.2 × 10−8 N/C .
15. The field of each charge has magnitude
= 3.6 × 10−6 N C .
The directions are indicated in standard format below. We use the magnitude-angle
notation (convenient if one is using a vector-capable calculator in polar mode) and write
with the proton on the left and moving around clockwise) the contributions to
E net as follows:
b E∠ − 20°g + b E∠130°g + b E∠ − 100°g + b E∠ − 150°g + b E∠0°g.
This yields c3.93 × 10 ∠ − 76.4°h , with the N/C unit understood.
(a) The result above shows that the magnitude of the net electric field is
| Enet |= 3.93×10−6 N/C.
(b) Similarly, the direction of E net is –76.4° from the x axis.
16. The net field components along the x and y axes are
Enet x =
q2 cos θ
Enet y = –
q2 sin θ
The magnitude is the square root of the sum of the components-squared. Setting the
magnitude equal to E = 2.00 × 105 N/C, squaring and simplifying, we obtain
q12 + q22 − 2 q1 q2 cos θ
With R = 0.500 m, q1 = 2.00 × 10− 6 C and q2 = 6.00 × 10− 6 C, we can solve this
expression for cos θ and then take the inverse cosine to find the angle. There are two
(a) The positive value of angle is θ = 67.8°.
(b) The positive value of angle is θ = − 67.8°.
17. The magnitude of the dipole moment is given by p = qd, where q is the positive
charge in the dipole and d is the separation of the charges. For the dipole described in the
p = 160
. × 10−19 C 4.30 × 10−9 m = 6.88 × 10−28 C ⋅ m .
The dipole moment is a vector that points from the negative toward the positive charge.
18. According to the problem statement, Eact is Eq. 22-5 (with z = 5d)
9801 πεo d
and Eapprox is
The ratio is therefore
Eact = 0.9801 ≈ 0.98.
19. Consider the figure below.
(a) The magnitude of the net electric field at point P is
Enet = 2 E1 sin θ = 2 «
¬« 4πε 0 ( d / 2 ) + r ¼»
( d / 2)
4πε 0 ª( d / 2 )2 + r 2 º 3/ 2
For r >> d , we write [(d/2)2 + r2]3/2 ≈ r3 so the expression above reduces to
| Enet |≈
4πε 0 r 3
(b) From the figure, it is clear that the net electric field at point P points in the − j
direction, or −90° from the +x axis.
20. Referring to Eq. 22-6, we use the binomial expansion (see Appendix E) but keeping
higher order terms than are shown in Eq. 22-7:
d 3 d2 1 d3
d 3 d2 1 d3
¨¨1 + z + 4 z2 + 2 z3 + … ¸ − ¨1 − z + 4 z2 − 2 z3 + … ¸¸
4πεo z2 ©©
Therefore, in the terminology of the problem, Enext = q d3/ 4πε0z5.
21. Think of the quadrupole as composed of two dipoles, each with dipole moment of
magnitude p = qd. The moments point in opposite directions and produce fields in
opposite directions at points on the quadrupole axis. Consider the point P on the axis, a
distance z to the right of the quadrupole center and take a rightward pointing field to be
positive. Then, the field produced by the right dipole of the pair is qd/2πε0(z – d/2)3 and
the field produced by the left dipole is –qd/2πε0(z + d/2)3. Use the binomial expansions
(z – d/2)–3 ≈ z–3 – 3z–4(–d/2) and (z + d/2)–3 ≈ z–3 – 3z–4(d/2) to obtain
2 πε 0 z 3 2 z 4 z 3 2 z 4
4 πε 0 z 4
Let Q = 2qd 2. Then,
4 πε 0 z 4
22. We use Eq. 22-3, assuming both charges are positive. At P, we have
Eleft ring = Eright ring
4πε 0 ( R 2 + R
2 3/ 2
Simplifying, we obtain
= 2¨ ¸
q2 (2 R)
4πε 0 [(2 R) 2 + R 2 ]3/ 2
23. (a) We use the usual notation for the linear charge density: λ = q/L. The arc length is
L = rθ if θ is expressed in radians. Thus,
L = (0.0400 m)(0.698 rad) = 0.0279 m.
With q = −300(1.602 × 10−19 C), we obtain λ = −1.72 × 10−15 C/m.
(b) We consider the same charge distributed over an area A = πr2 = π(0.0200 m)2 and
obtain σ = q/A = −3.82 × 10−14 C/m².
(c) Now the area is four times larger than in the previous part (Asphere = 4πr2) and thus
obtain an answer that is one-fourth as big:
σ = q/Asphere = −9.56 × 10−15 C/m².
(d) Finally, we consider that same charge spread throughout a volume of 4π r3/3 and
obtain the charge density ρ = charge/volume = −1.43 × 10−12 C/m3.
24. From symmetry, we see that the net field at P is twice the field caused by the upper
semicircular charge + q = λ ⋅ πR (and that it points downward). Adapting the steps leading
to Eq. 22-21, we find
Enet = 2 −ˆj
4πε 0 R
(a) With R = 8.50 × 10− 2 m and q = 1.50 × 10−8 C, | Enet |= 23.8 N/C.
(b) The net electric field Enet points in the −ˆj direction, or −90° counterclockwise from
the +x axis.
25. Studying Sample Problem 22-4, we see that the field evaluated at the center of
curvature due to a charged distribution on a circular arc is given by
4 πε 0 r
−θ / 2
along the symmetry axis
where λ = q/rθ with θ in radians. In this problem, each charged quarter-circle produces a
field of magnitude
1 2 2 |q|
| E |=
[sin θ ]− π / 4 =
r π / 2 4πε 0 r
4πε 0 π r 2
That produced by the positive quarter-circle points at – 45°, and that of the negative
quarter-circle points at +45°.
(a) The magnitude of the net field is
§ 1 2 2 |q|·
1 4| q | (8.99 ×109 )4(4.50 ×10−12 )
Enet, x = 2 ¨¨
= 20.6 N/C.
π (5.00 ×10−2 ) 2
4πε 0 π r 2
© 4πε 0 π r ¹
(b) By symmetry, the net field points vertically downward in the −ˆj direction, or −90°
counterclockwise from the +x axis.