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Halliday/Resnick/Walker 7e
Chapter 23 – Gauss’ Law
- The magnitude of the electric field produced by a uniformly charged infinite line is E = λ/2πε 0 r , where λ is the linear charge density and r is the distance from the line to the point where the field is measured. See Eq. 23-12. Thus,
( )( )(^ ) 12 2 2 4 6
2 ε 0 Er 2 8.85 10 C / N m 4.5 10 N/C 2.0 m 5.0 10 C/m.
λ = π = π × −^ ⋅ × = × −
- We combine Newton’s second law ( F = ma ) with the definition of electric field ( F = qE )
and with Eq. 23-12 (for the field due to a line of charge). In terms of magnitudes, we have (if r = 0.080 m and λ = 6.0 x 10 -6^ C/m)
ma = eE =
e λ 2 πεo r ⇒ a =
e λ 2 πεo r m = 2.1 × 1017 m/s^2.
- We reason that point P (the point on the x axis where the net electric field is zero) cannot be between the lines of charge (since their charges have opposite sign). We reason further that P is not to the left of “line 1” since its magnitude of charge (per unit length) exceeds that of “line 2”; thus, we look in the region to the right of “line 2” for P. Using Eq. 23-12, we have
E net = E 1 + E 2 = λ 1 2 πεo ( x + L/2 )
λ 2 2 πεo ( x − L/2 ) .
Setting this equal to zero and solving for x we find
x = λ 1 − λ (^2) λ 1 + λ (^2)
L
which, for the values given in the problem, yields x = 8.0 cm.
- (a) To calculate the electric field at a point very close to the center of a large, uniformly charged conducting plate, we may replace the finite plate with an infinite plate with the same
area charge density and take the magnitude of the field to be E = σ/ε 0 , where σ is the area charge
density for the surface just under the point. The charge is distributed uniformly over both sides of the original plate, with half being on the side near the field point. Thus,
6 4 2 2
6.0 10 C
4.69 10 C/m. 2 2(0.080 m)
q A
− = = × = × −
The magnitude of the field is
4 2 7 12 2 2
4.69 10 C/m 5.3 10 N/C. 8.85 10 C / N m
E
− −
×
× ⋅
×
The field is normal to the plate and since the charge on the plate is positive, it points away from the plate.
(b) At a point far away from the plate, the electric field is nearly that of a point particle with
charge equal to the total charge on the plate. The magnitude of the field is E = q / 4 πε 0 r^2 = kq / r^2 ,
where r is the distance from the plate. Thus,
( ) ( ) ( )
9 2 2 6 2
8.99 10 N m / C 6.0 10 C 60 N/C. 30 m
E
× ⋅ ×^ −
- According to Eq. 23-13 the electric field due to either sheet of charge with surface charge
density σ = 1.77× 10 −^22 C/m^2 is perpendicular to the plane of the sheet (pointing away from the
sheet if the charge is positive) and has magnitude E = σ/2ε 0. Using the superposition principle,
we conclude:
(a) E = σ/ε 0 = (1.77 × 10 −^22 )/(8.85 × 10 −^12 )= 2.00× 10 −^11 N/C, pointing in the upward direction, or
E = (2.00 × 10 −^11 N/C)jˆ.
G
(b) E = 0;
(c) and, E = σ/ε 0 , pointing down, or E = − (2.00 × 10 −^11 N/C)jˆ
G
- We use Eq. 23-13.
(a) To the left of the plates:
E = ( σ / 2 ε 0 )( ˆi)
G
− (from the right plate) +^ (^ σ^ / 2ε^0 )iˆ(from the left one) = 0.
(b) To the right of the plates:
E = ( σ / 2 ε 0 )ˆi
G
(from the right plate) + (^) ( σ / 2 ε 0 )( − ˆi)(from the left one) = 0.
(c) Between the plates:
( ) (^ )
22 2 11 2 0 0 0 12 2
ˆ ˆ ˆ 7.00^10 C/m ˆ ( i) i ( i) i 7.91 10 N/C i. 2 2 N m 8.85 10 C
E
− − −
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜^ × ⎟
= ⎜ ⎟ − + ⎜ ⎟ − = ⎜ ⎟ − = − ⎜ ⎟ = − ×
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎜^ ⋅ ⎟
⎜ × ⎟
G
- The field due to the sheet is E = σ 2 εο^.^ The force (in magnitude) on the electron (due to that field) is F = eE , and assuming it’s the only force then the acceleration is
Since the ball is in equilibrium the net force on it vanishes. The sum of the horizontal
components yields qE – T sin θ = 0 and the sum of the vertical components yields
T cos θ − mg = 0. The expression T = qE /sin θ, from the first equation, is substituted into the
second to obtain qE = mg tan θ. The electric field produced by a large uniform plane of charge is
given by E = σ/2ε 0 , where σ is the surface charge density. Thus,
0
tan 2
q mg
and
12 2 2 6 2 0 8 9 2
2 tan 2 8.85^10 C / N.m^ 1.0^10 kg^ 9.8 m/s^ tan 30 2.0 10 C 5.0 10 C/m.
mg q
− − − −
× × °
×
= ×
57. (a) We use meg = eE = e σ/ε 0 to obtain the surface charge density.
σ ε = =
× ×
×
= ×
− − −
m g − e
e
C 0
31 12 19
2
.. .
kg 9.8 m s C
N m (^) C m
c hb gd 2 i
(b) Downward (since the electric force exerted on the electron must be upward).
- Since the fields involved are uniform, the precise location of P is not relevant; what is important is it is above the three sheets, with the positively charged sheets contributing upward fields and the negatively charged sheet contributing a downward field, which conveniently conforms to usual conventions (of upward as positive and downward as negative). The net field
is directed upward ( +ˆj), and (from Eq. 23-13) its magnitude is
6 1 2 3 4 12 0 0 0
| | 5.65 10 N C.
E
− −
×
= + + = = ×
× ×
G
In unit-vector notation, we have E = (5.65 × 104 N/C) jˆ.
G
