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26.71. Model: Model the infinitely long sheet of charge with width L as a uniformly charged sheet.
Visualize:
Solve: (a) Divide the sheet into many long strips parallel to the y-axis and of width ∆x. Each strip has a linear
charge density λ = η∆x and acts like a long, charged wire. At the point of interest, strip i contributes a small field
r
Ei of strength
Ei =
2λ
2η∆x
=
4πε 0 ri 4πε 0 ri
By symmetry, the x-components of all the strips will add to zero and the net field will point straight away from the
sheet along the z-axis. The net field’s z-component will be
Ez = ∑ ( Ei ) z = ∑ Ei cosθ i
i
(
The distance ri = x + z
2
i
)
2 1/ 2
i
(
and cosθ i = z ri = z x + z
Ez = ∑
i
2η∆x
4πε 0 ( x + z
2
i
2
i
)
2 1/ 2
. Thus,
z
) (x
2 1/ 2
2
i
+z
)
2 1/ 2
=
2ηz
4πε 0
∑x
i
2
i
∆x
+ z2
If we now let ∆x → dx, the sum becomes an integral ranging from x = ∆L/2 to x = L/2. This gives
Ez =
2ηz
4πε 0
L/2
dx
ηz  1 −1  x  
η
=
tan
=
2
2
∫




x +z
z  − L / 2 2πε 0
2πε 0  z
− L/2
L/2
 −1  L 
−1  − L  
tan  2 z  − tan  2 z  


From trigonometry, tan −1 ( −φ ) = − tan −1 (φ ) . So finally,
Ez =
(b) As z → 0 m,
r
L
η
η
L
tan −1   ⇒ E =
tan −1   kˆ



πε 0
2z
πε 0
2z 
r
η π ˆ
η ˆ
π
L
⇒E→
k=
k
tan −1   →
 2z 
πε 0 2
2ε 0
2
This is the electric field due to a plane as you can see from Equation 26.26. We obtain this result because in the
limit as z → 0 m, the dimension L becomes extremely large. As z → ∞,
r
η L ˆ
1 2λ ˆ
L
L
⇒E→
k=
k
tan −1   →
 2z 
πε 0 2 z
4πε 0 z
2z
where we have used ηL = λ as the charge per unit length of the sheet. This is the electric field due to a long,
charged wire. We obtain this result because for z >> L, the infinitely long sheet “looks” like an infinite line charge.
(c) The following table shows the field strength Ez in units of η/ε0 for selected values of z in units of L. A graph of
Ez is shown in the figure above.
η
ε0
z L
Ez
0
0.25
0.50
1.0
2.0
3.0
4.0
0.5
0.35
0.25
0.15
0.08
0.05
0.04