27.70. Model: Model the infinitely long sheet of charge with width L as a uniformly charged sheet.
Visualize:
Solve: (a) Consider a point on the x-axis at a distance d from the center of the sheet of charge. (We’ll call this
distance d to begin with, rather than x, to avoid confusion with x as the integration variable.) Once again, let the
sheet of charge be divided into small strips of width Δ x. Each strip has a linear charge density λ = ηΔ x and acts
like a long, charged wire. Strip i is at distance ri = d − xi from the point of interest, so it contributes the small field
G
2λ ˆ
2ηΔx
Ei =
i =
iˆ
4πε 0 ri
4πε 0 ( d − xi )
G
In this situation all fields Ei point in the same direction. Their x-components all add to give a net field in the
+x-direction:
Ex = ∑ ( Ei ) x =
i
2η
Δx
∑
4πε 0 i ( d − xi )
We’ll replace the sum with an integral from x = −L/2 to x = +L/2, giving
L/2
2η
dx
2η
2η
2η ⎛ 2d + L ⎞
=
⎡ − ln ( d − x ) ⎦⎤ − L / 2 =
⎡ − ln ( d − L 2 ) + ln ( d + L 2 ) ⎦⎤ =
ln ⎜
⎟
4πε 0 − L∫/ 2 ( d − x ) 4πε 0 ⎣
4πε 0 ⎣
4πε 0 ⎝ 2d − L ⎠
L/2
Ex =
Now that the integration is complete, we can note that d really is the x-coordinate of the point of interest.
Substituting x for d and changing to vector form, we end up with
G
2η ⎛ 2 x + L ⎞ ˆ
ln ⎜
Enet =
⎟i
4πε 0 ⎝ 2 x − L ⎠
(b) If the point is very distant compared to width of the sheet of charge (x >> L), then the sheet of charge looks
like a line of charge with linear density (charge per unit length) λ = ηL. Write
Ex =
2η ⎛ 1 + L/2 x ⎞ 2η
ln ⎜
[ln(1 + L/2 x ) − ln(1 − L/2 x )]
⎟=
4πε 0 ⎝ 1 − L/2 x ⎠ 4πε 0
If x >> L, then L/2x << 1 and we can use the approximation ln(1 + u) ≈ u if u << 1. Thus
Ex ≈
2η ⎡ L ⎛ L ⎞ ⎤ 2η L
2λ
− ⎜ − ⎟⎥ =
=
⎢
4πε 0 ⎣ 2 x ⎝ 2 x ⎠ ⎦ 4πε 0 x 4πε 0 x
This is the field of a line of charge with λ = ηL, as we expected.
(c) The following table shows the field strength Ex in units of 2η/4πε0 for selected values of z in units of L. A
graph of Ex is shown in the figure above.
z L
0.75
1.0
1.5
2.0
3.0
4.0
Ex
2η
4πε 0
1.61
1.10
0.69
0.51
0.34
0.25