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想請問這題🙏，有點看不懂圖三的公式是從何而來，卡住了

EXAMPLE 21.11 FIELD OF A UNIFORMLY CHARGED DISK
A nonconducting disk of radius R has a uniform positive surface
charge density o. Find the electric field at a point along the axis of
the disk a distance x from its center. Assume that x is positive.
SOLUTION
IDENTIFY and SET UP: Figure 21.25 shows the situation. We rep-
resent the charge distribution as a collection of concentric rings of
charge dQ. In Example 21.9 we obtained Eq. (21.8) for the field on
the axis of a single uniformly charged ring, so all we need do here
is integrate the contributions of our rings.
EXECUTE: A typical ring has charge dQ, inner radius r, and outer
radius r + dr. Its area is approximately equal to its width dr times
its circumference 2πr, or dA = 2πr dr. The charge per unit area is
σ = dQ/dA, so the charge of the ring is do = o da = 2πor dr.
We use dQ in place of Q in Eq. (21.8), the expression for the field
due to a ring that we found in Example 21.9, and replace the ring
21.25 Our sketch for this problem.
Idr
dQ
X
P
dEx
dEx
radius a with r. Then the field component dEx at point P due to
this ring is
1
2πσrx dr
4π€0 (x² + ²)3/2
SOLUTION
Continued

To find the total field due to all the rings, we integrate dEx over r
from r = 0 to r= R (not from -R to R):
Ex
R 1
- S. A
=
4TTEO
(2πσr dr)x
(x² + x²)3/2
Ex
You can evaluate this integral by making the substitution t =
x² + ² (which yields dt = 2r dr); you can work out the details.
The result is
OX
260
O
2€0
2r dr
E. CR
4€00 (x² + ²)3/2
1
OX
+ R²
+
1]
1
(R²/x²) + 1
(21.11)
EVALUATE: If the disk is very large (or if we are very close to it),
so that R >> x, the term 1/√(R²/x²) + 1 in Eq. (21.11) is very
much less than 1. Then Eq. (21.11) becomes
E
σ
2€0
(21.12)
Our final result does not contain the distance x from the plane.
Hence the electric field produced by an infinite plane sheet of charge
is independent of the distance from the sheet. The field direction
is everywhere perpendicular to the sheet, away from it. There is
no such thing as an infinite sheet of charge, but if the dimensions
of the sheet are much larger than the distance x of the field point P
from the sheet, the field is very nearly given by Eq. (21.12
If P is to the left of the plane (x < 0), the result is the same
except that the direction of E is to the left instead of the right. If
the surface charge density is negative, the directions of the fields
on both sides of the plane are toward it rather than away from it.

dEx
=
-1.
2πσrx dr
4π€0 (x² + √²)3/2

電磁學

Answer 1

Alstormeria5

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