I start from 2f₁= max, derive to this expression {fy = may, 2) ap(x, y, z) dx (_ ap(x, y, z) ay
I start from 2f₁= max, derive to this expression {fy = may, 2) ap(x, y, z) dx (_ ap(x, y, z) ay
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Transcribed Image Text:Ⓒ start from ZF₁ = max, derive to this expressing
1... 2fy = may.
(2
(ap(x, y, z)
ах
ap(x, y, z)
ay
lax
=pay

Transcribed Image Text:H. Wis
to do the
Jame with
n
an
- Apply Newton's 2
ΣE
= maz
=
- dF₂ -dfg + df = maz
|d F₁ = ( P ( u, y, z _ dz) dm.dy
d F g = dmg = edt g = e.q_dady dz
dE
D(n,y, 2+² ) du dy
nd
Tayt or
Seress
expration
f(~₂9₁ 2+² dz) = f(mgy, z) +
- ma
ā
Law
P(2) eg.
The gradiant
Vf=of ↑
In
to Z-direction
- eg = ez
6 +
бра
LOPA
SP
on ² - dy î - dfk - egk= //
ду
J
of
of n
бу
de
dF3
dz
J +
dfs
dF,
и
- P(1,9, 7-1²) du.dy) - eg d m dydz + P(1,4, 2+0 2 ) dndy = @g.dm. dy.dząz
af (И, У, 7)
IZ
of A
->
7
fr
- Pin y 2) andy_ @P(1, 4, =) Andydz_ @dad de P(4,32) July - P(1,9,2), dady de 20., drodytes
арси, z)
Taz
dz
ap(2₂ y, z)
= egreaz
az
Vactor for me
ΣΕ
A
J
m
[ F₁₂ ₁ + E F y ĵ + [F₂ k=mani + may^³ + may Â
(-³P ) î + (- 3 P ) ^j + ( DP - eg) Â =(Can) î+ (Cay)û +(Ca_ ) Â
J
dy
Spont (m,y,zodz
du
DP-8A = Pa
scavar finction
N
+ (~4+3)
VdFg
df₂
+dz
back
+
ать
d Fu
da²
807
(Y,Z) (dz)
21
f(nyoz)
d
A
d
V = 3 ₂ ² + 2 y ³ + 3 = ²K
ач
↑
î
5
ins
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