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Samenvatting + Uitgewerkte examenvragen Elektriciteit

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  • Course
  • T1AEL1
  • Institution
  • Katholieke Universiteit Leuven (KU Leuven)

Behaalde 16/20 met deze samenvatting, hiervan 9/10 op het theoriegedeelte, de oefening was minder xd. Alles staat in volgorde van de lessen, op het einde alle (tot 2020) uitgewerkte theorievragen van vorige jaren.

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Document information

  • April 10, 2024
  • 26
  • 2021/2022
  • Other
  • Unknown

Subjects

  • elektriciteit
  • 1ste bachelor
  • industrieel ingenieur
  • mogelijke examenvragen
  • uitgewerkte examenvragen
  • groep t

Written for

  • Katholieke Universiteit Leuven (KU Leuven)
  • Industriële Ingenieurswetenschappen
  • T1AEL1
All documents for this subject (1)

1  review

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By: katleenkretzschmar • 9 months ago

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alex66

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Content preview

Alex Otten

16/20 waarvan 9/10 op het

Elektrisch
Veld theoriegedeelte




wet Coulomb veldlijnen
Fei = 2
99 [N) met n
=
4TE o (NM) )
Ingeleider : E = 0

Veld Oprand : El rand


E
=In
= E =

Fei no =




~
oon
bij geinduceerde ladingen Ebinnen O

= E =
4 .




7 [N/C]

Puntlading in Uniform Veld
t
v
↑
9 +- ↑ &
F E
.......
.... .
=
...

Vovvvvvvv
..... q m

E
= .
= .




-




↳a
(dankzij El
E Ea
x =
Xo + Voxt +
+ &
t =




Y =
Yo +
Voy .
t +
1ay .
E
-GE

E
x =
Vot
D
Y = -



1 9E.
.

=
= n
:

x (m)

Lading in veld vie dipool

Dipoolmoment =
9. <C m) .




9 Ex =
/E - 1 =
* %(E)
e y

e·

Se E+ ot COS+
'a
E + COST E
E
1 - =

E +
.
+ ..




/2
q
rel
vo
:

2
~
:
=
=

THE
-
-




E

! Bij :
Eto =

3 (N/C]

DiPool in uniform veld
uitwendig
E
I
iP Fi =
0
9 +
Ft 8 --




~
FusinO 7O
X I + Sino M = F+ Sino C
--qM
.




F-
& =
9 E Sino . . e
L
( .




(= M =
P .
E SinO . [N m] .




=
= x E

, W et
van Gauss (flux)
d
etto
=

Eo
Ag
↓
Elektrische Flux
door A

Vb .
Puntlading
E -

-
X
&

dA E >
- Et bolop E - C IA
GE .
da cos .
=
EdA
&
↳ .




Ag (IE) =
+
( e)
d
9 A
+ r El = Cover bolopp
9
↳ Gaussians =
E . 4 =
netto
= E =
-



CT
Bolpervlav Eo 4 Or




Extra vb .
Lijnlading
l
L




Ea =
+ + + + ++ + + + + + + + + + + + + + ++ + ++ + +
S
ladingsdichtheid


P ~
V

-dat
>


SOE
>
- -




SE da
>




=JT
e
to
S E dA
-




1
dA I
+
.




bottom


·
Mantel :
J .
A =
SE . dA =
ESdA
LENDEL (Symm + E ( =
)
=
E Girl
.




Boven/onder
J. SEACOS90"
·

: = = O




neto
>
-
= E. =


Eo

X
9 netto =
7 e
.
dus E =

CEOU buurt Puntlading (2)
-
In van

E seller dan
daalt
bij lijnlading (7)


Oneindig -
Vianne Plaat



~
E &

⑥ d =Ed . +
JE dAs +
SE da .




top bottom
t
+ne+
+
+
· ·
Mantel :
S .
d =
Sos90 = O
L




!
& +
t +
&
·
Boven + onder :
S = E dA =
GESdA = GET
V - >
-




r VE (ENdA) (Symm-
> E =< +e)

=
19 a >
-

=
netto
ofwel E =
9 netto

Eo 2Eo A
/ladingsdichtheid)
O
↑ netto = O .

A dus E = - In buurt van Vlanne Plaat is
25 afstand
E op eine
gelijk (uniform veld)

,Geladen holle ,
bot


= E Gaussians P .
Ag en Ag2
,




↑
* (
+
+ + A
·
Dinte bol verwaarloosbaar

Tr
+
1
+
Ly
-

t 2 -
-

3)
·
Pos . ladingen Storen elkaar af en verdelen zich Op Opervlah
T




+ +
+ Q
+

..
Y

Rondbol :




O W Symmetrie is Punt
groot in
·
V
.
. .
E even elk van GausoP .




>
-

D =


G
.d =
GEA
Gr
=

Ed
An
= E .
Yπr =

8
~
(E) (E =
+
< )
Q
(7) E =

4π So


Binnenbol :




·
Exact dezelfde rednering maar netto omsloten lading Q = 0


omdat gausopp volledig in de geladen bol zit .




- E = 0



* Niet in
bewijs

YTE
a
E
-




............
nodiga E
zelfde grafien bij
geleidende bol MAAR
volle

:



1

: S Bij volle insulator
I S
r R ↑
R

, Potential
Elektrische
VB-Va =

-d .

Nukunnen we Eoon in (/m) zetten




Vb
. Uniform veld

- x(m)
↳
B
VA =
V Va - =
Ov Wan de
..... y C ...
= - -.
- - - - - -




..............I S
"


di
+


-
[
·
VBc = Vc -


VB = -




SE - = -
-
4
,
2
m
.
dx = -
v

A B
L
I C
-
3
O i x /m) Vac =
Vc Va -
=
VAB + VBc = v

(conservatief weld)

Vb Holle geleidende bol
.




E
a
-
~ Want als je tussen en V bereuent krijg je 0

YTE dus Verandering
-




............ :
= O

D
-




: S
:
I S
r R r R

Vb . Puntlading

F
~
E E =
=%
4


B0
An
9
&
+ r
-

A
L ~I
V

↑B rB
B- 9
JE d) S 1 dr
SE
d
VB Va
-
=

=
-
- = .
.




4/1 r2
A PA
(1 del

:
-




1
.
9 .




F1 s
-Va = 30 Pl -




Stel we kiezen = 00 als referentie

dan is V = 0



9 in 3 dimensies
= VB =
D
E B D




Vb Stelsel .
Puntlading
b b
I I I I

+ & & -




+9 A B -


9 Bepaal V =
VB-VA
I I
d


Superpositie :
VV Va Lelfde referentie V(00) = O

9
Epot
1

Av 9
-




=
= .




(a-b


> Av vB va y3 (3 ab (5 a 8)
-
= =
+
-


+
-




1 2b-d
1= AV =
29.
·
.




43 (d b) b
-

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