Physics 1300 - Electrical Potential Field

Electric potential Energy
The energy related the electric
to
charge particle due to its location in
·




field depending on its
position
Similar where the
to
gravitational potential energy related
·




energy
w) the due to its the
mass is location in
gravitational field ~Ep mgy =




the
u =
qE y due
q Q# uniform field to plate
electric
of charge
:




proton located in a
uniform electric
field of magnitude ISONIC
created
by negatively charged plate Proton .
is 1 0 m
.

from the plate
a) what is electric
potential energy relative to plates
b) what direction released ? does /U) increase
is
of proton when or


decrease once released ?
0)
a) U =
gEy = (16 X10" ) ( (S0) (1 .




2 4
= . X 10- J
b) the downward towards theplate because of the
proton will move



negative charge on the plate and the potential energy will lower
,


the
to the plate as electric
field lines are moving further apart
Q #2
electron is located
a in a
uniform electric
field of ISONIC ,
1 Om.
away from plate
a) (v) relative to plate ?
b) What direction will it move ? (v) increase or decrease ?
a) U =
qEy = 1 6 X10"4)
. (150) (1 0) .




= -
2 . 4 x 10. j
b) electron will move
upward away from the
plate as the

plate has a
negative charge ,
the (U) will increase of the electric

field lines will move closer x

↳ the (U) will lower the
as it
gets further from plate

, Another
way of approaching conservation
of energywi uniform
electric
field is

Vi + Exel = Uzt Elez
U
1, electric
U2 :
potential energy
Ele , Ely Kinetic
energy =
-DER
Q#
electron starts 5 2 cm
from rest at .

from positively charged
plate . A
uniform electric
field or a
magnitude of 220 NIC is formed
.




what is the speed of the electron
before it hits the plate.
Let the plate be O
-
&


E = 220N/C LU =
Up-Vi >
-



g Y GE Yi
v = 0 OS2M
.
=

.4
= 1 6 x 10 C
q
.




electron " (0) ( 6 x 10 P) /220) 70 0S2)
4 11 A10
Reg
:
= -


mass
of = . .
.




10"0
1 0304 x
&
= -
.




SAMPLEQ#
positively charged plate produces
O En = -
OU
1 3 . X/03 NIC
. Electron is released OEK = -
1 . 8304 X 10. 18j
from rest at a
position that is DEL =
< mu 2


1 2 X 10
away from plate
. m . What

if hits the 2(1 8304x10
-

18)
is the
speed as plate ? vg
= .




& (1 X/0
-
3)
the that
Yo position
-


= 0 as is when the
°
electron hits the . 2
plate = .
00 x 10 m/S


gEy-gEY9) (1
-
ou =


= 10) -
/- 1 6
. x 10" . 3 x 10") / 1 2 x 10
.
m)
18
2 446 X10 J
-

= .




·Ex = - OU
%8
OEK = -
2 . 446x10 ju
I mu
2
OEK =


=
vq =
26 2:
.
446x10-8)
31
4 /1 X10
-

.




"
= 2 34/X10
.

mrs