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Physics II

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"
what
r
"
how
q
r
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UC
;
are
many electrons
=
n
✓h"%
=
↳
c
to
.
. .
106
=
missing
"
.
.
"
OR
.
¥eo
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e
=
of electrostatic force between two identical ions
magnitude
"
">
F
i
§
=
"
the distance
be
most
charge
one
of
a
:
÷
us
a
=
r✓É
n=§
electron
((
"
"
If pcirtice
↳
Fz
↳
-
41¥ ( 91€
=
9292
-
✗z
particles
all
(
)
1-3=0
where
( <→ gz
Ms
L
-
✗ 12
V52
=
✗
92
=
I
¥
at
a-
Fyi
# n°4
just
solve for ×
"
Magnitude
É
y=o
K
=
9¥¥#
'
Fili
=
,,
of net electrostatic force
'
Fy
;
=
cos
-
-09
on
particle
Sino
-
,
;^
113 ))
due to the other
.
diagonal one
¥
-
/NETT
Ñhl
particles
-
41 is the
-
Éy
+
,
''
Éyz
,
+
f-
43
4- K be attractive
1931
.%¥
iii.
laundry T Éyz
razz
DIRECTIONAL
coordinate
a
"
"
"
→
"
,
=
i
2
qz
.
he
1941
n
-
"
,
'
da
cancel , keep
*
so that
change flip
ratio
the
you get
they
nonzero
a
force
net
the direction of the
want
will be
force
accelerated in
Magnitude
F- ✗
the
of
41¥ ¥
=
electric
net
.
Ey
(19,1+1921-1931-1941)
.
cc
particles
four
9,
qz =3 e.
=
ÉNET
fixed in place
are
.
Orient
.
y
4T¥ (¥fz
=
-
.
.
axis to
the
41¥ ¥
=
( -19,1+1 qzl
'
of net electric field
Magnitude
.
's
@ Swore's
by particle
produced
field
particles that didn't
1-
center
1931
-
@ point
•É
-
to
due
the
a
high
Em e
-
Same
✗
Vo :O
;
V2
V02
Zax
+
,
What is $ I
on 4
→
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particles
different
question
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guns
due to the four other
particles
=81T%d- (&99%09
=
thus
v
✓Zai
=
=> m
Cm
=
-
✓
← mass of proton
=
zero
particle force
93
electric potential
Net
"
in
-
← electric field
=
=
produced
be
can
)
-
arctanl Ex
Image
.
"
speed protons
is
•••
-
other
✓
Beams of
-
If the net
"
I
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§
1941 )
P
cancel out
"
-0
{
.
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cc
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MAGNITODF-TVF-nei-ciitfnei-c.sn#-..n.a.a-/IfIs
arctanc.FI?-Y.?-,
ratio
kg
*
"
remain in place when released
y coordinates
S
✗
is to be located such that
✓
+
•
-
J Nd •
PM
d
d
<É•
u
0
*
t
*
"
×
=
Ia
the time it
f- at
=
shows a
L
=
rQ
Eapg
2
takes
distance
if -0
is
rod
.
F"
41T C- OR2 -0
The rod
.
.
forms
radians
;
-0
=
③
at
-
E
/
④
=
ZEO
I
✓Z2+Rz-
-
/
-
✓zz¥rz=
=
÷
=
+
Of circle
.
.
.
by what
is the
factor
magnitude
multiplied
25in (-0/2)
are
central
axis
.
.
.
is
the
magnitude
of
the
field equal to
electric
]
>
"
of the E
②
PARTICLE
along the
Z
•
=
E
-
✓
,
( 21T )
cc
what distance
10/21
a
-
expressed in
ZQ Sin (-0/2)
=
attain final speed
proton travels
a
nonconducting
proton to
a
2-
✓zz+pz-
¥
=
ZZ
=
¥
2-
2
+
÷ R2
=>
44
"
times
.
.
.
W
z2
Where can
room SKIP
good
while
the
yourself
"
dipole
an electric
moment
.
.
.
What is
potential
energy , magnitude
of
torque
-
U
(f)
(a) U
=
=
-
-
I É
-
I
.
É
.
=
-
pE
(b)
cos
I
=
.
,
If
an external
Ñx É
(c) W
Review Dot and Cross product
.
=
=
AU
-
[
=
(Ñ ;
( pi )
-
agent
how much work is done
(G)
W
*
.
you do the
-
É
.
"
what p is
multiplied by ,
in
this case
harm
"
Ñf ) É
( pf ) ]
doing
least
by agent
'
-
still
1.30×10 -3°C
"
-
m
most
"
spherical drop of
a
9
*V
>
=
41TEoR-
=
(b) R'
"3
=
=
C
to
q
'
✗
'
drop
the
radius of
.
18 99
potential surface
.
.
.
of the
drop
new
"
.
109N MYC2) ( q )
-
V
10-12
=
zq
=
=
4-4 Charges
2
EV
=
;
.
=>
41T€ ou
'
Identical
U
R
=
4¥ zy}p?_
"
✓
"
2
=
.
q
R
PC
water
Qa
are
Kf
i
×z+yz-
41T£ ☐ ✓
2213 V
>
fixed
on
Ki
Oi
=
+
x-axis
an
Of
-
Ki
=
.
.
energy of particle
Kinetic
.
?¥÷o ( ✓×zt+y;z-
+
.
at what value
.
.
✓×z+lyz
-
*
set
y
will the
origin @
✗
"
negative value
y momentarily stop
of
=
these values should be the same
(b)
Kf
=
Initial and final coordinates
0
kit Ui
Uf
=
>
=
Ki
ZQ a
+
41T£ ☐ ✓ ✗z
=
+
yiz
"TEO
mw →
✓×Zty
solve for
cc
yf
F- ✗
"
"
rod
nonconducting
V
2
:
.
.
(8.99×109) (z)
take V
.
0
=
@
infinity
.
What is V
ln[É]
(b)
along
1-
the rod's
bisector
Magnitude of
electric
field
@
if
-
the electric
V
potential is
=
5.50W yz2
"
how
y
=
=
Uf
(g)
(¥ 2)
2
-
=
41T C- OF
=HE×i+cEy5+u
IET
is required to set up the
much work
2
-5.50×2-2
=
*
W
four
work is
then there is
component
of
charge configuration
-
negative , meaning that
.
.
.
É
X
an
-
,
,
2✗
=
>
+
V2 y
'
'
(2×-29)
×
w/
=>
g)
Vz
-
E. F
0
✗
=L
✗
I
=
ZV
some
>
=
Mug
back
in
,
.
corresponding nonvanisning
a
particles
"
take
it would
Ey
V
ZV
=
E×
along
in the opposite direction
and the
-
point
I
direction ,
=
Ey
.
* If the electric potential increases
JI
dy
-
"
✗
dx
F-
is V @ P
superposition
V :O due to
d
((
(b) What
.
V , (✗ 2
are
amount to
disrupt
it
"
S@
infinitely apart
"
positive
.
rest
.
CC
The plates of
C
=
spherical capacitor
a
ab
41T C- o
.
.
.
(a)
ab
=
b. a
the
calculate
(b)
A
(8.99×109) ( b- a)
capacitance (b)
What must
be the
plate
area of
a
parallel
plate
capacitor
"
.
.
.
C(b
=
€0
[ pF ]
[ cmz ]
"
If
Cz
measured potential difference across the first
the
=
Vo
c
* capacitors in parallel
have the same potential
1
"
'
"
"
→
')
C,
"
1-
¥
1-1
°
C.
=
÷
.
=
Ci
=
c
9
=
,
/vs
(2)
(b)
Vf
=
U;
=
÷ Vi
% Avi
a
capacitor
⇐
final
⇐
Initial
:
Of
(a)
W
=
=
di
"
a
certain
-
V
Capitance
of this second capacitor
"
.
)
-
.
.
and
an
potential difference
air
energy
9
,
=
C
,
zzv
all
three capacitors
=
>
V
,
=
9
,
/c
=
> V
,
-
V
,
=
V2
=
>
92 Cz V2 [ uc ]
-
-
ea
MM
↳ %
cz=y%s
✓
i
-
=
df
filled separation
/ di
A
df_ U ; ⇐
Up
( par
what is the
s
↳N
=
stored
"
•
-
-
.
10-4
Ui
-
q
U
=
=
(F- max
-
Id
ZKEOA
103 ) KEOA
with
a
dielectric
I 2. OV
Cece ( ÷
to 3
-
=
final energy stored
parallel plate capacitor is filled
Czz
by graph
'
Zdi
(C)
.
>
cz + Cz
+
given
parallel plates in
g
C. ( Cz + C 3)
123
¢
(C
difference
-
,
cc
÷
capacitor drops to
.
.
.
Maximum
energy
that
can
be stored
"
•
c÷)
-
+
/
=
Cztcz
A"d
a
"
c
E°÷ ( "')
=
[
hrnnrrrrrrrrrrrrrrrrrrrrrrr
"
The
V
figure
V
=
+e
Vz
1-
,
V3
+
+
arrangements
.
Which
.
.
"
-
e
41¥%
=
ex
is
arrangement
off _•e
V4
4T¥
=
4
shows
•
the electric
A is in
-
-
↳
The
moves to a
charge
region
of
charge goes
If
higher
a
correct
is
↳ The
electric potential
to
closer
8 electric field lines
electric field
about
it
A
@
"
central
adding fractions
0
O O
•
common denominator
rewire
¥g ye
,
→
"
flies
0
point
away
#
Once the
the
.
"
rod
removed
the closer the electric field
charged object ,
a
object
near
"
what
=
,
,
.
zero
*
y,¥ ÷ 4¥ g- 4¥ g- t.gg#-q
.
fixed place
a
potential
-•e lose
"
charge object
PRACTICE EXAMS
lines
are
spaced
has an excess
.
wire
are
the
,
S
bail
Charge
+
4T
"
F- o
In terms of
@ Eo
Q
F-
"
41T¥ zYzz+¥-
=
Is their a
↳
have same
what
↳
"
42
=
C is
E
between
direction
halved
'
on
symmetry
a line of
,
@
1- dist
L
.
"
.
+•E
i
ET
=
is
"
particles
the
due
U
,
point p
to
opposite
doubled
wi , ,
where an erection
.
.
charge
so
Separation
.
no
be in
If
.
IT
,,
egg, ,, , gum
Same
Charge
between plates is
then
doubled
yes
-•3
Where
Zero
↳
.
a
the
°"
is
axis
the
E. P
"
to the left of
between the
"
the
charges
charges
-
"
WRONG about capacitor S capacitance
is
for
given
a
4
orientation
↳
E
2-
parallel plate capacitor
Isolated
↳
@
point
"
"
z
'
'
@
electric field
what is the
,
au
1,2
to
A- 03
=
,
capacitor
>
,
,
we
and
3
can store
as
much
charge
as
we
wish
by increasing
the
potential difference
btwn plates
.
Et
"
respectively
A- V2
•
-
.
•
cc
.
'
.
"
.
=
.
U
.
.
.
A
=
9
•
,
=
9192
I
+
-
K
qz
.
.
E?÷
.
electric
.
+
K
%÷
potential energy of
=
>
u
=
the
ER
EL
system
9?
-
=
Eb
a
A•
→
<
to
→
%
+
<→
•
•
B
C
es
•
•
to
si
LET
HEB
↳
The
change
"
k
=
D
in electric
potential energy of
the three
-
particle
system
is
"
positive
"
Bubble Problem
↳
F-
STOPPED @
33
:
=
Ey
=
✓
÷
EL
EL
Et
:
K
2£
*
1
→
KII
Ex
=
←
Er
=
dz
+
t
Er
F- b
=
=
K
-
-
K
¥
3£
lE×5+lEy¥?¥
,
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