lecture 10 - Announcement:
 Clickers:

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Unformatted text preview: Announcement:
 Clickers:

 75%
of
ques6ons
answered
per
session:
3
points
 Each
ques6on
answered
correctly:
1
more
point
 In‐class
Quiz:
Tuesday,
30
minutes,
3
ques6ons
Start
9
am
Sharp
 Handing
it
in
worth
2
clicker
points
 Each
ques6on
worth
2
clicker
points

 r q
 Q
 Coulomb’s
Law
 Electric
Field
 F= 1 Qq ˆ r 2 4πε 0 r F = qE E (r ) = 1Q ˆ r 2 4πε 0 r Gauss’s
Law
 Flux
over
a
closed
surface
 Φtotal Φtotal Qenclosed = ε0 ˆ = ∫ E ⋅ ndS Electric
Poten6al
Energy
on
charge
q:



U(q)


(J)
 








Work
done
by
electric
field
is
independent
of
path,
 



















































but
dependent
on
charge
 Electric
Poten6al:
V
=
U/q

(V)

 Example
:
Parallel
Plate
Capacitor
 σ=
1
µC/m2
 +σ
 a
 1
m
 
b 
 1
µC
 ‐σ
 1 × 10 −6 C / m Electric
field
between
parallel
plates
=
σ/ε0=



 = 1.1 × 10 5 N / C −12 2 8.85 × 10 ⋅ C / Nm Work
done
by
my
hand
=
qEd
=
 (1 × 10 −6 C ) × (1.1 × 10 5 N ) × (1m ) = 0.11Nm = 0.11J C σ=
1µC/m2
 +σ
 a
 1
µC
 1
m
 
b 
 ‐σ
 Poten6al
Energy
at
posi6on
a
 U(a)‐U(b)
=0.11
J
 Poten6al

 (U(a)‐U(b))/q
=
V(a)‐V(b)
=
0.11
J/1µC=1.1×105
V

 Problem
1:
When
a
nega6ve
charge
is
 released
and
moves
along
an
electric
field
 line,
it
moves
to
a
posi6on
of
 a.
 b. c.  d.  e. 
lower
poten6al
and
lower
 poten6al
energy.
 
lower
poten6al
and
higher
 poten6al
energy.
 higher
poten6al
and
lower
 poten6al
energy.
 higher
poten6al
and
 higher
poten6al
energy.
 
decreasing
magnitude
of
 the
electric
field.
 Example:
Q
sta6onary:
how
much
energy
does
it
take
to
bring
q
to
the
below
posi6on?
 R Q
 F= 1 Qq ˆ r 2 4πε 0 r At
distance
r
 Work
done
by
hand
=
Required
Energy


 q
 How
much
energy
does
it
take
to
move
it
an
infinitesimal
distance
in
at
distance
r?
 ˆ − drr q
 Q
 1 Qq 1 Qq ˆ ⋅ − drr = − ˆ Work = Force × Dis tan ce = r dr 2 2 4πε 0 r 4πε 0 r Worktotal 1 Qq 1 Qq =∫ − dr = 2 ∞ 4πε 0 r 4πε 0 R R ˆ x Q
 1Q V (x) = 4πε 0 x E= 1Q dV ˆ ˆ x=− x 4πε 0 x 2 dx X
axis
 r q
 Q
 Poten6al
Energy
at
R
:

 Poten6al
at
R:

 E= 1 Qq 4πε 0 r 1Q V (r ) = 4πε 0 r 1 Qq dV ˆ ˆ r=− r 2 4πε 0 r dr In
general,





 Gradient
Discussion:
Board
 Problem
2:
The
electric
field
in
a
region
of
space
is
given
by

 Ex
=
(3.0x)
N/C,
Ey
=
Ez
=
0,
where
x
is
in
m.
Points
A
and
B
are
 on
the
x
axis
at
xA
=
3.0
m
and
xB
=
5.0
m.
Determine
the
 poten6al
difference
VB
–
VA.
 a.  b.  e.  d. e. –24
V
 +24
V
 –18
V
 
+30
V
 
–6.0
V
 Problem
3:
Points
A
[at
(3,
6)
m]
and
B
[at
(8,
–3)
m]
are
in
a
 region
where
the
electric
field
is
uniform
and
given
by
E
=
12i
 N/C.
What
is
the
electric
poten6al
difference
VA
–
VB
?
 a. b. c. d. e. 
+60
V
 
–60
V
 
+80
V
 
–80
V
 
+50
V
 Problem
4:
Points
A
[at
(2,
3)
m]
and
B
[at
(5,
7)
m]
are
in
a
 region
where
the
electric
field
is
uniform
and
given
by
E
=
(4i
+
 3j)
N/C.
What
is
the
poten6al
difference
VA
–
VB
?
 a. b. c.  d.  e.  
33
V
 
27
V
 30
V
 24
V
 11
V
 Example:
Conduc6ng
Sphere
with
total
charge
Q
 Radius
:
R
 P
 r Example:
Uniformly
Charged
Sphere
with
total
charge
Q
 Radius
:
R
 P
 r What
about
a
non‐tradi6onal
objects?
 Conduc6ng
irregular
shaped
object
 Radius
b
 Radius
a
 ...
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