Download Notes & ConcepTests

ConcepTest #5: Assume that a strip of tape, length 2L, has
a uniform charge distribution and is oriented as shown.
What is the direction of the electric field at the point P?
L
1. Up
4. Right
2. Down
5. Into/Out of page
3. Left
6. Not enough info
P
-L
What is the electric field at the point P?
L
P
dQ

r
10.
(x,y)
-L

dQ
dE  k 2 rˆ
r
dEx  Ex   dEx
dEy  Ey   dEy
Take a nap, because you’re
probably exhausted!
ConcepTest #6: Which of the following best represents

r?
L
P
1.
yĵ
4.
x P î  yĵ
2.
x P î
5.
x P î  yĵ
6.
x P2  y 2
(xP, yP)
dQ
(xP, 0)
3.
x P2  y 2
(x,y)
-L
(0,y)

ˆ
ˆ
r  ( x P  x )i  ( y P  y ) j

r  ( x P  x )iˆ  ( y P  y ) ˆj  x P iˆ  yˆj
r  x P2  y 2
L
P
xP
dQ

r  x P î  yĵ

x P î  yĵ
r
r̂  
r
x P2  y 2

dQ
dE  k 2 rˆ
r

dQ r
dQ 
k 2 k 3 r
r r
r
(0,y)
-L
x P î  yĵ
dQ
k 2
x P  y 2 x P2  y 2

dE  k
L
dEx  k
P
xP
dQ

r  x P î  yĵ
(0,y)
-L
dQ
( x P î  yĵ )
2
2 3/2
( xP  y )
dQ
x
2
2 3/2 P
( xP  y )
dQ
dEy  k 2
( y )
2 3/2
( xP  y )
kx P dQ
Ex  dEx 
( x P2  y 2 ) 3 / 2


Ey  dEy 

 kydQ
 ( x P2  y 2 )3 / 2
dQ  dL
For uniform charge distribution, then
  constant; here  = Q/2L
L
P
xP
dQ
dQ  dA
dQ  dV
dL = dy

Ex  dEx 

Ey  dEy 
kx P dy
 ( x P2  y 2 )3 / 2
 kydy
 ( x P2  y 2 )3 / 2
-L
Limits? Look at integration variable
and charge distribution...
L
Ex 
kx P dy
L( x P2  y 2 )3 / 2
L
L
Ey 
kydy
L
 kx P 
L( x P2  y 2 )3 / 2
L( x P2  y 2 )3 / 2
L
 k
dy
ydy
L( x P2  y 2 )3 / 2
P
xP
dQ
-L
Think about This: Earlier we said that the y- component
of the electric field for this configuration will be zero due
to symmetry. Prove this explicitly by evaluating the
integral for the y-component of the electric field.
Some possibly useful integrals (not an exhaustive list!)
1

 2
u
u
du
 ln(u)

u
du




udu
2
2 1 /2
(u  b )
 sin udu   cos u
 cos udu  sin u
2
(u  b )

du
2
2 1 /2

ln
u

(
u

b
)
2
2 1 /2
(u  b )
udu
2
2 3 /2
(u  b )
du
2
2 3 /2
(u  b )
1
2
sin
u
cos
udu

sin
u

2
2 1 /2

1
(u2  b 2 )1 / 2
u
 2 2
b (u  b 2 )1 / 2

2
 sin udu 
u 1
 sin 2u
2 4
u 1
 cos udu   sin 2u
2 4
2
L
Ex  kx P 
dy
L( x P2  y 2 )3 / 2

2 k
L
x P L2  x 2
P
L
ConcepTest #7: Consider the case where L
gets very large compared to xP. Which of
the following best represents Ex?
P
xP
dQ
4.
2 kL
xP
2.
2 k
L
5.
2 k
xP
3.
2 k
L2
6.
2 k
x P2
1.
0
-L
Electric field of special charge distributions
“infinite” uniform line charge:
2 k
y
uniform disk of charge (on axis)

x

2 k 1 

2
2
x

R





points perpendicular to line, with
y = perpendicular distance from
line
points along axis, with
uniform ring of charge (on axis):
“infinite” uniform plane of charge:
kQr
( r 2  a 2 )3 / 2
points along axis, with a = radius
of ring; r = distance along axis
R = radius of disk; x = distance
along axis
2k
points perpendicular to plane