Transcript Exercise_6

Three charges are placed at the corners of a 2m X 2m square as
shown below. What charge must be placed at the fourth corner so
that the electric potential is zero at the center of the square (point
O)? How much work would be required to bring the fourth charge
from infinity and place it at its corner of the square?
q 3  30C
q4  ?
2m
q1  20C
O
2m
q 2  40C
 O    O,i   O,1   O,2   O,3   O,4  0
i
O,4  O,1  O,2  O,3 
kq i
i  r
O,i
kq1 kq 2 kq 3 
kq 4






rO,4
r
r
r
 O,1
O,2
O,3 
rO,1  rO,2  rO,3  rO,4  rO
k q   k q  q  q 
2
3
rO 4
rO 1
q 4  q 1  q 2  q 3 
q 4  20C  40C  30C
q 4  10C
Finding the potential at the fourth corner
prior to placing q 4 there :
4    4,i  4,1  4,2   4,3
i
kq i
 4,i  r
4,i
kq1 kq 2 kq 3
4  r  r  r
4,1
4,2
4 ,3
 q 1
q2
q 3 
4  kr  r  r 
 4,1
4,2
4,3 
r4,1  r4,3  2m
r4,2  2m 2  2.83m
2
6
6
6


9 N  m  20  10 C



40
10
C
30
10
C
 4  9  10




2
2m
2.83m
2m




C
4  9.78  10 volts
4
Wq  4   q 4  4     
4
Wq 4   10  10 C9.78  10 volts  0
6
4
4
Wq  4   0.98J
4
No work is required by an external force to
bring q 4 from infinity and place it at the fourth
corner...The electric field does the work.
How much electrical potential energy is stored in the completed
square of 4 charges?
The potential energy stored in the charge arrangement is equal to
the work required to bring each charge from infinity and place it at
its vertex.
Initially no charges are present so no work is required to place q1 at
its vertex.
2m
O
No Work
2m
O
2m
2m
q1  20C
Now there is an electric potential at the vertex where q2 is to be
placed.
 2,1 
2m
O
2m
q1  20C
2,1  2, 1
kq 1
r2,1
2

9 N  m 
6
9  10

20

10
C

2


C

2m
J
 2, 1  9  10 4
C
Wq 2 (2 )  q2  2 ,1   
J
Wq 2 ( 2 )  40  10 C9  10 C 
Wq 2 ( 2 )  3.6J
6
No work required.
4
3
2m
O
No Work
2m
O
2m
2m
q1  20C
q1  20C
q 2  40C
3   3,1   3,2
3 
kq 1
kq2

r3,1
r3,2
2
2


9 N  m 
6
9 N  m 
6
9  10







20
10
C
9
10
40
10
C



2
2




C
C
3 

2.83m
2m
3   1.16  105 J
C
Wq 3( 3)  q 3  3    
J
Wq3 ( 3)   30  10 C  1.16  10 C 
6
5
Wq3 (3)  3.48 J
No work required.
We have already shown that no work is required to place q4,
therefore no work is required to place any of the charges and the
charge distribution does not store any electric potential energy.
A capacitor is oriented so that its plates lie horizontally, negative
plate below the positive plate. The plates carry charges of +/-20C,
are separated by a distance of 4m and have an area of 0.1m2. An
object with a mass of 1.5kg and carrying a charge of 5C is ejected
vertically upward from the negative plate with an initial velocity of
25m/s.
A) Will it reach the upper plate? If not how far from the upper
plate will it stop?
B) If the capacitor were rotated so that its plates were vertical,
would the object reach the positive plate? If it reaches the positive
plate what will be its velocity?
A  0.1m
2
Part A
20C
?
4m
v i  25 m
s
 20C
m  1.5kg, q  5C
As the object moves between the plates of the capacitor the only
forces acting on it are the gravitational force and the electric force,
both of which are conservative. Therefore, during its motion the
total mechanical energy of the object remains constant. If we
choose the negative plate as the reference level for both the
gravitational and electric potential energies, the initial energy of
the object is totally kinetic.
As it rises the kinetic energy is converted into gravitational and
electrical potential energy. This will continue until either all of the
kinetic energy has been converted to potential energy or the
object reaches the upper positive plate. If it does not reach the
upper positive plate, there is a point where the total mechanical
energy is all potential. In either case the total mechanical energy
of the object remains constant.
Ei  E f
1 mv2  1 mv2  mgh  q
i
f
f
f
2
2
q hf
f   A
o
We can find the value of hf where the object stops, vf = 0. If this is
greater than the distance between the plates (4m) the object will
reach the upper plate.
=0
1 mv2  1 mv2  mgh  q
i
f
f
f
2
2
1 mv2  mgh  q
i
f
f
2
1 mv 2  mgh  qq h f
i
f
2
oA
qq 
1 mv2  

mg


h
i
2
 o A  f

1 mv2
i
2
h f  
qq 
mg 

 o A 

2
m
.51.5kg  25 s 
hf 
6
6
5  10 C 20  10 C


m
1.5kg 9.8 2 
2 
 s  
12 C
2
8.85  10

0.1m 

2

Nm 
hf  3.67m
The object does NOT reach the upper plate. It stops 0.33m from
the positive plate
Part B
 20C
m  1.5kg, q  5C
20C
?
v i  25 m
s
4m
A  0.1m
2
 20C
q
20C
Electric Field
vi
FE
FG
vy
Gravitational
Field
vx
As the object moves to the right two things happen:
1. Horizontally (FE is opposite to vx)
Kinetic energy is converted into electrical potential energy.
2. Vertically (FG is in the same direction as vy)
Gravitational potential energy is converted into kinetic energy.
Since both forces are conservative the total mechanical energy will
remain constant.
Horizontal Motion
Ei,x  Ef ,x
Ignore Gravitational Potential Energy
1 mv2  q  1 mv2  q
i,x
E ,i
f ,x
E,f
2
2
1 mv  q  1 mv2  q qx

f ,x
2
2
oA
2
i,x
0
Where x equals object' s horizontal displacement.
1 mv2  1 mv0 2  q q x
i,x
f ,x
2
2
oA
If when vx = 0 x > 4m the object will reach the positive plate.
1 mv2  q q x
i,x
2
oA
2  o A 
1
x  2 mvi,x 

 qq 
2

12
2 
C
2
8.85  10
2 0.1m  
1
m


x  1.5kg  25 
Nm
2
s 
6
6

5

10
C
20

10
C

 
 


x  4.15m  4m
The object reaches the positive plate.
At the positive plate:
1 mv2  1 mv2  q q d
i,x
f ,x
2
2
oA
Where d equals the distance between the plates.
vf ,x 
v f ,x 
qq d
mv  2  A
o
m
2
i ,x


6
6
2
 5  10 C20  10 C 4m 
m
1.5kg25 s   2 

2
12
2
C
8.85 10
2  0.1m  


N m
1.5kg
vf ,x  4.73 m
s
Time required to reach the positive plate:
m  4.73 m
25
vi  v f
s
s  14.9 m
v= 2

2
s
4m  .27s
t d

v 14.9 m
s
Vertical Motion
0
v f,y  v i,y  gt
vf,y
v f,y  gt


m
 9.8 2 .27s

s 
vf,y  2.65 m
s
E i,y  E f ,y
Ignore Electric Potential Energy
1 mv02  mgh0  1 mv2  mgh
i,y
i
f ,y
f
2
2
2
1
mghf   mvf ,y
2
2
v f ,y
hf   2g
2
m
 2.65 s 
hf   

m
29.8 2 
 s 
hf  .36m
Final Velocity
vf,x  4.73 m
s
vf,y  2.65 m
s
v f  v 2f ,x  v 2f ,y
2
2
m
m
vf   4.73 s    2.65 s 
vf  5.42 m
s
  360  tan
1
v f,y
vf,x
2.65 m
1 
s
  360  tan
4.73 m

s
  331
vf  5.42 m
s @ 331



