### Chapter 21- Examples

```Chapter 21- Examples
1
Problem
Two free point charges +q and +4q are
located a distance, L, apart. A third
charge is placed so that the whole
system is in equilibrium
Find the location, magnitude, and sign of
the third charge
2
Step 1: Draw IT!
+q
L
+4q
3
1
+q
2
L
+4q
3
There are 3 positions available for the new charge (marked in
blue).
Which position do you think is the most likely position and more
importantly why?
Positions 1 & 3 are untenable since by equilibrium, we mean the
force on each particle is zero. If 1, for example, then there will be
a repulsive force on +4q that cannot be overcome without creating
a strong attractive force on +q
4
Step 3: Now a free body problem
+q
2
+4q
L
x
F1
F2
Call the new charge, q0, and let it be distance
x from +q
The free-body diagram shows relationship of
the two forces, F1 and F2 on the new charge
qq0
4qq0
k 2 k
x
( L  x) 2
1
4

2
x
( L  x) 2
( L  x) 2  4 x 2
L  x  2x
L
x
3
5
Step 4: Use the other equilibrium
F4
+q
2
+4q
L
x
F1
The free-body diagram shows relationship of
the two forces, F1 and F4 on charge +q
qq0
4q  q

k
 0 where
x2
( L) 2
q0  4 q

2
( L) 2
L
9
9 q 0  4 q
k
L
x
3
4
q0   q
9
6
Problem
An electron is projected with an initial speed of v0, 1.6 x
106 m/s in the uniform field between the parallel plates
shown below. Assume that the plates are uniform and
the field is directed vertically downward and the field
outside the plates is zero. The electron enters the
field at a point midway between the two plates
If the electron just misses the upper plate as it emerges
from the field, find the magnitude of the electric field
2.0 cm
1.0 cm
e-
7
Step 1: Draw It
Obviously, the electron is 0.5 cm below the
upper plate when entering the field so, the
initial position is (0,0.5) and the final position is
(2.0, 0.0)
SI: (0,0.005) and (0.02, 0)
Also, v0 is actually in the x-direction so vinitial=
(v0,0)
2.0 cm
1.0 cm
e-
8
This reminds me of a projectile problem except it is kinda upside
down.
In this case, there is no acceleration in the x-direction only in the y
and since F=ma=eE then a=eE/m
Where e=charge of electron and m is the mass of the electron.
Since I can never remember the trajectory equation, then I must
solve for the time it takes the electron to traverse 2 cm and then
plug that into the an equation relating acceleration to distance in
the y-direction
2.0 cm
1.0 cm
e-
9
Step 3: Equation Time
2.0 cm
1.0 cm
e-
x  v0t  x0
0.02  (1.6 106 )t  0
t  1.25108 s
1 2
at  y0
2
y  y0  .005m
y
0.010  a(1.25108 s) 2
a  6.4 1013 m / s 2
m  9.11 1031 kg
and
e  1.602 1019 C
eE
ma
a
or
E
m
e
(9.11 1031 )(6.4 1013 )
E
19
1.60210
N
364  E
C
10
Problem
What is the magnitude and direction of the
electric field at the center of the square if
q= 1.0 x 10-8 C and distance, a, is 5 cm
+q
a
a
-q
-2q
a
a
+2q
11
Field Lines
+q
a
-2q
Free body diagram of + test
charge in center
2q
a
-2q
a
-q
q
-q
a
+2q
12
Need break down into x and y
components
First find the magnitudes of all the
electric fields
2q
E2 q  k 2
r
q
Eq  k 2
r
2
2
a2
1  1 
2
where r   a    a  
2
2  2 
13
For example, the field line from +2q
The line from +2q makes an angle of
135
1350 with the positive x-axis
 So
 E2qx=E2q cos (1350)=-E2q cos (450)

0


E2qx=-0.707*E2q
E2qy=E2q sin (1350)=E2q sin (450)

E2qy=0.707*E2q
14
And from -2q, (at 450)
E-2qx=0.707*E2q
 E-2qy=0.707*E2q

15
And –q and +q

-q makes angle of
2250
 E-qx=-0.707*Eq
 E-qy=-0.707*Eq

-q makes angle of
3150
 Eqx=0.707*Eq
 Eqy=-0.707*Eq
Now we add the x and y components, respectively
16
Ex and Ey
2q
E2 q  k 2
r
q
Eq  k 2
r
a2
where r 
2
E2 q  2 Eq  k
Eq  k
2
E x  E2 qx  E 2 qx  Eqx  E qx
E x  0.707* ( E2 q  E2 q  Eq  Eq )
Ex  0
E y  E2 qy  E 2 qy  Eqy  E qy
E y  0.707* ( E2 q  E2 q  Eq  Eq )
4q
a2
E y  0.707* (2 Eq  2 Eq  Eq  Eq )
E y  2 Eq  2 k
2q
a2
2q
a2

2q
E  E x2  E y2  2k 2
a
tan 
Ey
Ex
 Undefined
  tan1 (undefined)  900
17
```