Download E=0_example

For the charge distribution shown below:
1. At what distance from the origin would the electric field = 0?
2. At what distance from the origin would the electric field = 9.625x104 N/C?
E1
E1
E2
E2
+
0
q1
E1
1m
25mC
2m
E2
+
3m
4m
q2
5m
6m
40mC
The electric field can only equal zero in the region between the
two charges.
The point will be nearer q1 since it has the smaller charge.
E1
E2
P
+
0
q1
1m
x
+
2m
3m
4-x
E1  E2
4m
q2
5m
6m
k q1
k q2
 2
2
r1,P
r2,P
q1
q2
 2
2
r1,P
r2,P
q1
q2
2 
x
(4  x) 2
q1
q2
x  (4  x)
q1  (4  x) 
q2  x
4 q1  x q 1  x q 2
4 q1  x q 2  x q 1
4 q1  x q 2 
x
q1 
4 q1
q 2  q1
x
4 25  10 C
25  10 6 C  40  106 C
x
4 25
25  40
6
x  1.77m
2. At what distance from the origin would the electric field = 9.625x104 N/C?
E1
E1
E2
E2
+
0
q1
E1
1m
25mC
2m
E2
+
3m
4m
q2
5m
6m
40mC
To the left of q1 both fields are negative therefore there is no point
to the left q1 where the electric field could be 9.625x104 N/C .
Between the two charges (0 < x < 4m) the field due to q1 is
positive and the field due to q2 is negative. We found that at a
point 1.77m from the origin the electric field is zero. There could
be a point between the origin and 1.77m with the required
electric field.
To the right of q2 (x > 4m) both fields are positive so there is a
point to the right of q2 where the total field is 9.625x104 N/C .
0 < x < 4m
E2
q1
+
P
+
0
E1
1m
x
2m
3m
4m
q2
5m
6m
4-x
E1  E2  9.625  104 N
C
k q1
k q2
4 N


9.625

10
C
r12
r22
2
2


9 N  m 
6
9 N  m 
6
9  10

25

10
C

9

10

40

10
C



2 
2 


4 N
C
C


9.625

10
C
 4  x 2
x2
2
2


9 N m
6 
9 Nm
6 
259  10
 10 C 409  10
 10 C
2
2




4 N
C
C


9.625

10
C
4  x 2
x2
9.625  10 4 N
25  40 
C
x 2  4  x 2 9  109 N  m 2  10 6 C
C2
25  40  10.69m 2
x 2  4  x 2
5 
8
2
2  2.138
4  x
x
54  x 2  8x 2
 2.138
2
2
x 4  x
516  8x  x 2   8x 2
x 16  8x  x 
2
2
 2.138
80  40x  5x 2  8x 2  2.138
16x 2  8x 3  x 4
80  40x  3x 2  34.208x 2  17.104x 3  2.138x 4
2.138x 4  17.104x 3  37.208x 2  40x  80  0
y  2.138x  17.104x  37.208x  40x  80
4
3
2
Solve by graphical analysis to find values of x for
which y = 0.
Step 1. Use a spreadsheet to create a data table.
Step 2. Copy and paste data into graphing software such as
Cricket Graph.
Step 3. Plot graph of y vs. x.
Step 4. Adjust scales of axes to determine a more precise value of x.
Step 5. Continue adjusting scales until desired precision is
achieved.
x=1.2508m
x > 4m
E2
E1
+
+
0
q1
1m
2m
3m
4m
q2
Q
5m
6m
x
x-4
E1  E2  9.625  104 N
C
k q1
k q2
4 N


9.625

10
C
r12
r22
2
2


9 N  m 
6
9 N  m 
6
9  10

25

10
C

9

10

40

10
C



2 
2 


4 N
C
C


9.625

10
C
x  4 2
x2
2
2


9 N m
6 
9 Nm
6 
259  10
 10 C 409  10
 10 C
2
2




4 N
C
C


9.625

10
C
 x  4 2
x2
9.625  10 4 N
25  40 
C
x 2 x  4 2 9  10 9 N  m 2  10 6 C
C2
25  40  10.69m 2
x 2 x  4 2
5  8x 2  2.138
x 2 x  4  2
5 x  4 2  8x 2
 2.138
2
2
x  x  4
5 x 2  8x  16   8x 2
x  x  8x  16
2
2
 2.138
5x 2  40x  80  8x 2  2.138
x 4  8x 3  16x 2
13x 2  40x  80  2.1385x 4  17.104x 3  34.208x 2
2.1385x 4  17.104x 3  21.208x 2  40x  80  0
y  2.1385x  17.104x  21.208x  40x  80
4
3
2
Solve by graphical analysis to find values of x for
which y = 0.
Step 1. Use a spreadsheet to create a data table.
Step 2. Copy and paste data into graphing software such as
Cricket Graph.
Step 3. Plot graph of y vs. x.
Step 4. Adjust scales of axes to determine a more precise value of x.
Step 5. Continue adjusting scales until desired precision is
achieved.
X = 6.0003m
0
q1
+
P
+
2m
1m
3m
4m
q2
5m
1.77m
r1,P  1.77m
r2,P  2.23m
2
r1,P 2
 0.63
r    1.77m

2.23m
 2,P 
q 1 25  10 6 C
q 2  40  10 6 C  0.625  0.63
6m