Today 2/21
E-Field and Coulomb’s Law 18.4-5
Examples
HW:
“2/21 2-D E Field ” Due Wednesday 2/26
E = kQs /r2
Many Charge Example (Like HW)
Q1
Q2
Q3
Each Square is ro on a side, Q1 = -4qo, Q2 = +2qo, Q3 = -6qo
(See how this simplifies the math)
q1
Find the E field at Q1’s location.
k Qs
EQ2,x =
r2
k 2qo
(4ro)2
1 kqo
8 ro 2
k Qs
EQ3,x =
r2
k 6qo
(6ro)2
1 kqo
6 ro 2
4 kqo2
FOthers,q1 = q1Enet,x
Add vectors to get
Enet,x
1
6
1 kqo
8 ro 2
24 ro2
1 kqo
24 ro2
Up, (away
from Q2)
Down,
(toward Q3)
Down at the x
E = kQs /r2
Many Charge Example (Like HW)
Q1
Q2
q2
Q3
Each Square is ro on a side, Q1 = -4qo, Q2 = +2qo, Q3 = -6qo
(See how this simplifies the math)
Find the E field at Q2’s location.
k Qs
EQ1,x =
r2
k 4qo
(4ro)2
1 kqo
4 ro 2
k Qs
EQ3,x =
r2
k 6qo
(2ro)2
3 kqo
2 ro 2
Up,
(toward Q1)
Down,
(toward Q3)
5 kqo2
FOthers,q2 = q2Enet,x
2
2
r
o
Add vectors to get
3
1 kqo
5 kqo
Enet,x
Down at the x
2
4 ro 2
4 ro 2
E = kQs /r2
Many Charge Example
+qo
-2qo
+4qo
-3qo
What is the force on +4qo? (direction also)
Find the field at +4qo due to the “other”
charges. These are the “source” charges.
ro
E+q = kQs/r2 = k(qo)/(ro)2 = Eo
Direction?
Away from + so right
E-2q = kQs/r2 = k(2qo)/(2ro)2 = k2qo/2(ro)2= k(qo)/(ro)2 = Eo
E-3q = kQs/r2 = k(3qo)/(ro)2 = 3kqo/(ro)2 = 3Eo
E = kQs /r2
Many Charge Example
+qo
+4qo
What is the force on +4qo? (direction also)
E+q = Eo
E-2q = Eo
E-3q = 3Eo
ro Now add E-Field vectors
-2qo
-3qo
Ey = E-2q,y + E-3q,y
Ey = Eosin45 + 3Eo
Ex = E+q,x + E-2q,x
Ex = Eo - Eosin45
Ex = 0.29Eo
 = tan-1(0.29/3.7) = 4.5°
Ey = 3.7Eo
Fon +4q = (4qo)3.71Eo = 14.84qoEo
Etotal = 3.72 + 0.292 Eo = 3.71Eo Fon +4q = 14.84kqo2/ro2
same direction as Etotal