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Today 2/20
E-Field and Coulomb’s Law 18.4-5
HW:
“2/20 E Field 1” Due Monday 2/24
Next week’s lab: “Electrostatics”
Electric Forces
What’s the force on this
charge?
The net force caused by all the other charges!
Electric Field
What’s the Electric
Field at this location?
The net Field caused by all the other charges!
Electric Field “E”
A condition of space caused by the presence
of charges.
The field has both size and direction. (it’s a
vector)
Far away the field is weak and near charges
the field is strong.
A charge placed in an electric field will feel
a force.
Charges both cause and react to electric
fields
Force from the Electric Field
F = qE
Put charge q in an E field and it feels a force F
This actually serves to define what we mean by
the E field since we measure E with F and q.
Note: q is NOT the charge that is causing E !!
(more later)
Electric Field
F is in units of:
q is in units of:
Newtons (N)
Coulombs (C)
So E is in units of: Newtons/Coulomb (N/C)
E tells us how many Newtons of force are on q
for every Coulomb of charge in q.
F = qE
just like W
= mg
E and g are both field strengths and are vectors!
Electric Field
E and F are both vectors. How are their
directions related?
If q is a positive charge, then F points in the
same direction as E.
If q is a negative charge, then F points in the
opposite direction as E.
This is different than gravity where there is no
repulsion.
Example
At the location marked with an x, the electric
field is 2000 N/C and points right. What is the
electric force (size and direction) on a 6 x 10-6 C
charge that is placed at the x?
+q
F
E
F = qE = (2000)(6x10-6) = 1.2 x 10-2 N (to the right)
Example cont.
What if the charge were the same size but
negative?
-q
F
E
Same size F = 1.2 x 10-2 N (to the left)
What if a charge were placed somewhere else?
E
F = Who Knows?
Must know E at the new location, where q is.
Example
A charge of -5 x 10-8 C feels a force of 2.5 x 10-1 N
to the right. What is the electric field (magnitude
and direction) at the charge’s location?
-q
E
F
F = qE , E = 5 x 106 N/C
(to the left)
This is the field at the location of q. We know
nothing about the field at other locations.
Electric Field
Electric fields are caused by other charges in
other locations. It is very important to separate
the charges causing the field from the one feeling
the effects of the field.
Field at q’s
F  qE
Force on q due
to the field, E at
q’s location
The charge that is
feeling the force
location due to
other charges at
other locations
Electric Fields
A positive charge is placed as shown below.
What direction does the electric field point at the
x?
Qsource
E
A positive charge placed at the x would feel a
force to the right (repulsion), so the electric
field at the x must also points right. For +Q the
E field points away from Q.
Electric Fields
The electric field caused by a positive charges
points away from the positive charge.
E
The electric field caused by a negative charge
points toward the negative charge.
E
Coulomb’s Law
kQ source kQ s
The size of E at x is given by: E 
 2
2
r
r
Qs
E
r
k = 9 x 109 Nm2/C2
r is the distance from Qs to the x.
Example
A 2.5 x 10-6 C charge is placed as shown below.
What is the electric field at a point 5 cm to the
right?
5 cm
Which way does E point?
Qs
E = kQs /r2 = (9 x 109)(2.5 x 10-6) /(5 x 10-2)2
= 9 x 106 N/C
Two point charges +Q and -Q are fixed in place a
distance 2d apart as shown. What direction is the
electric field at the midpoint between the charges?
+Q
d
d
-Q
Student 2: “The electric field is given by E=kQs/r2 so
if do the calculation I get:
Enet = k(+Q)/d2 + k(-Q)/d2 = 0
So, the electric field is zero and has no direction.”
What do you think?
Example
What is the electric field strength at the
location of Q1 due to Q2?
0.20 m
Q1
Which charge do we care about?
Q2
Example
What is the electric field strength at the
location of Q1 due to Q2?
0.20 m
Q2
Which charge do we care about?
How does this change the problem?
Example
Note that the minus sign on Q2 only gives direction
of the E field. It does not matter in the equation.
1350 N/C
Q2 (-6 x 10-9C)
E
= kQs /r2
= (9 x 109)(6 x 10-9) /(0.20)2
= 1350 N/C
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)
Find the E field at Q1’s location.
k Qs
EQ2,x =
r2
k 2qo
(4ro)2
1 kqo
8 ro 2
Up, (away
from Q2)
k Qs
EQ3,x =
r2
k 6qo
(6ro)2
1 kqo
6 ro 2
Down,
(toward Q3)
Add vectors to get
Enet,x
1
6
1 kqo
8 ro 2
1 kqo
24 ro2
Down at the x