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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 for every Coulomb of charge in q.
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?
F
-
q E
Same size
F =
1.2 x 10 -2 N (to the left ) What if a charge were placed somewhere else?
Must know
E E F =
Who Knows?
at the new location, where
q
is.
Example
A charge of -5 x 10 to the right. What is the electric field (magnitude
E
-
q F
-8 C feels a force of 2.5 x 10
F = qE , E =
5 x 10 6 N/C (to the left) -1 N 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.
F
qE
Field at
q
’s location due to other charges at other locations Force on
q
due to the field,
E at q’s location
The charge that is feeling the force
Electric Fields
A positive What direction does the electric field point at the x?
Q
charge is placed as shown below.
source 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
The size of
E
at x is given by:
E Q s
kQ source r
2
E
r
kQ s r
2
k =
9 x 10 9 Nm 2 /C 2
r
is the distance from
Q s
to the x.
Example
A 2.5 x 10 What is the electric field at a point 5 cm to the right? -6 C charge is placed as shown below. 5 cm Which way does
E
point?
Q s E = kQ s /r 2 =
(9 x 10 9 )(2.5 x 10 -6)
/
(5 x 10 -2 ) 2 = 9 x 10 6 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=kQ s /r 2 if do the calculation I get: so E net = k(+Q)/d 2 + k(-Q)/d 2 = 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 Q 1 due to Q 2 ?
0.20 m Q 1 Q 2
Which charge do we care about?
Example
What is the electric field strength at the location of Q 1 due to Q 2 ?
0.20 m Q 2
Which charge do we care about?
How does this change the problem?
Example
E
Note that the minus sign on Q
2
only gives direction of the E field. It does not matter in the equation.
1350 N/C Q 2
(-6 x 10 -9 C)
= kQ s /r 2 =
(9 x 10 9 )(6 x 10 -9 )
/
(0.20) 2 = 1350 N
/
C
E = kQ s /r 2
Many Charge Example (Like HW)
Q 1
Each Square is r
o
on a side, Q
1
= -4q
o
, Q
2
= +2q o,
Q 3 (See how this simplifies the math)
= -6q
o
Find the E field at Q
1
’s location.
E Q2,x =
k Q r
2
s
k 2q (4r
o
)
o
2 1 8
kq r o o
2 Up, (away from Q 2 )
Q 2 Q 3
E Q3,x =
k Q s r
2 k 6q
o
(6r
o
) 2 1 6
kq o r o
2 Down, (toward Q 3 ) Add vectors to get E net,x 1 6 1 8
kq o r o
2 1 24
kq o r o
2 Down at the x