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Electrostatics #3
The Electric Field
HW #2, last page of handout
I. Electric Field: Day #1
Introduction:
When scientists try to study an unknown arrangement of charges, the
only tool available is to place a test charge near the distribution and
measure the force on the test charge. The downside to this technique is
that the test charge might cause a change to the distribution of charge.
A test charge is placed near an
object to measure the force on
the test charge.
Example of how the presence of a test charge may change
the distribution of the charges in the original object.
The electric field is a mathematical tool used by physicists to represent
the strength of electric forces available at a given location without the
physical interaction of a real charge at that location.
An imaginary positive test charge, q, is placed at some location near a
distribution of charge. A force F would be exerted on this test charge.
The electric field measures the force available at that point in space, but
represents it as the force per unit charge. In other words:
F
E
q
Units for the electric field:
force per charge =
newton N

coulomb C
Point Charges as Sources of Electric Fields:
The electric field created by a point charge (or spherical distribution of
charge) will point radially outwards from or inwards towards the
charge.
Direction of the Electric Field:
The electric field can then be mapped around some given charge
distribution (without altering the charge distribution due to
measurements).
Forces Caused by Electric Fields: The
force on any charged particle, q, that is
placed into an electric field, is given as:


F qE
If q > 0, the force is in the same direction as
the electric field. If q < 0, then the force on
the charge points opposite of the electric field.
The strength of the electric field around a point charge is given as
follows:
The force on the positive test charge, qo, is
given from Coulomb’s law.
F
k Q qo
r2
The strength of the electric field is found by dividing the
force on the positive test charge, qo, by the value of the
test charge:
F kQ
E
 2
qo
r
The electric field depends only on the properties of and the
distance from the point charge, Q.
Electrostatics #3
The Electric Field
2nd day towards this lesson
Electric Field: Day #2
Force on a test charge:


F qE
Electric field due to a point source charge:
{always true!}
F kQ
E
 2
qo
r
Example #1: A point charge with a mass of 1.40 kg and charge of +25.0
mC is placed into a constant electric field. a. If the strength of the
electric field is 863 N/C pointing due north, what is the force on the
charged particle?
F  qE


F  25.0 10 3 C 863 CN 
F  21.6 N
Since the charge interacting with the electric field is positive, the
force also points north (same as the electric field).
b. If the particle starts from rest, how long will it take the particle to
reach 15.0 m/s?
F  ma  a 
a
F
m
21.6 N
 15.4 m s 2
1.40 kg
v  vo
v  vo  at  t 
a
15.0 ms  0
t
15.4 m s 2
t  0.973 s
Example #2: An object with a net charge of 24 μC is placed in a
uniform electric field of 610 N/C, directed vertically. What is the mass
of the object if it “floats” in the electric field?
If the mass is to float, there must be an upward force to
balance gravity!
Flift  qE
Note that if the charge is positive and the
electric force points upwards, then the electric
field must point upwards as well.
Balance the forces:
mg
qE
m
g
mg  qE

24 10 C 610 

6
9.80 
m
s2
m  1.49 103 kg
N
C
Example #3: An electron is accelerated by a constant electric field of
magnitude 300 N/C. (a) Find the acceleration of the electron. (b) Use
the equations of motion with constant acceleration to find the electron’s
speed after 1.00 × 10−8 s, assuming it starts from rest.
ma  qE
qE
a
m

1.6022 10 C 300 

9.109 10 kg
v  vo  at
19
N
C
31


 5.28 1013 m s 2
 0  5.28 1013 m s 2 1.00 108 s
v  5.28 105
m
s

Example #4: An airplane is flying through a thundercloud at a height of
2000 m. (This is a very dangerous thing to do because of updrafts,
turbulence, and the possibility of electric discharge.) If there are charge
concentrations of +40.0 C at a height of 3000 m within the
cloud and

−40.0 C at a height of 1000 m, what is the electric field E at the
aircraft?
Note that the airplane is 1000 m below
the top charge and 1000 m above the
Qtop  40.0 C
bottom charge. Both electric fields
point downwards.
Etop 
kQ
Ebottom 
r2
Etotal  Etop  Ebottom
kQ
r
2
Qbottom  40.0 C
Etotal 
kQ
r
2

kQ
r2
Etotal 
Etotal

8.988 10

9 Nm 2
C2
kQ
r
2

kQ
r2
40.0 C  8.988 10
1000 m2
9 Nm 2
C2
40.0 C
1000 m2
N
Etotal  7.19 10
C
5
Field points downwards.
Example #5: A proton accelerates from rest in a uniform electric field
of 640 N/C. At some later time, its speed is 1.20 × 106 m/s. (a) Find the
magnitude of the acceleration of the proton.
F  qE
F  ma
qE
a
m
1.6022 10 C   640 


1.67 10 kg 
19
N
C
27
a  6.14 1010 m s2
Example #5: (b) How long does it take the proton to reach this speed?
v  vo  at
v  vo
t
a
t  1.95 105 s
Example #5: (c) How far has it moved in that interval?
x  vot  12 at 2
x  0 
1
2
 6.14 10
10 m
1.95 10 s 
5
s2
x  11.7 m
2
Example #5: (d) What is its kinetic energy at the later time?
KE  12 mv 2
KE  1.20 1015 J
Example #6: Each of the protons in a particle beam has a kinetic energy
of 3.25 × 10−15 J. What are the magnitude and direction of the electric
field that will stop these protons in a distance of 1.25 m?
2KE
KEo  mvo  vo 
m
1
2
2
 1.97 106
v 2  vo 2  2ax 
m
s
v 2  vo 2
a
2x
 1.56 1012 m s2
The magnitude of the force is given as:
F  ma
F  2.60 1015 N
Set this force equal to:
F  qE
2.60 1015 N
E

q 1.6022 1019 C
F
 1.62 104
N
C
Note that the force must point opposite to the motion to slow the
proton down. Since this is a positive charge, the electric field will
point the same direction as the force. The electric field thus points
opposite to the motion of the positive charge.
For a positive charge, the electric field will point in the direction of
motion in order to speed up the positive charge.
To slow down a positive charge, the electric field must point opposite
to the motion.
For a negative charge, the electric field will point opposite to the
direction of motion in order to speed up the negative charge.
To slow down a negative charge, the electric field must point in the
same direction as the motion.
Example #7: Each of the electrons in a particle beam has a kinetic
energy of 1.60 × 10−17 J. (a) What is the magnitude of the uniform
electric field (pointing in the direction of the electrons’ movement) that
will stop these electrons in a distance of 10.0 cm? (b) How long will it
take to stop the electrons? (c) After the electrons stop, what will they
do? Explain.
KEo  12 mvo 2  Fd  qEd
E  KE / qd
 1.60 X 1017 J / (1.60 X 1019 C )(0.1m)
 999 CN
Note that the force must point opposite to the motion to slow the
electron down. Since this is a negative charge, the electric field will
point opposite to the direction of the force. Thus the electric field
points in the direction the electron moves.
Example #7: {continued…} (b) How long will it take to stop the
electrons? (c) After the electrons stop, what will they do? Explain.
2KE
KEo  mvo  vo 
m
1
2
2
v 2  vo 2  2ax 
v 2  vo 2
a
2x
0  5.93 106 ms
t
1.76  1014 m s 2
 5.93 106
m
s
 1.557 1012 m s 2
 3.37 108 s
Even after the electrons stop, the acceleration is still present. The
12 m
electrons will accelerate in the opposite direction.
Thisis
same
as
 1.557
10the
s2
throwing a ball in the air. The ball comes back down after stopping at
the highest point.
Electrostatics #3
The Electric Field
HW #3,#4, and Coulomb’s Law
worksheet will be checked Tuesday.
Example #8: Three identical charges (q = −5.0 μC) lie along a circle of
radius 2.0 m at angles of 30°, 150°, and 270°, as shown in the
figure at right. What is the resultant electric field at the center of the
circle?
The three electric fields produced are
equal in size and symmetrically
spaced in angle. Because of the
symmetry, these three fields add to
zero.
Example #9: (a) Determine the electric field strength at a point 1.00 cm
to the left of the middle charge shown in the figure below.
Q2 
Q1 
E1 
E2
kQ
r2


E1 
E2 
E3 
8.988 10
9 N m 2
C2
E3 
9 N m 2
C2
2


8.988 109
N m2
C2
 1.50 10

6
C
2
2.00 106 C
 0.0300 m 
 1.35 108
N
C
points right
 0.0100 m 
r2
r2

6.00 106 C
 0.0200 m 
k Q  8.988 10


kQ
Q3 
2
 1.35 108
N
C
points left
 2.00 107
points right
N
C
Use (+) to indicate pointing right and (–) to indicate pointing left:
Etotal  E1  E2  E3
Etotal   1.35 108
N
C
8


1.35

10
 
Etotal  2.00 107
N
C
N
C
7


2.00

10
 
N
C

points right
(b) If a charge of −2.00 μC is placed at this point, what are the
magnitude and direction of the force on it?
F  qEtotal   2.00 106 C  2.00 107
F  39.9 N
points left
N
C

The force on a negative charge is
directed opposite to the electric field.
Example #10: Three point charges are aligned along the x-axis as
shown below. Find the electric field at the position x = +2.0 m, y = 0.
E2 
E1 
Q3 
Q2 
Q1 
E1
E2
k Q  8.988 10


9 N m 2
C2
k Q  8.988 10


E3 
9 N m 2
C2


8.988 10
9 N m2
C2
C
9
C
2
3.00 109 C
1.20 m 
 5.752 CN
points left
 5.00 10

9
2
 2.00 m 
r2
r2
 4.00 10
 2.50 m 
r2
kQ
E3 
2
 11.24 CN
points right
 18.73 CN
points right
Use (+) to indicate pointing right and (–) to indicate pointing left:
Etotal  E1  E2  E3
Etotal   5.75 CN    11.24 CN    18.73 CN 
Etotal  24.2 CN
points right
Example #11: A small 2.00-g plastic ball is suspended by a 20.0-cmlong string in a uniform electric field, as shown in the figure below. If
the ball is in equilibrium when the string makes a 15.0° angle with the
vertical as indicated, what is the net charge on the ball?
balance the forces by components:
T cos15.0  mg
and
Tx
T sin15.0  qE
T
T
F  qE
Ty
mg
Divide the two force equations:
qE
tan15.0 
mg
q
T sin15.0 qE

T cos15.0 mg
mg  tan15.0

q
E
3
2.00

10
kg   9.80 m s2  tan15.0

3
1.00

10

N
C

q  5.25 106 C  5.25C
Example #12: A positively charged bead having a mass of 1.00 g falls
from rest in a vacuum from a height of 5.00 m in a uniform vertical
electric field with a magnitude of 1.00 × 104 N/C. The bead hits the
ground at a speed of 21.0 m/s. Determine (a) the direction of the electric
field (upward or downward), and (b) the charge on the bead.
Start with kinematics to find the downward acceleration of the bead.
v 2  vo 2  2ax
v  vo
a
2x
2
2
21.0    0 


2  5.00 m 
m 2
s
2
the downward acceleration is greater than
normal gravity, so an additional (electric) force
is also pushing downwards on the bead.
(a) The electric field must be pointing downwards to make a
downward force on the positively charged bead.
a  44.1 m s2
take downwards as positive…
Fnet  ma  mg  qE
mg
qE
ma  mg
q
E
1.00 10

q
3
kg   44.1 m s2  9.80 m s2 
4
1.00

10

N
C

q  3.43 106 C  3.43 C
Concept Question #1: Consider the direction of the deflection of the
charged particle as it passes through the electric field. Is the charge
positive, negative, or neutral on this particle?
Concept Question #2: What is the total electric field at the center of the
circles?
Concept Question #3: What is the total electric force on the charge at
the center of the square?
Concept Question #4: What is the total electric field at the center of the
square?