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Electric Fields
• Review of gravitational fields
=
;
• Electric field vector
<=
/div>
• Electric fields for various charge
configurations
• Field strengths for point charges =
and
uniform fields
• Work done by fields & change in
potential energy
• Potential & equipotential
surfaces
• Capacitors, capacitance, & vol=
tage
drops across capacitors
• Millikan oil drop experiment
<=
/span>
• Excess Charge Distribution on a Conductor
Gravi=
tational
Fields: Review
Recall that surrounding any object with mass, or
collection of objects with mass, is a gravitational field. Any mass placed in a gravitat=
ional
field will experience a gravitational
force. We defined the field strength as the gravitational force per unit mass on any
“test mass” placed in the field: g =3D
F=
/
m=
. &nbs=
p;
g<=
/b> is a vector that points in the direction of the net gravitational force; its =
units
are N / kg. <=
span
style=3D'font-size:58%'> F is the vect=
or force
on the test mass, and m is the test mass, a scalar. g and =
F are always parallel. The strength of the field is independent of the test m=
ass.
For example, near Earth’s surface
mg / m =3D g
=3D 9.8 N / kg for any mass=
. Some
fields are uniform (parallel, equally
spaced fields lines). Nonuniform fields are stronger where the field lines=
are
closer together.=
![](3D"slide0002_image001.gif")
Earth
Earth’s surface
10
kg
98 N
![](3D"slide0002_image002.gif")
m
![](3D"slide0002_image003.gif")
F
uniform field
nonuniform
fie= ld
fie= ld
Electr=
ic
Fields: Intro
Surrounding any object with charge, or collection =
of
objects with charge, is a electric field.
Any charge placed in an electric field will experience a electrical force.=
We defined the field strength as the electric force per unit charge on any “test charge” placed in the field:
E =3D
F=
/
q=
. E is a vector that points, by definition, in the direction of the net electric force on a positive
charge; its units are N / C. F is the vector force on the test charge, and q is the test charge, a scalar. E and =
F are only parallel
if the test charge is positive. Some fields are uniform (parallel, equally=
spaced fields lines) such as the field on the left
formed by a sheet of negative charge. Nonuniform fields are stronger where the field lin=
es are
closer together, such as the field
on the right produced by a sphere of negative charge.
![](3D"slide0083_image004.gif")
-
![](3D"slide0083_image005.gif")
- - - - - - - - - - - - - -
![](3D"slide0083_image006.gif")
![](3D"slide0083_image007.gif")
![](3D"slide0083_image008.gif")
![](3D"slide0083_image009.gif")
![](3D"slide0083_image010.gif")
![](3D"slide0083_image011.gif")
![](3D"slide0083_image012.gif")
q
F
q
![](3D"slide0083_image013.gif")
F
uniform field
nonuniform
fie= ld
fie= ld
![](3D"slide0083_image014.gif")
+
![](3D"slide0083_image015.gif")
+
Overvi=
ew of
Fields
Charge,
like mass, is an intrinsic property of an object. Charges produce electric=
fields that affect other charges; masses produce
gravitational fields that affect other masses. Gravitational fields lines always point to=
ward
an isolated mass. Unlike mass,
though, charges can be positive or negative. Electric field lines emanate =
from
positive charges and penetrate=
into
negative charge.
We refer to the charge produci= ng a field as a field charge= . A group of field charges can produce very nonuniform fields. To determine t= he strength of the field at a particular point, we place a small, positive test charge in the field. We then measur= e the electric force on it and divide by the test ch= arge:
We refer to the charge produci= ng a field as a field charge= . A group of field charges can produce very nonuniform fields. To determine t= he strength of the field at a particular point, we place a small, positive test charge in the field. We then measur= e the electric force on it and divide by the test ch= arge:
For an isolated positive field charge=
, the
field lines point away from the field charge (since the force on a positive charge would=
be
away from the field charge). The
opposite is true for an isolated negative field charge. No matter how comp=
lex the field, the electric force on a test charge is =
always
tangent to the field line at that point.
The coming slides will reiterate these ideas and p=
rovide
examples.
E
=3D F / q.
![](3D"slide0086_image016.gif")
Electric
& Gravitational Fields Compared
N /<=
/span> C<=
/span>
N /<=
/span> kg=
SI units
/
/
=3D
=3D
q
FE
E
m
W
g =
;
Intrinsic Property
Force
Field strength
![](3D"slide0086_image017.gif")
Gravity:
Electric Force:
Field strength is given by per unit mass or force =
per
unit charge, depen=
ding
on the type of field. Field strength means the magnitude of a field vector. Ex #1: If a +10 C charge is placed i=
n an
electric field and exper=
iences
a 50 N force, the field strength at the location of the charge is <=
/span>5 N/C. The electric field vector is given by: E =3D 5 N/C, where the direction of this vector is parallel to the force =
vector
(and the field lines). Ex
#2: If a -10 C charge experiences a 50 N force, E =
=3D 5
N/C in a direction opposi=
te the
force vector (opposite the direction of the field lines).
![](3D"slide0059_image018.gif")
Electr=
ic
Field Example Problem
A
sphere of mass 1.3 grams is charged via friction, and in the process
excess electrons are rubbed onto it, giving the sphere a charge
of - 4.8
μC. The sphere is then pla=
ced
into an external uniform
electric field of 6
N/C directed to the right. The sphere is released from rest. What =
is its
displacement after 15 s?
(Hints on next
slide.)
![](3D"slide0059_image019.gif")
![](3D"slide0059_image020.gif")
![](3D"slide0059_image021.gif")
![](3D"slide0059_image022.gif")
![](3D"slide0059_image023.gif")
![](3D"slide0059_image024.gif")
E
-
![](3D"slide0087_image025.gif")
Sample
Problem Hints
![](3D"slide0087_image026.gif")
![](3D"slide0087_image027.gif")
![](3D"slide0087_image028.gif")
![](3D"slide0087_image029.gif")
![](3D"slide0087_image030.gif")
![](3D"slide0087_image031.gif")
q E
m g
E
![](3D"slide0087_image032.gif")
-
1.Draw
a vector as shown. Note that F E =3D q E=
,
by definition of E,
and that FE is to
the left (opposite E ) <=
span
style=3D'position:absolute;top:35.0%;left:8.42%;width:48.12%;height:5.25%'=
>since
the charge is negative.
2.Instead
of finding the net force (which would work), compute the acceleration
due to each force separately.
<=
span
style=3D'visibility:hidden'>3.
3. Find the displacement due to each=
force
using the time given and
kinematics.
kinematics.
4. Add the displacement vectors to f=
ind
the net displacement
vector.
vector.
![](3D"slide0021_image033.gif")
Drawing=
an E Field
for a Point Charge
![](3D"slide0021_image034.gif")
![](3D"slide0021_image035.gif")
+
![](3D"slide0021_image036.gif")
+
Let’s use the idea of a test charge to produ=
ce the
E field for an isolated positive field charge. We place small, positive test charges in t=
he
vicinity of the field and draw the force
vector on each. Note that the closer the test charge is to the field charg=
e,
the greater the forc=
e, but
all force vectors are directed radially outward from the field charge. At any point near the field charge, the fo=
rce
vector points in the direction of the
electric field. Thus we have a field that looks like a sea urchin, with fi=
eld
lines radiating outwar=
d from
the field charge to infinity in all direction, not just in a plane. The number of field lines drawn in arbitrary, but =
they
should be evenly spaced around the
field charge. What if the field charge were negative?
![](3D"slide0021_image037.gif")
![](3D"slide0021_image038.gif")
![](3D"slide0021_image039.gif")
![](3D"slide0021_image040.gif")
![](3D"slide0021_image041.gif")
Test charges and force vectors surrounding a field charge
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Isolated, positive point charge and its electric field=
Single Positive Field Charge
![](3D"slide0012_image042.gif")
+
![](3D"slide0012_image043.gif")
Thi=
s is a
2D picture of the field line=
s that
surround a positive <=
span
style=3D'font-size:83%'>field charge that is <=
span
style=3D'font-size:83%'>either point-like or spherically symmetric.
Not shown are field l=
ines <=
span
style=3D'font-size:83%'>going out of and into <=
span
style=3D'font-size:83%'>the page. Keep in mind that the field lines
radiate outwards because,=
<=
span
style=3D'font-size:83%'>by definition, an electric field vector points
in the direction of the =
force <=
span
style=3D'font-size:83%'>on a positive test charge.
The
nearer you get to the <=
span
style=3D'font-size:92%'>charge, the more
uniform and stronger <=
span
style=3D'font-size:92%'>the field. =
<=
span
style=3D'font-size:92%'>Farther away the
field strength gets <=
span
style=3D'font-size:92%'>weaker, as indicated
by the field lines =
becoming more
spread out.
![](3D"slide0013_image044.gif")
Single
Negative Field Charge
-
The field surrounding an isolated, negative point (or spherically symmetric) charge looks just like that of an isolated positive charge except the field lines are directed toward the field charge. This is because, by definition, an electric field vector points in the direction of the force on a positive test charge, which, in this case is toward the field charge. As before, the field is stronger where the field lines are closer together, and the force vector on a test charge is parallel to the field.
![](3D"slide0085_image045.gif")
Point
Charges of Different Magnitudes
+ 1
Let’s compare the fields on two separate iso=
lated
point charges, one with a charge
of +1 unit, the other with a charge of +2 units. It doesn’t matter h=
ow many field lines we draw emanating from the +1 cha=
rge so
long as we draw twice=
as
many line coming from the +2 charge. This means, at a given =
distance, the strength of the E fiel=
d for
the +2 charge is twice that for the +1 charge.
=
+ 2
Equal =
but
Opposite Field Charges
Pictured
is the electric field produced by two equal but opposite charges.
Because the charges are of the same magnitude, the field is symmetric.
Note that all the lines that emanate from the positive charge
land on the negative charge. Also pictured is a small positive charge
placed in the field and the force vector on it at that position. This
is the vector sum of the forces exerted on the test charge by each =
field
charge. Note that the net force vector is tangen=
t to
the field line. This is always the case. In fact, t=
he
field is defined by the direction of net force vectors on test
charges at
various places. The net force on a
negative test charge is tangent to the
field as well, but it points in the
opposite direction of the field.
(Continued on next slide.)
various places. The net force on a
negative test charge is tangent to the
field as well, but it points in the
opposite direction of the field.
(Continued on next slide.)
![](3D"slide0052_image047.gif")
![](3D"slide0052_image048.gif")
+
-
![](3D"slide0052_image049.gif")
![](3D"slide0052_image050.gif")
Link #3