FALL 2009 – 2010
PHYS-102 I4
Electrostatic Potential
NEBRASKA WESLEYAN UNIVERSITY
NAME: ___________________________
SECTION:
8am
10am
1pm
3pm
Electrostatic Potential Difference. Every particle that displaces, or moves from some initial position r1 to some
final position r2, moves along a path. If the particle is acted upon a conservative force such as gravity Fg, or the
Coulomb (electrostatic) force FE, then the path integral of that force along the path is minus the potential energy change.
Path integration means selecting a small path element dl (short segment of the path), multiplying it by the component of the
force lying along the path element’s direction (using the dot product formalism F·dl), and adding up all such contributions
along the entire path. The path integral is called the work done by the force. The “minus” sign for the potential energy
change comes in because when a conservative force does work, its potential energy decreases. Using these ideas, the
electrostatic potential energy change UE can be expressed as
U E  

FE  dl
(1)
path
The UE has the unit of J (like any potential energy change). In electrostatic theory, it is greatly preferred to work with the
concept of electric field E instead of force. To convert, we divide both sides of the equation by q, the charge on the particle
under consideration, to get
V  

E  dl
(2)
path
The quantity V is known as the electrostatic potential (which is short for the electrostatic potential energy per unit charge), or
just potential. It has units of J/C = V, or volt. For this reason, the potential V is also called the voltage, especially within the
context of electric circuits. Just like potential energy, it is only the change in potential energy that is physically relevant.
Imagine a positively-charged point particle, and several paths A, B, C, etc. as
shown in the accompanying figure. Notice that the paths have a direction, which
indicates the direction of dl, and a beginning and ending point. Predict whether V will
be positive, negative, or zero for the paths shown, and enter your response in the table
below, under the column titled “PREDICTION.” You will verify the responses using the
EM Field simulator as instructed in the next activity.
B
C
A
+
PATH
A
PREDICTION:
SIMULATION:
POSITIVE, NEGATIVE,
OR ZERO?
POSITIVE, NEGATIVE, OR
ZERO?
D
E
B
C
D
E
DEAN SIEGLAFF
CREATED: 28 JAN 2010
MODIFIED:
1 of 5
PHYS-102 I4
Electrostatic Potential
FALL 2009 – 2010
NEBRASKA WESLEYAN UNIVERSITY
The EM Field application offers us the ability to draw paths and assess the potential difference along those paths. By
following the instructions below, fill in the final column in the table above.



Launch the EM Field application from Start → All Programs → EM Field 6 (or find the short cut in the Course
Software folder).
ITEM
VALUE OR
CONDITION
Sources → 3D point charges
select
Presents palette of point charges to drag and drop onto work area.
Display → Show grid
select
Places a grid in space for convenience.
Display → Constrain to grid
select
Snaps dropped bodies to grid.
DESCRIPTION
Select a +8 charge and drop it onto the workspace.
Create the paths indicated in the previous activity.
What color does EM Field use to indicate a positive change in potential? A negative change in potential?
Let us say that a path is drawn from an initial point r1 to a final point r2. What is the condition upon these two points such
that a positive change in potential results? A negative change in potential results?
In the space below, draw three paths A, B, and C in the vicinity of the positively-charged particle. Let A be a straight line
segment. Let B be a curved line segment. Let C be a loop-dee-loop-ing path. All must have the same V.
+
DEAN SIEGLAFF
2 of 5
PHYS-102 I4
Electrostatic Potential
NEBRASKA WESLEYAN UNIVERSITY
FALL 2009 – 2010
What do all of your paths have in common?
Is the change in the electrostatic potential path dependent or path independent?
Deriving the Formula for the Potential of a Point Charge. As was probably noticed, the farther away from the
point charge one was, the more difficult it was to “pick up” any V. This of course is because E itself decreases as
the inverse square of the distance r from the point charge. As we get very far away from the point charge, or in other words
approach infinity, in any direction, the field and the potential difference for any path out that far will be zero. This makes
infinity a convenient reference point from which to measure the potential everywhere in space around the point charge.
In the space below, follow your instructor as he sets up the path integral from infinity to some finite radial distance R, and
shows how the potential for a point charge relative to infinity can be derived.
What is the formula for the potential V of a point charge, assuming V() = 0?
DEAN SIEGLAFF
3 of 5
PHYS-102 I4
Electrostatic Potential
FALL 2009 – 2010
NEBRASKA WESLEYAN UNIVERSITY
Electrostatic Potential and Potential Energy. Particle A has charge qA = 4 nC, and particle B has charge qA = 4 nC. Let A lie at a position rA  1m yˆ and B lie at a position rB   1m  yˆ . Model the system in the PhET
simulation Charges and Fields (found at http://phet.colorado.edu -> Play with sims -> Physics -> Electricity, Magnets, and
Circuits). Determine the electrostatic potential V at the five points indicated in the table below, assuming that the potential at
infinity is zero. Let the point of interest in each case be referred to as P. Do this by computing the potential at P due to each
of the charged particles A and B separately, then adding them (by the principle of superposition). The distance rPA is the
distance to point P from particle A. Use the space below to express some sample calculations. Be sure to express the
formula for the electrostatic potential V(r) at a distance r from a point charge. In the final column, compute the amount of
work it would take an external agent to move a q = 1 C charged particle from infinity to point P. No calculation allowed on
the last two rows. Use the space below for your work.
xP
(m)
yP
(m)
1.0
0
1.5
0.5
2.0
-1.0
1.5
-0.5
2.0
1.0
rPA (m)
VA (V)
rPB (m)
VB (V)
V = VA + VB
(V)
V (V) from
PhET Sim
Wext (J)
Why is computing the electrostatic potential easier than determining the electric field?
DEAN SIEGLAFF
4 of 5
PHYS-102 I4
Electrostatic Potential
FALL 2009 – 2010
NEBRASKA WESLEYAN UNIVERSITY
Finding Electric Field by Potential Difference. The change in the electric potential is defined as “minus the line
integral of the electric field” along some path of motion:
V    E  dl
(0.1)
If the path taken is short, then the electric field can be considered to be constant, and the integral expression can be simplified
to
 E  dl   E xˆ  E yˆ     x  xˆ   y  yˆ   E  x   E  y 
x
y
x
y
(0.2)
(remembering that xˆ  xˆ  yˆ  yˆ  1 and xˆ  yˆ  yˆ  xˆ  0 ). What this means is if we choose a small horizontal path (for
which y = 0), then the x component of the electric field would be just Ex = –V/x. Or if we choose a small vertical path
(for which x = 0), then the y component of the electric field would be just Ey = –V/y. Notice by this analysis, the electric
field can be thought of as having the unit V/m.
Using the PhET simulator digital potential meter, determine the electric field at the first three points identified in
the previous exercise (whose positions are repeated in the table below). Use horizontal and vertical paths that
cross the point of interest and have positive displacement x =y = 0.2 m. Remember that  means “final minus initial.”
The final V is from where the path ends. The initial V is from where the path begins. Then compare your computed electric
field with that using an electric field sensor.
xP
(m)
yP
(m)
1.0
0
1.5
0.5
2.0
-1.0
DEAN SIEGLAFF
V (V)
horizontal
path
V (V)
vertical path
Ex
(V/m)
Ey
(V/m)
by potential
difference
by potential
difference
canonical
E (V/m)
magnitude

(deg)

E (V/m)
canonical
magnitude by
simulation
(deg) by
simulation
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