THE POTENTIAL DIFFERENCE DUE TO CONTINUOUS CHARGE DISTRIBUTIONS
The potential from a continuous charge distribution can be calculated several ways. Each method should yield approximately the same result. First, we can use an integral method in which the potential dV from each element of charge dq is integrated mathematically to give a total potential at the location of interest. Second, we can approximate the value of the potential V by summing up several finite elements of charge  q by using a computer spreadsheet or hand calculations. Finally, we can use Gauss’ law to find the electric field along with the defining equation for potential difference to set up the appropriate line integral.
Again, let’s consider a relatively simple charge distribution. In this case we will look at a ring with charge uniformly distributed on it. We will calculate the potential on the axis passing through the center of the ring as shown in the diagram below. (Later on you could find the potential difference from a disk or a sheet of charge by considering a collection of nested rings.)
Diagram showing the distance between an element on a charged ring of radius r and point P along an axis passing through the center of the ring. Write an expression for the distance between the point P and an element dq as a function of x and a.
A ring of charge has a total charge of Q = 20 µ C (or 20  10
-6
C ). The radius of the ring, a , is 30 cm. We want to find the electric field,

E , at a distance of x cm from the ring along an axis that is perpendicular to the ring and passing through its center and the potential, V . Let’s begin by calculating the potential. Hints: Since the potential is a scalar and not a vector, we can calculate the potential at point P (relative to  ) for each of the charge elements  q and add them to each other. This looks like a big deal but it is actually a trivial problem because all the charge elements are the same distance from point P .
Activity: Estimate of the Potential from a Charged Ring a.
Divide the ring into 20 elements of charge  q and calculate the total V at a distance of x = 20 cm from the center of the ring using a spreadsheet program. Summarize the result below. Be sure to take a screenshot of your spreadsheet results for your blog.

Activity: Calculation of the Potential from a Charged Ring
We know that the limit of the sum is an integral. Since we were summing over the charge, this integral must be:
By following the steps below, you can use an integral to find a more exact value of the potential. a.
Show that V
 k e
 dq r
 k e
 dq x
2  a
2 b.
Show that k e
 dq x
2  a out of the integral).
2
 k e x
2  a
2
 dq (that is, show that x 2  a 2 is a constant and can thus be pulled c.
Perform the integration in part b. above. Then substitute values for the ring radius, a , the distance to point P , x , and the total charge on the ring, Q , into the resulting expression in order to obtain a more “exact” value for the potential.
d.
How does the “numerical” value that you obtained from the spreadsheet compare with the “exact” value you obtained in c?
Now let’s take a completely different approach to this problem. If we can find the vector equation for the electric field at point P due to the ring of charge, then we can use the expression

V

V x

V

  
 x


E ds as an alternative way to find a general equation for the potential at point P .

Activity:

V from a Ring using the E-field Method a.
Show that the electric field at point P from the charged ring is given by

E

 x
2  a
2

3
2 x
   x
2  a
2

3 x

Hints: 1. There is no y -component of the E -field. Why?
2. cos  = x x
2  a
2 b.
Use the equation 
V

V
X

V

  
 x


E ds to show that

V
 x 2  a 2
. c.
Use the equation you verified in part b. to find a numerical value for  V .
d.
How does the result compare to that obtained previously?

A 16.0 cm long, horizontal rod has a uniform charge density of 2.70  C/cm. Taking the left end of the rod to be the origin:
(A) Determine the expression for the electric potential for any point in the first quadrant, and find the magnitude of the potential at the point (10.0 cm, 15.0 cm).
A positive line charge creates an electric potential in the space surrounding the line charge.
:
To find the electric potential at some point we imagine that the rod is divided into infinitesimally that act like point charges. Then the differential potential at any point will be:
Assume that you have some point (a,b) where you would like to find the potential. Write an expression for r as a function of (x, a, b)
For a uniform line charge dq =  dx along the rod. By integrating along the length of the rod we can add up the contributions of "all" of the infinitesimal point charges. Use this to rewrite the integral from a function of dq as a function of  dx. Be sure to apply appropriate limits of integration.
Now substitute the definition of r as function of x, a and b.

There are several ways that you could evaluate this integral. If you have a program like Wolfram Alpha or Maple, you could get the symbolic solution directly. For example using wolfram Alpha you get the following:
Use this indefinite integral to evaluate the definite integral found above.
Now evaluate this for the point 10cm, 15 cm.
The potential at the point (.10 m, .15 m) is (note that x - L = .10 m - .16 m = -.06 m)
One way to see if our answer is in the correct ball park is to imagine that we replace the line charge with a point charge at the center of the rod. What is the equivalent magnitude of the point charge, given that it has to be equal to the entire charge of the rod?

The magnitude of the charge is Q =

L = (2.70x10
-6 C)(.16 m) = 4.32x10
-7 C.
What is the distance from the point charge model to our desired point at (.1m , .15 m)?
The distance is given by
.
Now calculate the potential due to this point charge:
The potential at (a,b) would be,
This is not a bad approximation for the potential.
(B) Determine the work it would take to move a -22.0  C charge from a point (10.0 cm, 15.0 cm) to a point on the x axis 12.0 cm to the right of the end of the rod along a straight line between the points.
The work done to move a charge Q is just the product of the charge and the potential difference between it starting point and ending point.
Since work is path independent, it will not matter along which path the charge is move.
The external work depends upon the potential difference between the two points. What was the potential at the point (10.0 cm, 15.0 cm)?

We already found the potential at the first point to be V
1
= 24668 V. Now use the general equation we determine to find the potential at the second point, (.16m+.12m, 0) = (.28m, 0)
The work is given by points.
use this definition to find the work to move the charge between the two
Since the work done against the electric field is independent the path along which charge is moved, it will always take 89.8 mJ of energy no matter how the charge is moved between the two points.
Also observe that if the charge were positive instead of negative, the external work would have been a negative value. This would means that the charge would have gained energy going between the two points.
EQUIPOTENTIAL SURFACES
Sometimes it is possible to move along a surface without doing any work. Thus, it is possible to remain at the same potential energy anywhere along such a surface. If an electric charge can travel along a surface without doing any work, the surface is called an equipotential surface .
Consider the three different charge configurations shown below. Where are the equipotential surfaces? What shapes to they have? Hint: If you have any computer simulations available to you for drawing equipotential lines associated with electrical charges, you may want to check your guesses against the patterns drawn in one or more of the simulations.

Activity: Sketches of Electric Field Lines and Equipotentials a.
What path could you move along without doing any work—that is,

•

is always zero? What is the shape of the equipotential surface? Remember that in general you can move in three dimensions. b.
Find some equipotential surfaces for the charge configuration shown below, which consists of two charged metal plates placed parallel to each other. What is the shape of the equipotential surfaces? c.
Find some equipotential surfaces for the electric dipole charge configuration shown below. d.
In general, what is the relationship between the direction of the equipotential lines you have drawn (representing that part of the equipotential surface that lies in the plane of the paper) and the direction of the electric field lines?

 An equipotential surface is a surface upon which the potential has the same value at all points on that surface.
 In two-dimensional plots, equipotentials form closed lines that look like the contours on a relief map. The contours indicate heights of potential energy rather than heights of distance.
 In two-dimensions, the electric potential V ( x,y ) can be pictured as a surface where the potential is plotted along the z -axis. Horizontal slices at equal distance of potential along the vertical axis of the potential produce an equipotential contours when viewed from above along the axis of the potential.

In this lab you are going to measure the electric potential at various points in space (on a sheet) due to a source of electrical potential difference (power supply).
Equipment:
Cork board
Conductive paper* marked with various patterns of conductive paint.
Power supply
Voltmeter
Various pins and wire connector
Excel worksheet.
Red on power supply
Black on power supply
*Note: although the paper is called “conductive” it is not a true conductor.
Procedure:
Secure the conductive paper to the cork board with four pins. Connect each terminal of the voltage source to each of the marks of silver paint on the conductive paper.
Part 1:
V
20
COM V
Set the potential difference of the voltage source. To do this, touch (or attach) the ground terminal of the multimeter
(labeled “COM”) to the black terminal of the power supply. and touch or connect the other terminal of the multimeter to the red terminal of the power supply. Set the multimeter to measure up to 20 volts DC (V ). Adjust the power supply to
15.0 volts.
Question: When you set the power supply such that the potential difference between the two bits of metallic paint on the conductive paper is 15 volts, what physically does that quantity (15 volts) mean?
Measure the potential difference (voltage) between the two metallic paint marks on the paper. Record the value
_________.
Measure the potential difference (voltage) between two points on the lower potential conductor: ___________.
Measure the potential difference (voltage) between two points on the higher potential conductor: ___________.
What do your last two measurements tell you about the amount of work required to move a charge along a conductor?
Explain.
Part 2:
Connect the COMMON voltmeter lead to the low potential metallic paint mark. Use the other voltmeter lead to measure the voltage at 1-cm intervals along the line joining the two paint marks, starting on the low potential paint mark and ending on the higher potential paint mark. (There are more spaces on the data table than you will need.)
Create a table in excel such as the one on the following page.

Questions:
How much work would you have to do to move 1 coulomb of positive charge from a) x = 0 cm to x =3 cm? _________ b) x = 4 cm to x =6 cm? _________ c) x = 5 cm to x =2 cm?? _________
Position
(cm)
0
1
Voltage
(V)
∆ V /∆ x
(V/cm)
∆ V /∆ x
(V/m)
How much work would you have to do to move 32  C of charge from
2
3
4
5
6
7 a) x = 0 cm to x =3 cm?? _________ b) x = 4 cm to x =6 cm? _________ c) x = 5 cm to x =2 cm? _________
8
9
10
How much work would you have to do to move –25 millicoulombs of charge from a) x = 0 cm to x =3 cm?? _________ b) x = 4 cm to x =6 cm? _________ c) x = 5 cm to x = 2 cm? _________
11
12
13
14
15
Take your Position and Potential columns copy them to LoggerPro.
Make a graph of Potential ( y-axis ) vs. Position (x-axis, in meters ) .
We know that for gravity, the potential energy of each kilogram of mass decreases as the mass travels in the direction of the gravitational field, and increases as it travels opposite the direction of the gravitational field. That is, the direction of
Recall that the direction of the electric field at a point is defined to be the direction of the force experienced by a positive test charge at that point. A positive charge will naturally go “downhill”, from higher to lower potential energy, along the direction of the electric field.
Question: Looking at your data, is the direction of the electric field in your experiment in the +x-direction or is it in the
–x-direction? Explain how you can tell.

TO DO : Calculate the ratios ∆ V /∆ x between each pair of points (that is, from x = 0 cm to x =1 cm , from x = 1 cm to x
=2 cm , from x = 2 cm to x =3 cm , etc.). Fill in the rest of the table on the previous page.
Suppose that you have a region where the electric field is uniform and has a magnitude of 10 N/C toward the left.
E = 10 N/C x = 0
(origin)
If you want the potential energy of + 1 coulomb of charge to increase by 10 Joules, do you need to move the charge toward the right or toward the left? Justify your answer.
How far and in what direction do you have to move the coulomb of charge to increase its potential energy by 10 Joules?
Show your calculation below:
How far and in what direction do you have to move the coulomb of charge to increase its potential energy by 20
Joules? Show your calculation below:
Calculate the ratio ∆ V /∆ x for the answers you came up with above:
Suppose now that you have a region where the electric field is uniform and has a magnitude of 5 N/C toward the left.
E = 5 N/C x = 0
If you want the potential energy of + 1 coulomb of charge to increase by 10 Joules, do you need to move the charge toward the right or toward the left? Justify your answer.
How far and in what direction do you have to move the coulomb of charge to increase its potential energy by 10 Joules?
Show your calculation below:

How far and in what direction do you have to move the coulomb of charge to increase its potential energy by 20 Joules?
Show your calculation below:
Calculate the ratio ∆ V /∆ x for the answers you came up with above:
What seems to be the relationship between the electric field strength (in Newtons per Coulomb) and the ratio ∆V/∆x?
Justify your answer
Show that the units Volts/meter are equivalent to the units Newtons/Coulomb.
Going back to your data from your lab, in which locations are/is the electric field greatest?
Suppose that you have two points, A and B, in a region where there is no electric field. Will you have to do any work to move a charge from A to B? Explain why or why not?
Suppose instead that there is a very strong electric field pointing from A to B. Which one of these points, A or B, is at a higher potential? Explain, referring explicitly in your explanation to the definition of potential.

Below is shown a graph of potential vs. distance from a point charge, located at the origin:
Potential vs. distance from a point charge
45000
40000
35000
30000
25000
20000
15000
10000
5000
0
0 0,05 0,1 0,15 0,2 0,25
Distance from the charge (m)
Is the point charge positive or negative? Explain how you can tell:
0,3 0,35 0,4
e
A infinitely long line-charge has a linear charge density of 14.0

C/m. A 33.0 gram particle with a charge of -
3.50

C is 2.80 m from the line-charge. Assume the line-charge is in outer space were the effects of gravity can be neglected.
(A) What is the initial force on the particle? Does force depend upon the mass of the charge?
(B) If the line-charge has a cylindrical shape with radius R, determine an expression for the potential V(r) around the line-charge . If the radius of the line-charge is 10.0 cm, what is the voltage at the initial location of the charge?
(C) If the charge particle is moving at an initial speed of 5.60 m/s directly outwards from linecharge perpendicular to line-charge, what is the maximum distance the charge can get from the line-charge?
(D) How fast would the charge particle have to move in order that the charge could orbit the linecharge in a circular orbit that is perpendicular to the line-charge?

Sketch and Process :
A positive charge particle in moving perpendicular away from a positive, infinite line-charge.
:
Part A is an application of Newton's 2 nd Law for a charged particle in the electric field created by an infinite line-charge.
The generic equation for the electric field due to an infinite line-charge is where the origin is at the location of line-charge and the point charge is on the x-axis.
The electric force on the point charge q is F = q E, and since it is the only force acting on the point charge.
(Gravity is turned off)
We can find the potential in part B using the fact that the electric field is the negative of the gradient of the electric potential. Moreover, the E-field is only radially in the xy-plane so that we can express this in the xyplane as,
Some care is needed to determine the constant C.
Part C : When the point charge reaches its closest approach to the line-charge, the point charge will come to a halt and turn around. We can use the observation that v = 0 when x = x min
to solve part (B). The best way to solve this problem is using the conservation of energy.
The electrical potential energy is simply PE = qV. Better yet we can rearrange the conservation of energy equation to look like,

The advantage is that it allows us to not have to worry about the constant them C when we integrate the electric field to find the potential,
The constant C will cancel out when we calculate

V.
Part D: In order for a particle to move in a circular orbit at a constant speed there must be a force acting on the point charge towards the line-charge that equals the centripetal force.
Since the sign of the point charge is negative in part D the electric force will be towards the positive linecharge and can be set equal to the electric force.
A) Find the Electric force on the point charge initially.
Placing our coordinate system so that the particle is located on the positive x-axis initially , the electric field due to an infinitely long line-charge is
Note that we can write the units of the E-field as either V/m or N/C. Here N/C make the units of the next calculation easier to see that it is a force. However, in the next part of the problem V/m are more appropriate.
The Electric Force on a point charge is easily found by multiplying the value of the E-field by the magnitude of the charge,
Since q and

have opposite signs the force is towards the line-charge. Alternately, the negative sign in our answer means that the force is in the opposite direction as the E-field. Since E is outwards then the force must be towards the origin.

The electric force on the particle does not and will never depend upon the mass of the particle. The electric force depend upon its charge. However, the particle's acceleration will depend upon the particle's mass since F
= ma .
B) Find an expression for the potential V(r) in the xy-plane and find the potential 2.80 m from the line-charge.
Since the electric field is the negative of the gradient of the electric potential we can find the electric potential by integrating the electric field which we already know - the E-field of an infinite line-charge.
Where C is an arbitrary constant which determines the distance (call it r o
) were we will set the potential to be equal to zero - the zero level for the potential energy.
Thus
A reasonable choice is to make the potential at the surface of the line-charge (of radius R ) to be the PE's zero energy reference level.
.
We could chose any distance to be the zero-point except at infinity.
Note that this solution is negative for all values of r larger than R, and that V approaches minus infinity as one gets farther and father from the line-charge. This means that the actual value (as a number) of the potential at some point is not very meaningful by itself. It is always measured relative to some reference point.
When determining potential differences the choice of the constant is a mute point since it will cancel out during the subtraction.

The potential difference will always be the same no matter what reference level we choose.
The potential at 2.80 m is
C) Find the farthest distance that the point charge can reach from to the line-charge before it turns around if the particle has initial outwards speed of 5.60 m/s.
Placing the particle's initial location on the x-axis, we can solve this problem using the conservation of energy.
The electrical potential energy can found from PE = qV since we know the electric potential V for infinite line-charge from part B.
Observer that the radius of the line-charge drops out of the solution since it was chosen as the zero of the electrical potential energy. This is verification that we could have chosen any distance as the zero level.
Plugging in the values given,

D) Find the speed needed by negative point charge to orbit the line-charge in a circular orbit perpendicular to the linecharge.
Setting the electrical force due to the line-charge on the point charge equal to the centripetal force needed by a particle to move in uniform circular motion,
What is interesting is that the speed does not depend upon the distance of the point charge from the linecharge. This means that for same size charge and mass the orbital speed is the same no how close or how far away the particle's orbit.
More interesting is that a negative charge will orbit the line no matter what the initial speed and direction of the charge. In this case the orbit is not a circle and probably not even closed.
If the particle had any velocity parallel to the line-charge, then the particle would move at a constant speed parallel to the line-charge since the E-field is zero in that direction. The 3-dimensional motion of the charge would be "ellipse-like helical" motion.