Lesson - Potential Energy and Potential In A Non-Uniform Electric Field
The applet simulates the motion of a charged particle in a non-uniform electric field, displays the particle's potential and kinetic energies, and displays the electric potential at the particle's location and at other selected locations.

Prerequisites

Students should be familiar with the concepts of charge, electric field, and electric field lines, and Coulomb's law of electrostatics and Newton's second law of motion.

Learning Outcomes

Students will be able to review the definition of the electric field, will be introduced to the concepts of electric potential and potential energy in an electric field, and the relationship between electric field and electric potential and the parallel relationship between electric force and electric potential energy. They will learn to apply these concepts in the situation where the electric field is due to a single point charge.

Instructions

Students should know how the applet functions, as described in Help and ShowMe.

The applet should be open. The step-by-step instructions in the following text are to be done in the applet. You may need to toggle back and forth between instructions and applet if your screen space is limited.


Contents

spaceLaws, Theorems, Definitions
spaceNewton's Second Law of Motion
spaceCoulomb's Law of Electrostatics
spaceDefinition of Electric Field and Electric Field Lines
spaceDefinition of Electric Potential and Equipotential Lines
spacePotential Energy and Energy Conservation
spacePotential and Potential Energy in the Electric Field Due to a Point Charge

spaceMotion of a Charged Particle in a Uniform Electric Field

spaceElectric Potential and Equipotential Lines

spaceElectric Potential and Potential Energy

space

Laws, Theorems, Definitions
space

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Motion of a Charged Particle in an Electric Field Due to a Point Charge

Task 1. Under "Laws, Theorems, Definitions" above, read the section "Definition of Electric Field and Electric Field Lines".

Exercise 1. RESET the applet. Make the following settings:

You may find it easier to move the particle to its initial position if you display the grid. The particle's position coordinates are displayed in the Data box.

Click the Initial button. Select the Trace toggle button, and PLAY the motion. PAUSE the motion at some point.

Describe the trajectory. In which direction relative to that of the electric field does the trajectory curve? Therefore, in which direction is the force acting on the particle? Hint: The acceleration vector is in the direction of the force, according to Newton's second law. The applet lets you display the acceleration vector.

Explain your observations in terms of the theory under "Definition of Electric Field and Electric Field Lines".

Answer. Figure 2 below shows the trajectory (in blue), with the particle paused on it and the velocity and acceleration vectors displayed.

Trajectory with Data and Energy Boxes

Figure 2

For a positive source charge as in the present Exercise, the electric field points radially away from the source. By Equ.(5), for a test particle with positive charge as in this Exercise, the force acting on the test particle is in the direction of the electric field, i.e., also radially away from the source. This is a repulsive force. The direction of the acceleration vector and the shape of the trajectory confirm the direction of the force.

Question 1. For the settings in Exercise 1, what is the particle's acceleration when the particle is at the point shown in Figure 1, i.e., at (x,y) = (66.0, 0.9) m? Calculate the value of the acceleration from basic principles.

Answer. When force expression (5) is inserted in the Newton's second law equation (1), one obtains an equation involving the particle's acceleration vector a. Solving this equation for vector a gives

vector a = (q/m) vector E .space(28)

We need to calculate the electric field vector at (x,y) = (66.0, 0.9) m. The electric field vector vector E is in the direction of the position vector vector r and therefore proportional to vector r. Thus,

vector E = c vector rspace(29)

where c is a factor such that the magnitude of cvector r is equal to E = kQ/r2, where Q > 0. Thus,

cr = kQ/r2 space(30)

whence

c = kQ/r3 .space(31)

Inserting Expression (31) for c into Equ.(29) gives

vector E = kQ/r3 vector r .space(32)

At (x,y) = (66.0, 0.9) m, r has the value

r = (x2 + y2)0.5 = (66.02 + 0.92)0.5 = 66.006 m.space(33)

Inserting the given values into Expression (32) for vector E gives

vector E = [8.988×109×0.003 / 66.0063] (66.0, 0.9) = (6188, 84.39) V/m .space(34)

Inserting this value for vector E and the given values for q and m into Equ.(28) gives for the acceleration of the test particle at (x,y) = (66.0, 0.9) m the value

vector a = 0.002/3×(6188, 84.39) = (4.126, 0.056) m/s2 .space(35)

Compare this value with the one shown in the Data box. The values are consistent within the number of significant digits displayed by the applet.

Exercise 2. Continuing from Exercise 1, click REWIND and change the value of the charge of the test particle from q = 2 mC to q = -2 mC.

PLAY the motion, and observe the trajectory. Which way does the trajectory curve relative to the electric field vector? Explain your observation in terms of basic theory. You should find the trajectory to look like the trajectory shown in grey in Figure 2 above.

Calculate the acceleration vector at a point where you paused the motion, and compare your result with that shown in the Data box.

Exercise 3. Continuing from Exercise 2, i.e., with q = -2 mC, click REWIND and set the source charge to Q = -3 mC. Also try the combination Q = -3 mC and q = 2 mC.

PLAY the motions, and observe the trajectories. Which way do the trajectories curve relative to the electric field vector? Explain your observation in terms of basic theory.

Without further calculation, predict what the acceleration vectors should be if you if you pause the motion at the same point as before (this may be possible only approximately), and verify your predictions by comparing them with the values shown in the Data box.

Exercise 4. Continuing from Exercise 3, investigate the motion for other values of the two charges and the mass, and other values of the initial position and velocity.

Explain your observations in terms of basic theory, and calculate the acceleration in each case at a point where you pause the motion.

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Electric Potential and Equipotential Lines

Task 1. Under "Laws, Theorems, Definitions" above, read the section "Definition of Electric Potential and Equipotential Lines".

Exercise 1. RESET the applet.

Select the Coordinates button Coordinates, and click in the applet window to set a field point marker. With the marker, the (x,y)-coordinates of the point and the electric potential V at that point will be displayed.

Click the Field Line button Field Line and the Equipotential Line button Equipotential Line to display the electric field line (in green) and the equipotential line (in mustard color) through the point.

Drag the field point, and observe how the field line and equipotential line (which is a circle in this case) move along with the point. Based on these observations, draw a pattern of electric field lines and equipotential lines in your Notebook and explain why the field lines and equipotential lines form the pattern you observe.

Answer. The electric field lines are straight lines going from the source to infinitiy in a radial direction. The force on a charged test particle at a given field point is in this direction, and therefore so is the electric field at that point. The electric field line through a given point has the direction of the field vector at that point.

The equipotential line at a given field point is perpendicular to the electric field line at that point. Therefore the equipotential "lines" are concentric circles whose centers are at the source point. Such circles are perpendicular at any point to the radially directed field lines. An example of such a line pattern is shown in Figure 3 below.


      Field Lines and Equipotential Lines

Figure 3

Question 1. Figure 4 below shows two equipotential circles, with radii of 100 m and 110 m, respectively. The potential on the 100-m circle is equal to 2.70×105 V. This is for a source charge Q = 3 mC.

Field Lines and Equipotential Lines

Figure 4

Calculate the electric potential on the 110-m circle using Equ.(10) from the section "Definition of Electric Potential and Equipotential Lines". This calculation will be approximate because the 10-m distance from one circle to the other is not an infinitesimal distance. Still this 10-m distance is "small" enough because the electric field does not vary much in magnitude over this distance. For more of an explanation, see Exercise 2 below.

Answer. We need the magnitude E of the electric field on the 100-m circle. According to Equ.(24), it is equal to

E = k|Q| / r2 = 8.988×109×0.003 / 1002 = 2,697 V/m.space (36)

Using dr = 10 m and using value (36) for E, Equ.(10) for the change dV = V(110 m) - V(100 m) implies

V(110 m) = V(100 m) - E dr = 2.70×105 - 2,697×10 = 2.43×105.space (37)

This value is quite close to the value of 2.45×105 shown in Figure 4 above. Confirm this value with the applet by placing a field point marker at a distance of 110 m from the source.

Exercise 2. Calculate the magnitude of the electric field at a distance of 110 m from the source and determine by what percentage it differs from the magnitude of the field at 100 m from the source that was calculated above. You will find that the two values don't differ by much. This justifies the use of Equ.(10) in the preceding calculation. Compare the percentage by which the two electric field values differ to the percentage error in result (37).

Exercise 3. Using the value of the magnitude of the electric field at r = 110 m calculated in Exercise 2, calculate the change in potential in going from a point at r = 110 m to a point at r = 120 m. Use the result and the value of the potential at r = 110 m to predict the potential at r = 120 m. Compare the result to the value given by the applet.

Comment. One can continue in this manner, calculating potential changes in small steps along a path from one point P to a distant point P' and adding the potential changes, and in that way obtain the potential change for two points P and P' quite far apart. This is the integration process referred to at the beginning of the section "Potential and Potential Energy in an Electric Field Due to a Point Charge". To get an exact result, one has to go to the limit of infinitely many such steps, each of vanishing size. This limit summation can be performed symbolically. The result is Expression (26) for the electric potential.

Exercise 4. Reset the applet, and observe the value of V at the particle's location. V is displayed in the Data box. Also, note the length of the V-column in the Energy box. The test particle's charge q should be equal to the default value of 2 mC.

Then vary the charge of the test particle, and observe the value of V displayed in the Data box and the length of the V-column in the Energy box while you are changing the charge. Explain your observations in terms of basic theory.

Answer. Neither the electric field nor the electric potential depend in any way on the properties of the test particle. They characterize the state of space without the test particle. Therefore, changing the charge of the test particle has no effect on the value of the electric potential at the particle's location.

Exercise 5. Continuing from Exercise 4, vary the value of the source charge Q, including negative values. Observe the value of V in the Data box and the length of the V-column in the Energy box, and explain your observations in terms of basic theory.

Answer. This time the electric potential does not stay the same. When the source charge changes, the electric field changes. See, e.g., Equ.(24) or Definition (4) of the electric field, in conjunction with Coulomb's law. When the electric field changes, the potential must change according to the defining relationship between potential and field described under "Electric Potential and Equipotential Lines". Note that V is positive for positive Q and negative for negative Q, in the case simulated by the applet.

Exercise 6. This Exercise is about the concept of an electric field vector as an indicator of the direction of most rapid decrease of the electric potential and the rate of this decrease.

RESET the applet. Display two equipotential circles that are fairly close together, as in Figure 5 below. Choose the two radii to be 100 m and 120 m.

Two Equipotential Circles

Figure 5

Place the particle at a point on the inner circle, as in Figure 5, and click the Initial button Set Initial Position and
     Energy.

Drag the particle to a point on the outer equipotential circle while the displacement vector is displayed. Drag the particle along the outer circle, and observe the magnitude of the displacement vector. In which direction is it the least? Therefore, in which direction does the potential decrease the fastest? Remember that the change in potential in going from one equipotential circle to another is constant, so that the direction of greatest rate of change of the potential is the direction in which the displacement vector is shortest.

Estimate the magnitude of the electric field at r = 100 m from the potential change indicated in Figure 5. The inner equipotential circle has a radius of 100 m. Then obtain a better estimate by using the data in Figure 4, where the two equipotential circles are only 10 m apart, instead of 20 m.

Answer. If you have chosen the particle's initial position as in Figure 5, the direction straight down is the direction of fastest decrease of the potential. Note that the potential on the outer circle is less than on the inner circle, so that the potential is decreasing in the downward direction. You can verify that the electric field at the location of the particle in Figure 5 is downward by displaying the particle's acceleration vector. The acceleration is in the direction of the force acting on the particle and, for positive test charge, this force is in the direction of the electric field, according to Definition (4). Therefore, the acceleration points in the direction of the electric field as long as the test particle has positive charge.

To calculate the magnitude E of the field, we use Equ.(9) with |dV| = |2.25×105 - 2.70×105| = 0.45×105 V and dr = 120 - 100 = 20 m. Thus,

E = |dV/dr| = 0.45×105/20 = 2,250 V/m.space (38)

This value is not particularly good, as a comparison with the exact value of 2,697 V/m shows. This value is calculated in Equ.(36). However, using a smaller value for dr, namely, the smaller dr of 10 m from Figure 4, gives a noticeably better result:

E = |dV/dr| = |2.45×105 - 2.70×105| / 10 = 2,500 V/m.space (39)

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Electric Potential and Potential Energy

Task 1. Under "Laws, Theorems, Definitions" above, read the sections

Exercise 1. RESET the applet.

Set the particle's initial position and velocity at time t = 0 to

Fix these initial values by clicking the Initial button Set Initial Position and
      Energy. This will also adjust the "initial" energy bar in the Energy box to be identical to the "final" energy bar.

You may find it easier to move the particle to the initial position if you display the Grid. The particle's position is displayed in the Data box.

Make sure the parameters of the system are still set to

Display the particle trace, and STEP into the motion until you are at time t = 6.00 s and the particle is at (x,y) = (57.0, -12.6) m, as shown in Figure 6 below. The Energy box in Figure 6 shows what the energy bars should look like at this moment. The particle is moving on the trajectory that starts out in blue.

Trajectory with Data and Energy Boxes

Figure 6

In your Notebook, describe the changes you observe in the particle's potential energy PE, kinetic energy KE, and total energy E as you step through the motion. Discuss if the changes are what you expect from basic theory. Also describe the changes in the electric potential V at the particle's position as you step through the motion, and explain them in terms of basic theory.

Answer. The potential V as a function of r is described by Equ.(26). According to this equation, if Q > 0, the potential increases as r decreases and decreases as r increases. Stepping through the motion should show this behaviour. You will need to go beyond t = 6.00 s to see the behaviour for increasing r.

Since the test particle's charge is positive, Definition (12) implies that the particle's potential energy behaves like the potential: it increases with decreasing r and vice versa. Since the total energy is conserved, the kinetic energy must decrease as the particle gets closer to the source, by as much as the potential energy increases. The particle slows down as it approaches the source. It speeds up as it moves away from the source. This makes sense intuitively because the Coulomb force is repulsive in this case.

Question 1. Click REWIND. The Data box will display the electric potential at the particle's start position as 1.91×105 V. Calculate the particle's potential energy at this point. The test particle's charge is equal to 2 mC.

Answer. Definition (12) of the potential energy, "PE = qV", gives the value PE = 0.002×1.91×105 = 382 J. Compare this value with the one shown in the Data box. The discrepancy of 1 J between the two values is due to round-off error.

Question 2. Continuing from Question 1, change the charge of the test particle from q = 2 mC to q = -2 mC. What are the values of the electric potential at (x,y) = (100, -100) m and of the particle's potential energy at this point now?

Answer. The electric potential is still equal to 1.91×105 V. Its value does not depend on any particle properties. However, the particle's potential energy has changed to -382 J, in accordance with Equ.(12). (The Data box shows -381 J.)

Exercise 2. Continuing from Question 2, with q = -2 mC, STEP through the motion until you are at time t = 6.00 s. What does the trajectory look like this time?

Observe the changes of the same quantities as in Exercise 1, and record in your Notebook if there are any differences compared to what you observed in Exercise 1. Explain the differences in terms of basic theory.

Answer. The trajectory is now curving towards the source charge. Again, in accordance with Equ.(26), the electric potential at the test particle is increasing as the test particle is approaching the source and decreasing as it is moving away from the source. Also, more fundamentally, as the particle is approaching the source, it is moving against the electric field and therefore into regions of higher electric potential.

The potential energy, however, has the opposite behaviour since it is equal to the potential multiplied by a negative constant. Thus, the potential energy decreases as the test particle moves closer to the source. Also, since the force is attractive in this case, we would expect the particle to be gaining kinetic energy and therefore losing potential energy as it approaches the source.

Question 3. Continuing from Exercise 2, click REWIND and change the test particle's charge back to q = 2 mC.

Make sure the particle's initial position and velocity at t = 0 are still equal to

Otherwise click RESET, and set these initial conditions. Lock the initial conditions by clicking the Initial button.

Make sure the parameters of the system are set to

Display the particle trace, and STEP into the motion until you are at time t = 9.00 s and the particle is at point P' with coordinates (x,y) = (85.3, 15.6) m.

Calculate the particle's kinetic energy at P'.

Answer. The answer can be obtained by using energy conservation. The calculation is broken up into four segments.

Electric Potential V at P'

By Equ.(26) and using the value r = 86.7 m given in the Data box (check this value by calculating it from the given x,y coordinates), the potential at P' is equal to

V(P') = kQ/r = -8.988×109×0.003 / 86.7 = 3.11×105 V.space(40)

This agrees with the value displayed in the Data box.

Potential Energy PE at P'

According to Definition (12) of the potential energy, we only need to multiply the electric potential at P' by the charge q to get the potential energy of the test particle at P',

PE(P') = qV(P') = 0.002×3.11×105 = 622 J.space(41)

We could also have used Equ.(25) for PE directly.

Energy E

The energy of the particle in the electric field is conserved. Therefore, the energy at P' is equal to the energy at the start point P. The latter can be calculated from the given initial conditions.

The kinetic energy at the start point P is equal to

KE(P) = (m/2)v2 = (m/2)(vx2 + vx2) = 3/2×((-12)2 + 182) = 702 J. space(42)

To calculate the potential energy at the start point, we need the distance r of the start point from the source. This can be calculated from the x,y coordinates of the start point to be 141.42 m. Verify this value. Substituting it and the other given parameters into Equ.(25) gives

PE(P) = kQq/r(P) = 8.988×109×0.003×0.002 / 141.42 = 381.3 J. space(43)

Adding the two energies (42) and (43) gives the total energy E both at P and at P',

E = KE + PE = 702 + 381.3 = 1,083.3 J .space(44)

This value agrees with that in the Data box.

Kinetic Energy KE at P'

The kinetic energy at P' can be obtained by subtracting from the total energy the potential energy at P',

KE(P') = E - PE(P') = 1,083.3 - 622 = 461.3 J. space(45)

Verify this value with the applet.

Exercise 3. Do more calculations analogous to those done in Question 3, by evaluating V, PE, E, and KE at some other point P' on the particle's trajectory and for points P' on other trajectories corresponding to other values of the particle's charge and mass and other values of the electric field.

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