The following Activities are for the 'Charged Particle in Capacitor' applet. Make sure you know how the applet functions by consulting Help and ShowMe under Applet Help on the applet's Help menu.

ACTIVITY 1. The purpose of Activity 1 is to predict the kinetic energy of a charged particle upon impact of the particle on one of the plates of a parallel-plate capacitor after the particle has been released from rest at the other plate. Any forces other than the electric force are ignored.

RESET the applet. Display the Data box. Set the following parameters and the following initial conditions at t = 0 for the charged particle:

• electric field: E_y = -50 V/m;
• particle charge: q = 3 C;
• particle mass: m = 3.0 kg;
• drag the particle so that it touches the top plate;
• set the particle's velocity to 0.

With these settings, the direction of the electric field and of the electric force acting on the particle are downward. Display the force vector to verify this.

Make the following measurements with the applet:

(a) At t = 0, record

• the particle's y-coordinate (measured at the particle's center),
• the force acting on the particle.

(b) PLAY the motion until the particle hits the bottom plate. At this point, record the particle's

• y-coordinate (measured at the particle's center),
• kinetic energy KE,
• speed v.

Calculations

(a) The particle's y-coordinate when the particle is at the top plate depends on the size of your applet window. The taller the window, the more widely the plates are separated, and the larger the particle's y-coordinate at the top plate.

For the following sample calculations, let us assume y(0) = 185.0 m.

The force acting on a charged particle of charge q in an electric field is given by the expression

= q.(1)

For the values set above,

F_x = 0andF_y = 3×(-50) = -150 N.(2)

Compare Results (2) with the values given in the Data box. Note that the x-component of the force is zero, which means that the force is parallel to the y-axis. The minus sign in the value of F_y means the direction of the force is the same as that of the negative y-direction.

(b) When the particle is at the bottom plate, it is at y = 15.0 m. This is true independently of your window size.

Since the particle is moving parallel to the y-axis, the work W done on the particle in moving from the top to the bottom plate involves only the change Dy in the y-component of the particle's displacement. Thus, since the force is constant,

W = F_yDy = (-150)×(15.0 - 185.0) = 25,500 J.(3)

By the work-kinetic energy theorem, the particle's kinetic energy KE should change by W when the particle is moving from the top to the bottom plate. Therefore, since the particle's kinetic energy is zero at the top plate, at the bottom plate we have

KE_bottom = W = 25,500 J.(4)

Equating Value (4) to the expression for the kinetic energy, KE = (m/2)v², and solving for v, gives for the speed of the particle when the particle reaches the bottom plate:

v_bottom = (2KE/m) = (2×25,500/3) = 130.4 m/s.(5)

Repeat these calculations with the value of the displacement Dy that applies in your case, and compare the results with those given by the applet.

ACTIVITY 2. The purpose of Activity 2 is to repeat the Experiment from Activity 1, but use potential energy and energy conservation to calculate the particle's kinetic energy upon impact on the bottom plate.

Explanation of Potential Energy. The calculations in Activity 1 were based on a view point in which the particle is the system. This system's energy is entirely kinetic energy. The energy can be changed by work done on the system, which is energy transfered to the system from outside the system.

Alternatively, one can consider a larger system, consisting of the particle and the electric field. For this larger system, it is possible to invent a potential energy. This potential energy belongs to both the particle and the field, although, for brevity's sake, it is common to refer to it as the electric potential energy of the particle. The other form of energy the system has is the particle's kinetic energy.

As the particle has kinetic energy, so does the electric field have energy. However, since the electric field is assumed not to change, the field's energy does not change either. Thus, the total energy of the particle-field system is equal to the sum of the electric potential energy and the kinetic energy of the particle plus some constant that is not important and that we will take equal to zero.

Taking the particle and field to be our system, we can describe a change in the kinetic energy of the particle as a transformation of potential energy of the system into kinetic energy of the system such that the sum of kinetic and potential energy, the total energy of the system, is conserved.

For this to be true, the potential energy must be defined so that the loss in potential energy is equal to the gain in kinetic energy, i.e.,

-DPE = DKE.(6)

Since DKE is equal to the work W done by the field on the particle, the potential energy must be then defined so that

-DPE = W.(7)

Eq.(7) is the general definition of potential energy change. Note that this way one obtains the change in potential energy. Definition (7) determines the potential energy up to an arbitrary additive constant in the potential energy which one can choose to have a convenient value.

The work W is given by Eq.(3). Using Eq.(1) to substitute for the force gives

W = F_yDy = qE_yDy(8)

Substituting this into Eq.(7) gives for the change in the potential energy of the particle in the electric field of the capacitor

DPE = -qE_yDy.(9)

Let us denote the potential energy when the particle is at elevation y by PE(y). With this notation, Eq.(9) can be written more explicitly as follows.

PE(y) - PE(0) = -qE_y(y - 0).(10)

Taking PE(0) = 0, which we are free to do, Eq.(10) reduces to

PE(y) = -qE_yy.(11)

This is the equation used by the applet to calculate the potential energy.

Investigations

Exercise 1. RESET the applet. Display the Data and Energy boxes. Set the Electric Field slider to -50 V/m, i.e., set E_y = -50 V/m. Don't change any of the other settings, i.e., keep the charge and mass at 3 C and 3.0 kg, respectively, and keep the velocity at 0.

With E_y < 0, as in the present setting, Eq.(11) can be written

PE(y) = q|E_y|y, if E_y < 0.(12)

This expression for the potential energy is analogous to the expression "mgy" that applies to a particle in the gravitational field of the earth if the y-axis is chosen to point upward as here.

Drag the particle up and down, and observe that the values of the potential energy in the Data box increase and decrease and that the height of the potential energy column in the Energy box increases and decreases, respectively, just like for a particle in the earth's gravitational field.

Exercise 2. Move the particle to y = 100 m, and calculate the potential energy using Eq.(12). Compare the result with that shown in the Data box.

Exercise 3. Drag the particle to the upper plate. As in Activity 1, let's assume that puts the particle at y = 185 m. The particle's potential energy at this point is

PE(y) = q|E_y|y = 3×50×185 = 27,750 J.(13)

If for your window size the value of y when the particle is at the upper plate is different from 185 m, calculate the potential energy in your case and verify your result against the value displayed in the Data box.

PLAY the motion until the particle hits the bottom plate. Observe the column on the far right in the Energy box, and observe the decrease in potential energy and corresponding increase in kinetic energy so that the sum of the two remains conserved. REWIND the motion and PLAY it once more, or STEP through the motion.

Calculate the potential energy when the particle is at the bottom plate, with y = 15.0 m. The answer is PE(15 m) = 2,250 J. Using both this value and value (13), or the corresponding value for your window size, use Eq.(6) to calculate the increase in kinetic energy during the fall from the upper to the lower plate. The answer should agree with that found in Activity 1 and recorded in Eq.(4), i.e.,

DKE = KE_bottom = 25,500 J,(14)

or should agree with the value corresponding to your plate separation.

Exercise 4. Repeat the calculations and observations done in Exercise 3, but with a different vertical drop Dy.

ACTIVITY 3. The purpose of this Activity is to perform similar observations and calculations as in Activity 2, but this time for a particle performing a kind of projectile motion, not a vertical drop.

Exercise 1. RESET the applet. Set E_y = -50 V/m, and keep the charge and mass at 3 C and 3.0 kg, respectively.

Display the Data and Energy boxes. Drag the particle to the bottom plate at x = 100 m, set the particle's speed to v = 200 m, and the velocity direction to q = 30^o relative to the positive x-axis. Display the velocity vector and the particle's path.

PLAY the motion. You should obtain a parabolic trajectory, first rising, then falling. REWIND the motion, and PLAY it again. Observe the energy columns. Observe that while the particle is rising, the potential energy is increasing and the kinetic energy decreasing, and that the opposite is true when the particle is falling. Is the kinetic energy zero when the particle is at the top of its path?

Exercise 2. Use the applet to determine the particle's kinetic energy approximately when the particle is at the top of its path. Do this by PLAYing the motion and stopping it as nearly as you can at the peak of the trajectory. STEPing the motion may help. Record the value of the kinetic energy from the Data box when the particle is near the peak.

Also determine the particle's y-coordinate at the peak using the value in the Data box.

Exercise 3. Calculate the kinetic energy at the bottom and at the peak.

Calculation of KE. At the bottom plate,

KE_bottom = (m/2)v² = (3/2)×200² = 60,000 J.(15)

The particle's x-component of velocity is constant because the x-component of the electric field and therefore the x-component of the force acting on the particle are zero. Thus, when the particle reaches the peak of its trajectory, its speed v_peak is not zero, but equal to the x-component of the particle's velocity. (The y-component of velocity is zero at the peak.) Thus, the particle's kinetic energy at the peak is equal to

KE_peak = (m/2)×v_x² = (m/2)×(v cos 30^o)²

= 3/2×(200×0.8660)² = 45,000 J.(16)

Exercise 4. Calculate the particle's elevation y_peak at the peak of the trajectory using energy conservation.

Note on Potential Energy. Eqs.(11) and (12) for the potential energy still apply even though the present motion involves changes in x while the motion in Activity 2 did not. The reason is that Eq.(8) for the work W done on the particle by the field applies whether the particle's displacement has a non-zero x-component or not. The general expression for W in two dimensions is

W = D = F_xDx + F_yDy.(17)

In the present case of the capacitor, with the electric field parallel to the y-axis, the x-component of the force exerted on the particle is zero. Therefore, whether Dx is zero or not, the general expression (17) for the work done on the particle reduces to

W = F_yDy = qE_yDy.(18)

Eq.(18) is identical with Eq.(8). Therefore, Eq.(8) for W applies in all cases, no matter what motion the particle performs, and the theory developed in Activity 2 for vertical motion applies to general projectile motion also. In particular, Eq.(11) for the potential energy applies along with Eq.(12) for the special case of a downward directed electric field, E_y < 0.

Calculation of the particle's elevation at the peak. The calculation is based on energy conservation. Energy conservation involves the potential energy at the peak which in turn involves the particle's elevation at the peak. Thus, energy conservation provides an equation for y_peak.

The potential energy at the bottom plate has already been used earlier,

PE_bottom = q|E_y|y_bottom = 3×50×15 = 2,250 J.(19)

Combining values (15) and (19) gives for the total energy E the value

E = KE_bottom + PE_bottom = 60,000 + 2,250 = 62,250 J.(20)

There is no need to use the subscript "bottom" with the energy because the energy is conserved and therefore has the same value everywhere.

Caution: Here, the symbol "E" denotes the total energy, not the magnitude of the electric field.

Equating the sum of the kinetic and potential energy at the peak to the energy value (20) gives the equation

KE_peak + PE_peak = E = 62,250 J.(21)

Substituting Value (16) for KE_peak and Expression (12) for the potential energy into Eq.(21) gives

45,000 J + q|E_y|y_peak = 62,250 J(22)

y_peak = (62,250 - 45,000)/(q|E_y|)

= 17,250/(3×50) = 115.0 m. (23)

Exercise 5. Repeat the preceding measurements and calculations for a different initial velocity.

ACTIVITY 4. The purpose of this Activity is to investigate potential energy as a function of charge, electric field, and position.

The general expression for the electric potential energy of the particle inside the capacitor is given by Eq.(11), derived earlier,

PE(y) = -qE_yy.(11)

Exercise 1. RESET the applet. Display the Data and Energy boxes. Keep the applet at its default settings, which are

• E_y = 50 V/m,
• q = 3 C,
• m = 3.0 kg,
• = 0.

Move the particle, and observe the values of PE and the height of the blue potential energy column.

Because of the minus sign in Eq.(11), PE decreases as y increases and increases as y decreases, in proportion to y. The "height" of the potential energy column is measured from the zero level, which is indicated by the thin black line across the energy columns.

Exercise 2. Keep the particle's position fixed, and vary the particle's charge q. The potential energy varies in proportion to q. Given the values of the other parameters as listed in Exercise 1, the potential energy is negative if q is positive and is positive if q is negative.

Exercise 3. Keep the particle's position fixed and set the charge to q = 3 C. Vary the electric field, i.e., E_y. The potential energy varies in proportion to E_y. It is negative for positive E_y and positive for negative E_y.

Change q to - 3 C. Again vary E_y. Now the potential energy is positive if E_y is positive, and negative if E_y is negative .

The potential energy is given by Eq.(11). Any of the three factors on the right can be negative, in general, although the applet is set up so that y > 0 at all points between the plates. Thus, the sign of PE in the applet is determined by the signs of q and E_y alone.

Exercise 4. For a given setting of q, E_y, and y, calculate the potential energy. Compare your result with the value in the Data box.