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 use energy conservation 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 cond itions 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.

Note on Potential Energy

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

= q.(1)

The electric field in the capacitor is parallel to the y-axis, which means that its x-component is zero. In this case, Eq.(1) written out in components amounts to

F_x = 0andF_y = qE_y.(2)

For the values set above,

F_y = 3×(-50) = -150 N.(3)

Compare Result (3) with the value for F_y given in the Data box. The electric force has this value no matter where the particle is and no matter how the particle is moving. 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.

The potential energy PE corresponding to force (2) is a function of the particle's y-coordinate,

PE(y) = -qE_yy.(4)

This expression is analogous to the expression "mgy" for the gravitational potential energy of a particle of mass m in the earth's gravitational field whose magnitude is g.

If the electric field is directed downward like the gravitational field, E_y is negative and Eq.(4) can be written

PE(y) = q|E_y|y, if E_y < 0.(5)

Expression (5) for the electric potential energy is exactly of the same form as the expression "mgy" for the gravitational potential energy.

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 the particle when it is at the top, or anywhere else, is given by Eqs.(2) and (3).

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

Energy conservation amounts to

KE_bottom + PE_bottom = KE_top + PE_top.(6)

Substituting the given values and using Eq.(5) this gives the following equation for the unknown KE_bottom.

KE_bottom + 3×50×15 = 0 + 3×50×185(7)

KE_bottom = 3×50×(185 - 15) = 25,500 J.(8)

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

ACTIVITY 2. The purpose of this Activity is to perform similar observations and calculations as in Activity 1, 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, that the opposite is true when the particle is falling, and that the total energy is constant throughout the motion. 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.(9)

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.(10)

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

Calculation. 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 is equal to

PE_bottom = q|E_y|y_bottom = 3×50×15 = 2,250 J.(11)

Combining values (11) and (9) gives for the total energy E the value

E = KE_bottom + PE_bottom = 60,000 + 2,250 = 62,250 J.(12)

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 (12) gives the equation

KE_peak + PE_peak = E = 62,250 J.(13)

Substituting Value (10) for KE_peak and Expression (5) for the potential energy into Eq.(13) gives

45,000 J + q|E_y|y_peak = 62,250 J(14)

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

= 17,250/(3×50) = 115.0 m. (15)

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

Exercise 6. Repeat the preceding measurements and calculations for different values of the charge, mass, and electric field.