Varying degrees of uniformity of the electric field are worth considering. Parallel plate capacitors have a nearly perfect uniformity (except at the edges). Spherical and cylindrical conductors have fields which are highly concentrated near the conductor. But they are the same in all radial directions. A pointy conductor has charge and field concentrations which are high near the pointy areas. The purpose the Pointy Conductor simulation is to demonstrate this non-uniformity near sharp points. A complete worksheet is available to guide students in this exploration. Note the following:

The E-fields are concentrated near points, and equipotentials are closely-spaced

The E-field strength is proportional to the density of the E-lines

The charge density at a spot on the conductor surface is proportional to the E-field there

The voltage gradient is proportional to the E-field strength

Pointier shapes have larger non-uniformities and higher E-fields.

In order access the oblong conductor website, use the following information, and choose

Study the E-field at a Point

Student Worksheet Downloadable Word Doc

We used the following word document as a lab procedure in the PHY112 and PHY131 classes.

Obtain the output of basic circuits using Bode plots

What are Bode Plots? Bode plots sketch the AC amplitude and phase at a chosen node in a circuit or control loop, as a function of frequency. They show, for example, the high or low frequency cutoffs of a filter circuit, or the positions of resonance peaks.

Why Bode Plots? Getting the response of simple networks is very easy and intuitive using this engineering technique. They provide a fast way to visualize the frequency response of basic networks with a minimum of complex algebra.

This post shows a simple progressive explanation of reactance and impedance. Step 1 is to plot reactances for basic circuit elements: resistors, capacitors, and inductors. Step 2 is to define and use the impedance concept. The final step in most textbooks is to rotate the impedance vector as a phasor. The discussion here departs from that approach and switches to s-space (as used in Laplace transforms). It is possible to do inverse transforms on s-space functions to obtain circuit response to transients, but that is not the purpose here. Rather Step 3 in this ‘lab experiment’ is to determine the poles and zeroes of the transfer function of the circuit. The result is an immediate sketch of the output amplitude and phase of any chosen node in the circuit as a function of frequency.

Simple Example

This simple example is a high pass filter consisting of an inductor and a resistor.

Step 1: The resistance is R and the reactance is just ωL, where ω = 2π*frequency. These two components of impedance can be plotted as perpendicular vectors on the complex plane:

In the complex plane, R is a real quantity, and the reactance of the inductor is a positive imaginary (often written jωL, where j =√(-1) ). If the impedance were purely real, as in the case of a resistance, then the current and voltage have the same phase. If the impedance were purely inductive – a positive imaginary – then the current would be 90 degrees out of phase with the voltage. In fact considering the voltage as a real V_{0 }cos (ωt), and dividing by a positive imaginary, jωL, the resulting current is a negative imaginary. Thus the current vector lags 90º behind the voltage vector as these two phasors rotate counterclockwise on the complex axes. Thus the voltage ‘leads’ current in a pure inductor. The current is –(V_{0} /ωL) sin (ωt).

Step 2: In this simple example, the net impedance Z = (R ^{2} + (ωL)^{2} )^{½} . The net impedance is indicated in the above graph by the dashed arrow. The phase angle between voltage and current is the angle between Z and the real axis in the figure, thus φ = tan^{-1} (ωL/R). The voltage leads the current by the angle φ.

Power is not dissipated in the reactance L because the current and voltage are 90º out of phase in the inductor. Therefore the entire power dissipation in the circuit is in the resistance R. Since the current in the circuit is I_{rms} = V_{rms}/|Z|, the power in the circuit, equal to the power in the resistor, is I_{rms}^{2} R = (V_{rms }/ |Z| )^{2} R = V_{rms}I_{rms} cosφ.

Step 3: The above circuit is a simple voltage divider, where the voltage V_{in} is divided by the network to give V_{out }= V_{in } jωL/Z . These are complex numbers, dependent on frequency. The Bode plot allows a quick graphical analysis of the high pass filtering effect. Divide by V_{in } and take the Laplace transform of both sides, substituting the variable ‘s’ for jω:

H(s) must be put in the form of a quotient of two polynomials in s, and preferably the polynomials are factored. Then directly determine the zeroes in the numerator: s = 0 in this case; and the zeroes in the denominator: s = –1/τ. Use this information to generate the Bode plots. Go to the Bode plotting website: Bode website

Suppose for example that R = 10 and L = 25 mH. Then τ = .0025 sec. The zero of H(s) is at s = 0, and the pole is at s = -400 radians/sec. Entering this info in the Bode website obtain the following result:

The Bode plot is the Asymptotic plot. It is a quick approximation to the Real plot, and there are simple rules for drawing it. It shows that the transmission of the filter is very low at small frequencies, and becomes flat for frequencies above ∼400 rad/sec. The phase plot is also available:

The phase plot shows there is a 90 degree phase shift at low frequencies due to the inductor. But at high frequency, where the impedance of the inductor becomes much greater than that of the resistor, the phase tends toward zero degrees as if the inductor were an open circuit.

Worksheets for Students

We provided a few editable worksheets to download for students in PHY131 (Physics Science and Engineering students).

Bode plot of an RC bandpass filter, which has both low frequency and high frequency cutoffs: only intermediate frequencies are transmitted – Bandpass Circuit

The purpose of the solenoid simulation is to let the user explore the magnetic field inside and surrounding a solenoid. A complete worksheet is available to guide students in this exploration. Note the following:

B-field strength is proportional to the current in the solenoid (I amps), and the number of turns per meter (N_{TPM}) in the solenoid coil. The exact formula for a long solenoid is B = μ _{0}I N_{TPM} , where μ _{0} = 4 π × 10 ^{-7 .}

The B-field strength is proportional to the spatial density of the B-lines: that is, a lot of lines close together means a strong B-field

The B-field is strong and very uniform inside the solenoid, and very weak outside.

The direction of the solenoid field is determined by the “Right-Hand-Rule” and the solenoid produces North and South poles similar to a magnet.

In order access the dual current-loop website, use the following information:

Student Worksheet Downloadable Word Doc

We used the following word document as a lab procedure in the PHY112 class. However, it can be used with more advanced or engineering-level classes, because the concepts are significant. Document:

Explain why the |B| at the end of a long solenoid is very close to half the B-field magnitude inside the solenoid?

The solenoid produces a uniform magnetic field which is very strong inside the solenoid. How is this analogous to the capacitor and the electric field?

There is much similarity between B-fields produced by a bar magnet and a solenoid. Is there an underlying similarity between magnets and solenoids that would explain this similarity?

Do adjacent turns in the solenoid coil attract one another or repel?

The purpose of the current loop simulation is to let the user explore the magnetic fields surrounding one or two current loops. A complete worksheet is available to guide students in this exploration. Note the following:

The number of B-lines is proportional to the current in the loops (I amps).

The B-field strength is proportional to the spatial density of the B-lines: that is, a lot of lines close together means a strong B-field

B-lines are complete: that is, they close on themselves and do not terminate on any “magnetic charges”. “Magnetic charges” do not exist.

The B-field is stronger near a wire carrying current.

In order access the dual current-loop website, use the following information:

Student Worksheet Downloadable Word Doc

We used the following word document as a lab procedure in the PHY112 class. However, it can be used with more advanced or engineering-level classes, because the concepts are significant.

The magnetic field is a force field: The force on a current is perpendicular both to current vector and the B-field line. The magnitude of force is proportional to the current acted upon and to the spatial density of the B-field lines.

All magnetic field lines are complete loops. They appear to terminate on the poles of a permanent magnet. But actually, they do not stop, but penetrate into the magnet and continue right through it.

If the spacing between two loops carrying parallel currents is 1/2 the loop diameter, this is called a “Helmholtz Pair”. This configuration gives a rather uniform B-field in between the loops.

When the two loops are near to each other, and the currents are circulating in the same direction, the magnetic fields in between them tend to add.

The purpose of the parallel plate capacitor simulation is to let the user explore the almost uniform electric and potential fields in a parallel plate geometry. A complete worksheet is available to guide students in this exploration. Note the following:

The fields are very uniform in between the plates

The number of E-lines is proportional to the charge in Coulombs (C)

The E-field strength is proportional to the density of the E-lines

The voltage gradient is proportional to the E-field strength

Very low density E-field extends outside the parallel plates, called the “fringing field”

The “fringing field” located at the capacitor ends is non-uniform

In order access the capacitor simulation website, use the following information:

Student Worksheet Downloadable Word Doc

We used the following word document as a lab procedure in the PHY112 class. However, it can be used possibly in PHY101 as well, because questions are qualitative.

The electric field is a force field: The force on a positive charge has direction parallel to the electric field lines. The magnitude of force is proportional to the density of the E-field lines.

All electric field lines start on the positive plate and terminate on the negative plate.

The “fringing field” is a weak E-field found at the open ends of the capacitor. The field is weak and gets weaker the farther one gets from the charged plates. However, if you study the fields at the highest level of charge, and at the very corner of the metal plates, you will see that the E-field is especially large right there. This is due to an accumulation of charge at the corners of the plates.

The purpose of the dipole field simulation is to let the user explore the electric and potential fields surrounding one or two electric charges. A complete worksheet is available to guide students in this exploration. Note the following:

The number of E-lines is proportional to the charge in Coulombs (C).

The E-field strength is proportional to the density of the E-lines

Equipotential (Voltage) rings surround each charge.

The gradient in voltage is proportional to the E-field.

In order access the electric dipole simulation website, use the following information:

Student Worksheet Downloadable Word Doc

We used the following word document as a lab procedure in the PHY112 class. However, it can be used possibly in PHY101 as well, because questions are qualitative.

The electric field is a force field: The force on a positive charge has direction parallel to the electric field lines. The magnitude of force is proportional to the density of the E-field lines.

All electric field lines start on a positive charge and terminate on a negative charge.

If the positive charge is bigger than the negative charge, or if the negative charge is zero, some E-lines appear to go out to infinity. This is true in principle, but in real life, the lines do terminate on some distant negative charge. There do not appear to be any concentrations of positive or negative charge at any location in the universe — the charges are matched with each other over fairly short distances.

The ISS is visible to the naked eye, so students can measure the motion, then analyze it as an exercise in angular velocity and orbital dynamics. Measure the angular velocity and use this to calculate the speed of the space station in orbit. Equipment required is completely free!

Basic Idea of the Experiment

Fingers held at arm’s length are calibrated to provide an angular measure. As the space station crosses each finger on the hand of an up-stretched arm, record the time on a stopwatch or cellphone stopwatch App. This gives the angular velocity. Combine this with the distance up to the space station to get the ISS velocity.

Materials

Ruler to measure width of hand and distance from eye to hand.

Find out a good date and time and spot the ISS. The above website – Spot the Station – shows

dates and times

starting direction and elevation where the shuttle first becomes visible

the ending of visibility direction and elevation

the maximum elevation for the traversal.

The ISS will be visible only in the morning and the evening, because for most of the night, the Earth casts its shadow over the shuttle orbit. The shuttle travels from generally West to generally East; therefore it is very brightly illuminated soon after it rises in the morning, and a little less brilliant when it rises in the evening, due to the relative positions of the sun, the ISS, and the observer.

Take measurements when the ISS is near its highest elevation. Hold the hand still above your head with the palm and fingers bent horizontal. As the ISS reaches the pinky finger, call out the start. Then call out as it crosses the three narrow spaces between fingers, as well as when it emerges past the index finger.

Hint: holding the hand absolutely still is difficult. It may help if there is a star or tree branch above the hand to help you steady your hand.

Measure the average angle in radians subtended by the four fingers. This is just the average width of the fingers divided by the distance from the hand to the eye.

Calculations

Let the finger width be w, and the distance from the eye to the finger be r. Then the angular displacement of any object whose image crosses the finger is Δθ = w/r. Let Δt be the time to cross over a finger. Then calculate the angular velocity as ω = Δθ/Δt = w/(rΔt).

To find the velocity of the ISS, multiply the angular velocity by the distance R from the eye to the ISS: v = ωR. A slight complication is that the distance from the observer to the ISS depends on the elevation angle. If the ISS passes directly overhead, so maximum elevation = 90ᵒ, then R is just the typical distance between the Earth and the space station, R = 249 +_ 5 mi = km. If the maximum elevation angle of the ISS is θ, then R = 249/sin θ.

Example:

On the morning of March 30, 2015, in Phoenix, AZ, USA, the table in the Spot the Station website shows that the ISS first became visible 10ᵒ above the horizon, reached a maximum elevation of 77ᵒ, and travelled for ≈ 6 minutes before disappearing at an elevation of 12ᵒ.

Selecting a date to view the International Space Station

Holding my hand with fingers straight but slightly apart, so I could see narrowly between my fingers, my hand-width is 8.0 cm. Hence each finger averages 2.0 cm width. The average time to cross a finger was 3.0 s.

When I hold my bent hand above my head, the fingers are 33 cm above my eyes. Therefore the angular velocity of the space station appeared to me as ω = (2.0 / 33 * 3.0) = 0.0202 radians/sec. If the ISS had passed directly overhead, the velocity would just be ω*R, with R = 249 miles. But since the elevation was 77ᵒ instead of 90ᵒ, the perpendicular distance between the object and me (the observer) was actually the hypotenuse of a triangle whose height was 249 miles and whose angle was 77ᵒ. Therefore R = 249 / sin 77ᵒ = 256 miles. Converting to kilometers, R = 256 mi *1.61 km/mi = 411 km, and v = 0.0202 * 411 = 8.3 km/s. This is in reasonable agreement with look-up values – 7.6-7.7 km/s.

Discussion

All low orbits — including the space station– move at about 5 miles (or 8 km) per second. Since the space station is about 250 miles above the Earth, the angular velocity of motion past an earthbound observer is about 5 /250 = .02 radians per second.

It is exciting to realize that while you are viewing the ISS for 6 minutes, it is moving across the country a distance of 300 s x 5 = 1500 miles or 2400 km. It is exciting to think that when the ISS is just going out of your view, it is passing over some far-away place that is 750 miles distant!

Measure the force between a charged object and a metal sphere. Then use the force and distance to estimate the charge using Coulomb’s Law.

Basic Idea of the Experiment

Place a lightweight metalized ball on an electronic balance. The scale registers any changes in the force on the ball. A charged rod held over the ball induces an opposite charge in the ball, because the ball is grounded to the scale. It is not a bad approximation to assume the charges are equal and opposite point charges, separated by the distance between the top of the ball and the bottom of the charged rod. Use Coulomb’s law to calculate the amount of charge. Vary the distance to show the inverse square law.

Students hold negatively charged rod over a metalized sphere resting on an electronic balance

Materials for Coulomb’s Law Experiment:

Scales: Electronic balances having 0.1 or 0.01 gram sensitivity are now available at low-cost — in the range of $25. The balance pan is actually grounded by the internal circuitry, and intentional grounding seems unnecessary.

Metalized ball: Use styrofoam balls from craft stores. Wrapped in aluminum foil, these weigh several grams.

Plastic rod: PVC plumbing pipe holds a good charge and costs pennies per linear foot. We used 1″ OD pipe.

Fur for rubbing the pipe: Common scientific supplies. If these seem overpriced, an old wool sock works fine.

Experimental procedure:

Place ball on balance pan and press the tare button to zero the balance. Rub the end of the plastic rod with fur or wool, and hold it about 4 cm above the ball. Record the weight reading.

Discuss with students why the reading is negative — that is, the force between the positive ball and negative rod is attractive. If conditions favor static charge, the ball may roll around the balance pan — making the attraction obvious.

Charge the plastic rod once, and quickly get the weight readings for a sequence of different distances between ball and rod — say, 3 cm , 6 cm and 9 cm. Students will be able to demonstrate a rapidly decreasing attraction as distance increases.

Calculations and Discussion

The scale reads in grams. Convert to kilograms and multiply by g = 10 m/s^{2} to get the force in Newtons:

F =kq^{2}/r^{2}

Solve the force equation for q. For example, if the scale reads -0.5 gram for a distance of 4 cm, this gives q = [(.0005 x 10 x .04^{2}/(9 x 10^{9} ) ]^{1/2} = 30 nC †. A few tens of nanocoulombs is about as much charge as one produces in the classroom, while the class van der Graaf machine accumulates a couple of hundred nC.

Discuss the inverse power law for force dependence on distance for fixed amount of charge. If the weight loss is multiplied by r^{2}, for the same charge and different r’s, the product should be constant. This proves the Coulomb force is an inverse square law.

Related Questions

The configuration used in this experiment is not unlike a charge cloud hovering over a mountain top. Suppose the cloud contains 1.0 C of + charge, and this induces an equal and opposite charge on the mountain top. Assuming the cloud is 1000 m above the mountain, what is the Coulomb force between the cloud and the mountain?

Exercise: Draw in the E-field lines between the plastic rod and the metal sphere, showing where these lines originate and terminate, and showing the direction they point.

[Advanced topic] Note it is only an approximation to assert that the charge induced in the aluminum ball is equal to the charge on the plastic pipe. This assertion is probably true within an order of magnitude. Therefore the calculation of ± q on either the ball or the rod would be accurate within half an order of magnitude.

A steel ball rolls along a level track at a known speed. It collides with a neodymium magnet at the end of the track. The velocity of the attached masses is measured after the collision.

Basic Idea of the Experiment

Students give the ball a consistent x-velocity by rolling it down an incline from a fixed height they choose. They figure out its x-velocity by seeing how far from the table edge the ball lands on the floor. Then they put the magnet in the path of the ball — at position M in the drawing — and repeat the process so that the ball and magnet hit and attach as they launch off the table. Again they find the new distance the pair goes before hitting the floor to measure the velocity after collision.

Materials

Lab table

A light, stable ramp — we used some old tracks from toy “Hot Wheels” taped to a ring stand.

A 16 mm steel ball bearing, weighing ≈ 16 grams

Neodymium magnet — ours was 18 mm diameter x 9 mm tall and weighed 19 grams. But other choices are possible and could be interesting!

To measure the velocity of an object such as the steel ball just before rolling off the table alone, or the steel ball after sticking to the magnet, students must:

Figure out the time of flight by measuring the height of the table and then using h = ½ g t ^{2}

Measure the horizontal distance, on the floor, between the edge of the table and the landing point X.

Student Worksheet Downloadable Word Doc

We used the following word document as a lab procedure in the PHY121 class. However, it can be used in PHY111 or possibly in PHY101 as well.

The worksheet asks for predictions about conservation of momentum and energy, and guides the student through the data collection process. Finally the following questions are posed:

Conclusion: Are the initial and final momentum equal? Is this what you expected?

Are the initial and final kinetic energy equal? Is this what you expected?

How much heat is produced in the collision? This heat is what percentage of the original kinetic energy?

Author:

Terryl Fender, Prof. of Physics, South Mountain Community College, Phoenix, AZ