Experiment 2: Field Mapping/ Capacitors
Part 1 - Electric Field Mapping
BACKGROUND: A DC power supply charges
two electrodes printed on a sheet which is coated
with a poor electrical conductor. The current
through this coating is large enough to give a
voltmeter something to measure, but small
enough that conditions are approximately static.
You will use the voltmeter to find several curves of constant potential. The direction of the field
vector, ⃑ , is perpendicular to these equipotentials. (With gravity, water runs in the direction that the
elevation changes fastest. Similarly, ⃑ points in the direction that the potential changes fastest). So,
you will sketch field lines by drawing them perpendicular to the equipotentials.
PROCEDURE:
1. Hook up two wires, one from each electrode on the sheet, to the DC terminals of a power supply.
(Red and black.) Hook up two more wires: One from minus on the digital voltmeter to minus on the
power supply, and one from plus on the meter to the probe. (The thing with the black handle.)
2. With the probe touching the sheet’s positive electrode, set the power supply’s knob so that the
meter reads 6.0 V.
3. Move the probe around the sheet to find points where the potential is 1.0 V. Plot enough on the
answer sheet to accurately sketch a curve of constant potential all the way across the picture.
(Maybe 2 or 3 cm apart; closer where the curve bends sharply.)
4 Repeat, obtaining equipotentials at 2, 3, 4 and 5 V.
5. Draw in some approximate lines of force by sketching smooth curves that cross all
equipotentials at right angles. The edges of the printed electrodes are your
equipotentials for 0 V and 6 V. Include arrowheads, and label which are the positive
and negative electrodes. Draw enough to show the field in all areas of the picture.
(Since all you did was to draw a picture, there is no conclusion to draw when you write your
discussion for this part.)
Part 2 – Parallel Plate Capacitors
Two parallel plate capacitors are known to have the same voltage between their plates. The plates
are twice as far apart in one capacitor as in the other. You will predict how the capacitors’ charge is
affected by the distance, then take measurements to see if you are right.
Theory: Combine the formula for the capacitance of parallel plates with the formula that defines
capacitance in general to obtain an expression relating charge to the distance between the plates.
According to this, what should doubling the distance (at the same voltage) do to the amount of
charge stored?
Experiment: A digital voltmeter connected to a known capacitance makes a good “charge meter:”
From the capacitance printed on the capacitor and the voltage on the meter, you can find q from C =
q/V. Since the capacitance of the parallel plates is small, a large voltage is needed to deposit a
measurable amount of charge. Therefore, a Van de Graff static generator capable of producing
around 200, 000 V will be used to charge the system.
Cautions: Although its voltage is quite high, the Van de Graff can only throw a small amount of
charge through you, so it is not normally dangerous. However,
1. Anyone with a pacemaker should keep back from it.
2. Electronic devices such as computers, cell phones or calculators should be at least a meter
away, due to the sensitivity of miniaturized parts to sparks. (If you follow the directions, sparks
are only possible between the capacitor plates. But, I never quite know what students will do.)
3. A shock from the Van de Graff is not otherwise dangerous, but doesn’t feel good at all, so
you might want to avoid it.
Get two identical capacitors around 25 F, and
connect this circuit. A wire goes from the top of the
Van de Graff to one plate of CW. Another wire goes
from the other plate of CW to the 25 F. And a third
wire goes from the other side of the 25 F to the
bottom of the Van de Graff. Likewise, use three more
wires to make the other path from the top to the bottom of the Van de
Graff, through the other two capacitors.
- To attach a wire to the top of the Van de Graff, lift off the dome,
and attach an alligator clip to the dome’s support. For the bottom,
there is a grounding tag near the power cord on the green ones; clip
it to the metal base if you have the other one.
- The capacitors CN and CW are parallel-plate capacitors you will make out of metal plates
which stand on the counter. To start with, put the plates of CN about 1 mm apart and CW about
2 mm apart. (Don’t invest time in carefully measuring this - it’s not going to work anyhow.)
- You must observe correct + and – polarity with the 25 F capacitors.
Once all of that is done, put two wires on each
meter, and run those to opposite sides of the 25
F capacitors. Set the meters to a scale which
will read thousandths of a DC volt.
Notice that both CN and CW will have essentially
the same voltage across them, as you assumed under “theory”: Both have one plate connected to
the top of the Van de Graff, and both effectively have the other plate connected to the bottom of the
Van de Graff. (There is a fraction of a volt across the 25 F capacitors, but compared to thousands
of volts from the Van de Graff, this is nothing.)
Have the instructor approve your wiring, then try to charge the capacitors by running the Van de
Graff. Turn it off. Describe what happened.
Touch the ends of a wire to the parallel plates, one end to each plate, to discharge them. If you
leave them charged, there could be a nasty surprise if you touch them.
A dielectric between the capacitor plates will solve the problem. Old transparencies work pretty
well. Put one sheet between the plates of CN and two sheets in CW.
Both meters should read zero when you start. If not, discharge capacitors through a wire until they
do. (Both the parallel plate capacitors, and the 25F capacitors.)
Run the Van de Graff until the meters read something like a tenth of a volt. As soon as you turn off
the Van de Graff, the readings begin to slowly drop, as the capacitors discharge through the
voltmeters. Take your readings soon after turning off the Van de Graff, at the same time.
Calculations: As will soon be covered in class, capacitors connected in series have the same charge.
So, calculate the charge on each 25F capacitor from the measured voltages. These numbers are
also the charges on the parallel plate capacitors.
Conclusion: How do the charges you determined compare to your prediction?
PHY 122
Part 1.
Report on Experiment 2
Part 2:
Show how to obtain expression relating Q and d:
So, twice the distance should mean ____________ (how many times?) the charge.
What was the problem with air between the capacitor plates?
VN = _______________, VW = ______________
Calculate QN and QW: