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CAPACITANCE and DIELECTRICS

In class we examined the theoretical nature of a capacitor ; there emerged a relationship between charge, voltage, and capacitance. We can experimentally investigate this relation and perhaps verify that

Q = CV

(1)

We may proceed by requiring each of the variables be held constant in turn, varying one of the others, and measuring the third with the electrometer.

The capacitance of a parallel plate capacitor is

C = o A/d

(2)

in a vacuum, where _o is the permittivity of free space, A is the plate area, and d is the plate separation. One can insert various materials between the capacitor plates, hold Q constant, measure changes in V, and, in fact, calculate the dielectric constant of a given material.

NOTE: It is assumed that the student understands the theory of capacitors connected in series and parallel.

APPARATUS:

Figure 1. Parallel Plate Capacitor and Electrometer

Determination of Capacitance of Electrometer

Prior to using the apparatus to determine the above relationships, you must FIRST estimate the capacitance of the wires and meters. Use the experimental capacitor and connect it to the electrometer with the low capacitance (unshielded) cable provided. It is important to remember that the electrometer and cable add approximately 35 pF capacitance in parallel with the experimental capacitor. (See Figure 1). In order to measure the electrometer's capacitance at the probe end of the unshielded cable, obtain a low leakage (polystyrene or air dielectric) capacitor of known value (about 100 picofarads). (Check the Appendix for the proper identification of the disc capacitors. The first two digits should be 10, indicating the first digits of the actual value. The third digit should read 1, indicating the number of zeros following. Thus the label should indicate 101 followed by a K, (tolerance of 10%). This identifies the capacitor as 100 pF. (One hundred picoFarads). This is for C₂. Connect the known capacitor across a known voltage V₂ (about 30 volts). Disconnect the power supply and reconnect the capacitor across the electrometer input cable. C₁ is the capacitance of the meter and cables combined. Take care not to discharge the capacitor by grounding it in any way. Note the voltage V, indicated by the electrometer. V₂ = 30 volts This is the value of voltage across the combined system of the meter and wires (C₁) and the known capacitor (C₂). They are in parallel, so the charge from C₂ is now shared across both C₁ and C₂. The capacitance C₁, can then be calculated from the known capacitor's value C₂, and the known charging voltage V₂ (See Figure 1 below for the formula).

	C2 = 100pf = Q / V2
	Q = C2V2
	Ceq = C1 + C2
	= Q / V = Q / V1 = C2 V2 / V1
or	C1 = C2 (V2 - V1) / V1	(4)

Figure 2. Known Capacitor used to measure capacitance of wire and meter. A proof plane, power supply, and sphere will also be used for transferring charges and maintaining voltages. The Faraday ice pail is necessary to measure charge.

PROCEDURE:

A. Measuring Capacitance Variables:

There are several ways in which one can hold one of the variables (capacitance, voltage, charge) constant, vary another, and measure the third. You should get a linear relationship, e.g. C = (1/V) Q where (1/V) is the slope of the line.

Two of these ways are minor variations of the others and hence do not represent a significantly different method of performing the experiment.

1. V Measured, Q Variable, C Constant:

   Connect one of the aluminum spheres to the green binding post of the power supply and set the power supply to 1000 VDC. Since the electrometer should be grounded, momentarily depressing the "zero check" button will remove any excess charge from either of the capacitor plates.

   With an initial plate separation of about 2 mm., use the proof plane to transfer charge from the sphere to the ungrounded capacitor plate (i.e., the plate connected to the red electrometer lead). The charge is transferred merely by touching the proof plane to the sphere and then to the capacitor plate. If the proof plane is touched to the sphere and capacitor in the same manner each time, then equal amounts of charge will be transferred each time. Why is it sufficient to add charge to only one plate?

   Adjust the electrometer's sensitivity so that each charge transfer results in a measurable deflection of the meter needle. Make several charge transfers and record the electrometer reading each time.

   How does the potential vary with charge? Using Excel to plot the results. What is the constant of proportionality? You are plotting "chunks" of charge versus voltage.

   CAUTION: Take care to place the charged sphere and power supply sufficiently far from the electrometer and capacitor so that the plates are not charged by induction or affected in any other way.

2. V Measured, C Variable, Q Constant:

   Connect the electrometer to the experimental capacitor with the black lead fastened to the fixed plate. With an initial plate separation of about 2 mm., charge the plates by momentarily connecting the power supply (set at about 6 volts) across the plates. Adjust the electrometer's sensitivity so that the initially charged plates represent a meter reading of about 1/5 scale. i.e., set the Electrometer on the 30V scale. This gives lots of room for the meter to change values. Because the capacitance is determined strictly by the geometry of the plate system, once a known voltage is applied across the plates, the charge is also known. In fact, it could be calculated.

   Increase the plate separation and note the electrometer's reading at various separations. How does the potential vary with capacitance? Again, use Excel to plot the results. You should plot both V vs. C and V vs. d, the plate separation. Discuss the differences and the relative usefulness of these two plots.

B. Determination of Dielectric Constants:

To determine dielectric coefficients, it would be ideal to just slip a sheet of dielectric between a set of charged capacitor plates and note the resulting potential changes. However, there are experimental problems with such a procedure. Sliding the dielectric between the plates can generate a significant static charge. Hence, it is best to proceed as follows:

With the electrometer connected across the plates of the experimental capacitor, raise the end opposite the stationary plate by setting it on a block about 3 cm. high.

With the plates set at about 3 mm., charge the plates to about 4/5 full scale; i.e., charge to about 24 V with the Electrometer set at 30 V. Record the reading. Carefully increase the separation of the plates enough to insert the dielectric sheet (without sliding it against either plate during the insertion process) and lean it against the stationary plate. Return the plates to the original separation and record the new electrometer reading across the plates.

Now, pull the plates apart again and remove the dielectric sheet carefully. Again, be sure not to allow the sheet to slide against either plate. Return the plates to the original position and check that the electrometer reading agrees with the original reading. (Caution: before using the phenolic dielectric sheet, it may be necessary to breathe on it in order that the moisture in your breath will leak off any residual charge.)

The calculations necessary to determine the dielectric constant are involved, but straightforward. First we assume that the dielectric fills the entire space between capacitor plates. Referring to the diagrams, we proceed:



Figure 3. Effect of added Dielectric Plate.

Q_i' is the charge on plates of capacitor C_i (i = 1,2.) V_o and V₁ are the voltages indicated by the electrometer before and after the plate is inserted respectively.

We can write:

(Q₁ + Q₂) = (C₁ + C₂) V_o and (Q₁' + Q₂') = (C₁ + C₂') V₁

but	Q1 + Q2 = Q1' + Q2' (total charge is constant)	(5)
hence	(C1 + C2 ) Vo = (C1 + C2' ) V1

Now, remember that by definition:

C' = C (eg. 17.32 of text)

(6)

We can write:

	= [ C1 (Vo - V1 ) + C2 Vo ] / C2 V1	(7)
	(C1 comes from calibration)

is the dielectric coefficient of the system with airspace and the dielectric. In the actual experiment, the dielectric does not fill the entire space between capacitor plates. Therefore we need to alter our calculations slightly. Recall that the electric field between two charged plates is constant at each point between the plates. The small electric field in the dielectric is also constant and acts to reduce the capacitor's electric field. Hence the reduction depends linearly on the thickness of the dielectric. With this in mind, let be the coefficient calculated above, and let ' be the actual dielectric coefficient of the dielectric plate placed between the two plates. If d is the plate separation and t is the thickness of the dielectric. You can calculate the actual coefficient given the experimentally determined one from equation (7).

This is similar to a homework problem 17.61 (Not assigned in class.) Hint: Consider two capacitors in series!

We can write:

1 / Ceq = 1 / Cair + 1 / Cdielectric

(8)

so that:

1 / (A/d) = 1 / A / (d - t) + 1 / ' ( A/t )

(9)

then:

d / = ( d - t ) / 1 + t / '

Solving this we get:

' = t / [ d - ( d - t ) ]

Do the algebra yourself. Submit this as part of your write up you should get the same result. The value of ' is that of the dielectric plate, itself is the value of the plate and the air together. Check ' with the Handbook of Chemistry and Physics. How close do the values agree? Discuss.

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