Virtual capacitor experiment

CAPACITORS

OBJECTIVES:

· To understand how a parallel plate capacitor works.

· To determine the dielectric constant for virtual paper used as the dielectric in virtual capacitor.

· To learn how capacitors connected in series and in parallel behave.

· To find equivalent capacitance for a complex combination of virtual capacitors.

EQUIPMENT:

Computer with Internet access.

INTRODUCTION AND THEORY:

Very common and important components in modern electrical devices are capacitors. They are what makes computer memory work. They are used with resistors in timing circuits, they occur as filters and coupling elements in every radio and TV set, they can eliminate ripples or spikes in DC voltages, they are used in flash units in photography.

A capacitor is considered a passive electronic element because it does not actively affect electrical currents in the circuit nor produce electrons or energy. Instead, it is used to store electrical charge (and hence electrical energy). A capacitor consists of a pair of conductors separated by a non-conductive material (or region) called dielectric which effectively insulates the two conductors.

The schematic symbol of a capacitor has two parallel vertical lines (representing the conductors) set a small distance apart, and two horizontal connecting wires:

                                                                         capacitor symbol

Any two conductors separated by an insulator (including vacuum) form a capacitor. When a voltage (potential difference) is applied across the capacitor, a charge is transferred to and from the conductors and an electric field develops in the dielectric. This field stores energy and produces a mechanical force between the conductors which can be released when needed. When a capacitor is charged the two conductors hold equal in magnitude but opposite in sign amount of charge so the net charge on the capacitor as a whole remains zero.

The capacitor’s ability to hold an electric charge is characterized by capacitance C defined as the ratio of the magnitude of the charge Q on either conductor to the magnitude of the potential difference ∆V between the conductors:

                                                                                                                                                 [Eqn. 1]

The SI unit of capacitance is one farad (1 F). One farad is equal to one coulomb per volt.

                                               1 F = 1 farad = 1 C/V = 1 coulomb/volt

The greater the capacitance C of a capacitor, the more charge Q can be stored on either conductor for a given potential difference ∆V and hence greater the amount of stored energy. One farad is a very large unit and we rarely see capacitors this big. In many applications the most convenient subunits of capacitance are the millifarad (1mF = 10-3 F), the microfarad (1µF = 10-6 F), the nanofarad (1nF = 10-9 F) and the picofarad (1pF = 10-12 F).

The value of the capacitance depends only on the shapes and sizes of the conductors and on the nature of the insulating material between them. The simplest way of making a capacitor is to build a unit with two parallel conducting fully overlapping sheets, each with area A, separated by a thin layer of air of thickness d.

Figure 1. Design of a parallel-plate

                 capacitor.

Air

                                              File:Parallel plate capacitor.svg

This arrangement is called a parallel-plate capacitor and its capacitance approximately (when d is small compared to the other dimensions and the field fringing effect around the periphery provides a negligible contribution) equals:

                                                                    C0 = , [Eqn. 2]

where is the permittivity of free space = 8.854-12 F/m. Please, note that A represents the area of overlap of the conducting surfaces.

The capacitance of a parallel plate capacitor is directly proportional to the area A of each plate (or in general to the area of overlapping) and inversely proportional to their separation d. Plates with larger area can store more charge. Similar effect takes place for the plates being closer together – according to Coulomb’s law when d is smaller the positive charges on one plate exert a stronger force on the negative charges on the other plate, allowing more charges to be held on the plates for the same applied voltage and this constitutes capacitor with larger C.

In commercial capacitors the layer of air is replaced by dielectric material such as Mylar, rubber, mica, waxed paper, silicone oil etc. If the space between the conducting plates is completely filled by the dielectric, the capacitance increases by the factor , called the dielectric constant (or relative permittivity) of the material which characterizes the reduction in effective electric field between the plates due to the polarization of the dielectric.

Figure 2. Charged parallel plate capacitor:

- air filled (top),

- with dielectric (bottom).

                                                  http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imgele/diel3.gif

In general the capacitance of a parallel plate capacitor with a dielectric can be expressed as:

                                                           C = = , [Eqn. 3]

where is now the permittivity of the dielectric = κ0 .

Capacitors are manufactured with only certain standard capacitances and working voltages. If a different value is needed for a certain application, one has to use a combination (series, parallel or mixed) of those standard elements connected in such a way to achieve the required equivalent capacitance.

For a series connection, the magnitude of charge on all plates will be the same. Conservation of energy dictates that the total potential difference of the voltage source will be split between capacitors and apportioned to each of them according to the inverse of its capacitance. The entire series acts as a capacitor smaller that any of its components individually.

 = ∆VSource = ∆V1 + ∆V2 + ∆V3 + ….. = + + + …

  = + + + …. [Eqn.4]

                    Figure 3. Two capacitors in series. (a) Schematicillustration. (b) Equivalent circuit diagram.

Capacitors are combined in series mostly to achieve a higher working voltage.

In a parallel configuration all capacitors have the same applied potential difference. However, their individual charges might not necessarily be equal because the charge is apportioned among them by size (C). Since the total charge stored in such combination is the sum of all the individual charges, the equivalent capacitance equals the sum of the individual capacitances and is always greater than any of the single capacitance in this arrangment.

QTotal = Q1 + Q2 + Q3 + …. = C1∆V + C2∆V + C3∆V +….. = (C1 + C2 + C3 +…)∆V

  CEQ = C1 + C2 + C3 + … [Eqn. 5]

                Figure 4. Two capacitors in parallel. (a) Schematicillustration. (b) Circuit diagram. (c)Circuit

                                 diagram with equivalent capacitance.

PART I. Parallel plate capacitor.

Open the PhET Interactive Simulations web page http://phet.colorado.edu/en/simulation/capacitor-lab and download the Capacitor Lab.

The two parallel plates connected to the battery constitute the simplest capacitor. You can change the parameters of this capacitor by placing the cursor on the green arrows and moving it along the indicated directions (up and down to modify the plate separation, sideways to alter the area of plates).

From the right side tool bar menu select all options except the electric field detector. With the mouse left click grab the red test lead of the voltmeter and place it on the upper plate. Next, grab the black test lead and move it to the lower plate. The voltmeter is now ready to measure the potential difference ∆V applied to the capacitor. The potential difference between the capacitor plates will depend on the position of the vertical slider on the battery which allows for adjustment of the battery output.

1. “Design” your own air-filled parallel plate capacitor with plate separation d = 7 mm and having the plate area A of 250 mm2. Apply approximately 1.00 V to it. If the bar graphs displaying the capacitance, the plate charge and the energy go out of range, click the magnifying glasses icons positioned next to them. Capture the screen and paste it into MS Word file – you will have to attach this screen shot to your lab report.

                                      Figure 5. Custom air filled capacitor – example.

              For your custom capacitor calculate its capacitance (based on its dimensions, ), the

              stored charge ( = and the stored energy (). How do the calculated

              values of C, Q and E compare with the corresponding ones shown in the simulation – calculate

              discrepancy?

2. Investigate the effect of changing the plate separation and the plate surface area (one parameter at a time) on the capacitance, the amount of stored charges and energy. In your statements quote values extracted from the simulation displays.

3. To your custom capacitor designed in step 1 add a dielectric.

Press the “Dielectric” tab in the simulation window and fully insert between the plates a custom dielectric of dielectric constant 5.0. The dielectric is aligned perfectly with the plates when the offset = 0 mm. Adjust the separation and plate area to match your settings in step 1. Apply the same voltage as you had in step 1 and record the values of capacitance, charge and energy displayed by the simulation. Capture the screen and paste it into MS Word file – you will have to attach this screen shot to your lab report.

                              Figure 6. Custom capacitor with dielectric – example.

By how much are these values different from the corresponding numbers registered for your custom air-filled capacitor? Do you measurements validate the statement that inserting a dielectric between the plates increases the capacitance by a factor equal to its dielectric constant?

4. Now, using your custom capacitor, you will conduct an experiment to determine the dielectric constant of paper. In this part you have to be extremely careful about the units!

In the Dielectric “Material” window select paper from the drop down menu. As you slide the dielectric out of the capacitor, take readings of the capacitance C and the offset ∆x for 5 positions of the dielectric material in between the plates. In Logger Pro program make a plot of C (y-axis) vs. ∆x (x-axis). Apply linear fit to your set of data and from the intercept of this linear fit find the dielectric constant of the paper along with its error. How does your experimental value of the dielectric constant for paper compare with the known value of 3.5.

Attach the graph from Logger Pro to the lab report.

Hint.

The capacitor with partially inserted dielectric can be looked at as two capacitors connected in parallel:

- the first one air-filled of plate area= , where symbolizes the dielectric offset measured from the left edge of plates, and represents the width of plates but, since the original plates are square in shape, is equal to (A is your custom plate area),

- the second one filled with dielectric material of plate area =

If both capacitors have the same plate separation d, we can describe their capacitances respectively as:

                               = = and = =

Hence, the equivalent capacitance for the two capacitors in parallel after rearranging some terms comes to:

                                        + [Eqn. 6]

It is apparent that eqn. 6 is a linear function of (same as with intercept

b = .

Then , and if we know the uncertainty in the intercept from the Logger Pro fit, we can propagate it into the error in dielectric constant: .

PART II.Capacitors in parallel – sharing charges.

Download the Circuit Construction Kit (AC + DC), Virtual Lab from the PhET website: http://phet.colorado.edu/en/simulation/circuit-construction-kit-ac-virtual-lab .

Construct the circuit shown in the figure below.

By right-click on the circuit elements adjust their values to the following parameters:

V0 = 9 Volts, C1 = 0.1 F, C2 = 0.1 F, C3 = 0.05 F.

To add the voltmeter, select it from the tools menu on the right hand side panel.

V0

C3

C2

C1

W1

W3

W2

Step 1. Charging capacitor C1 (initially discharged) → keep switches W2 and W3 open; close switch W1. Connect the voltmeter leads across C1 and measure voltage V1. Is it the same as V0? Calculate the charge Q1 stored on capacitor C1, Q1 = C1V1. What is the magnitude of positive charges deposited on C1? What is the magnitude of negative charges deposited on C1?

Step 2. Sharing the charge stored on C1 with capacitor C2 (initially discharged) → keep switch W3 still open; open switch W1 and close switch W2. Measure voltage V2 across capacitors C1 and C2 and calculate the amount of charges stored on each of them, Q12= C1V2 and Q2 = C2V2 . Is the measured voltage V2 different then V1? Why? Knowing the values of capacitance C1 and C2 and the initial charge Q1 stored on C1, calculate the theoretically expected voltage V2Theory after capacitor C2 got connected in parallel with C1. Does your measured voltage V2 agree with the calculated theoretical value of V2Theory?

Step 3. Sharing the charge stored on C2 (from step 2) with capacitor C3 (initially discharged) → keep switch W1 still open; also open switch W2 and then close switch W3. In this configuration the charge stored in step 2 on C2 will be distributed between C2 and C3. Following the procedure from step 2 measure the voltage V3 across C2 and C3, calculate the charges deposited on individual capacitors and find analytically the theoretical voltage V3Theory. Does your measured voltage V3 agree with the calculated theoretical value of V3Theory? Which capacitor stores less charges and why?

Step 4. Distribution of charges between three capacitors in parallel (all capacitors initially charged) → keep switch W1 open and the switch W3 closed. Capacitor C1 should still hold the charge from step 2 (Q12), while capacitors C2 and C3 should have the amounts of charges determined in step 3 (Q23 = C2V3, Q3 = C3V3). Close the switch W2. Measure the new voltage V4 across capacitors C1, C2 and C3. Prove analytically (by clear calculation) that the measured voltage agrees with the theory.

Hint: The total initial charge in this case is the sum of charges Q12 + Q23 + Q3. This amount of charge is conserved throughout the experiment and should equal the total final charge stored on all three capacitors after closing the switch W2 and equalizing the voltage to V4Theory.