Objective:
The objective of this experiment is to verify the exponential behavior of capacitors during charging and discharging processes.
Equipment:
A computer with the Internet connection, a calculator, paper, and pencil
Theory:
A capacitor is a passive electric device that stores electric energy. A parallelplates capacitor is made of two parallel metallic surfaces, each of area A, separated by an insulation layer of thickness d, and it has a capacity of
where C is the capacity in Farads, A the area of each plate in m^{2}, d the insulation (dielectric) thickness in (m), and ε_{o} the permittivity of free space (vacuum) for electric field propagation expressed in F/m that reads Farads/meter.
The factor, κ , pronounced kappa denotes the dielectric constant, and depends on the material of the insulation layer. The capacitance C does not depend on the material of the plates. The dielectric constant ε_{o} is related to Coulomb's constantk by
Figure 1
In Fig. 1, at t = 0, the capacitor is uncharged. As soon as the key in the circuit is closed, electrons flow from the negative pole of the battery toward the lower plate of the capacitor. They distribute over the lower plate, making it negative. At the same time, the repelled free electrons of the upper plate flow toward the positive pole of the battery. This causes the upper plate to become positively charged. This process does not happen suddenly. It takes some time. The current is greatest to begin with, and decreases as charges accumulate on the plates. At the beginning the capacitor is empty; therefore, the voltage across it is zero, but as more and more charges build up on its plates, its voltage keeps increasing. The voltage across the capacitor V_{C} asymptotically approaches the battery voltage V_{Bat }.
During the charging and discharging processes, the voltage across the capacitor and the current through it follow the following exponential equations:
(Charging) Battery in Circuit

At t = 0,
V_{C }= 0
I_{C} = V_{B} / R

As t → ∞ V_{C} = V_{Bat.} _{ } I_{C }= 0


(Discharging) With battery removed, the initial capacitor voltage is V_{o}= Q_{o}/C making the initial current I_{o} = V_{o}/R

At t = 0,
V_{C }= Q_{o }/C
I_{C} = Q_{o }/(RC)

As t → ∞
V_{C} = 0 _{ } I_{C }= 0

It is a good idea to examine the values in the third and fourth columns by once setting t = 0 and once t → ∞ in the appropriate equations. Note that the chargevoltage formula for a capacitor is Q = CV. These exponential variations will be observed in this experiment.
Procedure:
You may need to add the following Website to your Java exception list: http://www.phy.ntnu.edu.tw/. To do this, follow the path (Windows operating system),
Start → All Programs → Java → Configure Java → Security (use High) → Edit Site List … → Add → Type in the site URL (http://www.phy.ntnu.edu.tw/).You will notice circuits as shown in Figures 2 and 3 when you click on the following link.
Figure 2
As shown above. the twoway switch must be put in position 1, for the charging process. Doing this puts the battery in the circuit.
For discharging, as shown below, the twoway switch must be put in position 2 in which case the battery is excluded.
Figure 3
Now, click on http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=31.0 .
Start the applet, and with the mouse move the blue bar (on the left) up and down to set a battery voltage. You may then close the circuit for charging or discharging at the twoway switch.
Part I: Charging
1) As you see the resistance in the circuit is 100.0 kΩ that means R = 100,000Ω. The capacitance is 100.0μF, or C = 100.0x10^{ 6}F. Calculate the time constant τ = RC of this circuit. Record the value of the time constant, τ.
2) Now that you know τ = 10.00sec, go to the next step.
3) Click on the twoway switch a few times to observe the charging and discharging processes. If you hold the left click down on any part of the figure, you can temporarily stop the experiment and take readings. This eliminates the need for helpers as explained below. You will see that at intervals of 5 seconds, you will need to take readings. You can easily selfteach yourself the way it works.
4) Reset the circuit. At the top, set the time to zero. First choose a battery voltage of 10.00 volts by using the blue bar beside the battery. Start the circuit for charging. Have one helper to read V_{C }, the voltage across the capacitor, and the another helper read I_{C }, the current through the capacitor. While the circuit is on, V_{C} can be read just beside the capacitor, and I_{C} can be read at the top of the circuit below the moving yellow dots that represent the current. Your job is to carefully announce "time" at equal intervals of 5 seconds. Your helper job is to read the voltage or current as soon as he/she hears you announcing the word "time." and record it in the chart provided under "Data."
Note that the values of V_{C} and I_{C} at the very beginning (t = 0) are known or can be easily calculated. At t = 0, the capacitor is empty and therefore its voltage is zero. This is already recorded in the chart.
You need to calculate I_{C} at t = 0 and record it in the chart where it says V_{B}_{ }/R . At t = 0, all of the battery voltage drop across the resistor. Divide the battery voltage you have chosen by R = 100,000Ω to calculate the initial current (current att = 0).
{If you choose V_{B} = 6.00volts, for example, then the initial current will be (6/100,000)Amps or 60.0x10^{6}Amps, or 60.0μA. This is just an example.}
5) If you chose the battery voltage to be 10.00volts, then the initial current will be 100.0μA. Calculate the initial current and record it. When you are ready to turn the circuit on (t =0), you just say "Ready, set, go." Every 5s, when you say time, the helper that is watching the voltage, reads the voltage to his/her best judgment and writes it down, and the other helper does the same for current. This should continue for 60 seconds. After 50 seconds, some time between 50 to 60 seconds, you may want to click on the "stop" button and end the charging process because the capacitor is almost fully charged.
Part II: Discharging
6) After 50 seconds of charging, it means after 5 timeconstants (if τ = 10.00s), the capacitor is almost fully (99 percent) charged and its voltage almost equates the battery voltage. Now you need to prepare the group for thecapacitor discharging process. The final V_{C} of the charging stage becomes the initial V_{C} of the discharging stage. Record V_{C}_{ } in the chart. You may use this voltage to calculate the initial current and write it in the chart as well.
When you announce "Ready, set, go", you already must have the values of V_{C} and I_{C} at t = 0 as in Part I. Again at 5s intervals, announce "time" for your helpers to record the values they read. Continue for another 60 seconds and record the values of capacitor voltage and current in the second chart.
Graphs:
Graph the following: For charging: V_{C} versus t and I_{C }versus t, and
for discharging: V_{C} versus t and I_{C} versus t.
Data:
Given:
V_{B} = the battery voltage (to be set at the starting time of charging)
V_{oC} = the initial capacitor voltage (to be read at the start of discharging)
Measured:
Charging: Note that the value of I_{C} at t = 0 is V_{Bat }/R.
t (sec)  0  5  10  15  20  25  30  35  40  45  50  55  60 
V_{C} (volts)  0  
I_{C }(Amps)  V_{B}/R 
Discharging: Note that the value of I_{C} at t = 0 is V_{oC }/R.
t (sec)  0  5  10  15  20  25  30  35  40  45  50  55  60 
V_{C} (volts)  V_{oC}  
I_{C }(Amps)  V_{oC}/R 
Calculations:
For the charging part: I_{o} = V_{B} /R.
For the discharging part I_{o} = V_{oC} /R.
Comparison of the Results:
The 4 graphs may be compared with the corresponding graphs in your text or course material.
Conclusion: To be explained by students.
Discussion: To be explained by students.