Abstract: and d is the plate separation. The

Abstract:      This experiment illustrates how capacitance and stored energy are effected by the construction of the capacitor.  The Phet simulation, Capacitor Lab 2.02, is used to investigate various aspects of capacitors and the resulting effects.  A capacitor is constructed using an anode a cathode and a dielectric material.  The dielectric will help influence the capacitance and the stored energy of the capacitor.  A dielectric place half way in/out of the capacitor will be drawn into the capacitor as it pursues a location of lesser potential energy.  At the same time the dielectric will be pulled toward the capacitor because more work is required to force the dielectric out of the capacitor.?__________________________________________________________________________Introduction:    In this lab I studied capacitors including observing for trends in capacitance and stored energy.  A capacitor is constructed of two identical metal plates and a dielectric material in the center.  According to Rex and Wolfson (2010), “Capacitance depends only on the configuration of the two conductors.”  This statement can be proved by looking at the equation for capacitance.C=(?_0 A)/dwhere ?0 is a constant, A is the area of the identical plates, and d is the plate separation.  The equation proves that capacitance is solely reliant on capacitor construction.  Furthermore, the capacitance for a parallel plate capacitor will increase linearly in relation to plate area; and capacitance will decrease as 1/d in relation to the plate separation distance.  Here is some data to support my statements:Charge (V)    Area (mm2)    Separation (mm)    Capacitance (F)    Plate Charge (C)    Stored Energy (J)99    100    10    8.90E-14    8.70E-14    4.30E-1499    149.1    10    1.32E-13    1.30E-13    6.40E-1499    202.6    10    1.79E-13    1.77E-13    8.70E-1499    252.5    10    2.24E-13    2.21E-13    1.09E-1399    301.5    10    2.67E-13    2.63E-13    1.30E-1399    347.9    10    3.08E-13    3.04E-13    1.50E-1399    400    10    3.54E-13    3.49E-13    1.72E-13 Charge (V)    Area (mm2)    Separation (mm)    Capacitance (F)    Plate Charge (C)    Stored Energy (J)99    100    10    8.90E-14    8.70E-14    4.30E-1499    100    9    9.80E-14    9.70E-14    4.80E-1499    100    8    1.11E-13    1.09E-13    5.40E-1499    100    7    1.26E-13    1.25E-13    6.10E-1499    100    6    1.47E-13    1.45E-13    7.20E-1499    100    5    1.76E-13    1.74E-13    8.60E-14     The energy stored on a capacitor can be expressed mathematically as:U=Q^2/2Cwhere U is the stored or potential energy, Q is the charge, and C is the capacitance.  Therefore, half of the work done to move the charge from one plate to the other is stored as stored energy (Hyperphysics (n.d.).  Stored energy, potential energy, and work are equivalent in this instance.__________________________________________________________________________Methods:    The Phet simulation, Capacitor Lab 2.02, will be used to complete this lab experiment.  The simulation will open up in the introduction tab; it is necessary to select the dielectric tab.  The following features should be enabled:    Plate Charges    Capacitance    Stored EnergyNext, set the battery voltage to a practical value, and then disconnect the battery.  Take note of any changes in the capacitance as the dielectric is slid in between the charged plates.  Slide the dielectric back out and note any changes.  Determine how the dielectric would react if it were moved half way out/in of the capacitor.  Would the dielectric be pulled into the capacitor or would it be forced out of the capacitor?    Figuring out exactly what the battery’s voltage is may be a little challenging.  Using the electric field detector initially can help with determining the voltage of the battery.__________________________________________________________________________Results:    Prior to beginning this experiment I hypothesized that the dielectric would be pulled into the capacitor.  I performed the experiment as discussed in the methods section.  I was able to verify and set the battery voltage to 99V (before disconnecting) by utilizing the electric field detector.  I also left the area and the separation values at their default positions, 100 mm2 and 10mm respectively.  Here is the data I collected:Charge (V)    Area (mm2)    Separation (mm)    Offset (mm)    Capacitance (F)    Stored Energy (J)99    100    10    20    8.90E-14    4.30E-1499    100    10    15    8.90E-14    4.30E-1499    100    10    10    9.00E-14    4.30E-1499    100    10    5    2.66E-13    1.40E-1499    100    10    0    4.43E-13    9.00E-15      The data I have collected along with the graphs I produced help to support my preliminary supposition.  Similar to a gravitational system, where mass will always fall, seeking a lower potential energy state, the dielectric will be drawn into the capacitor, which will result in a lower stored energy, or potential energy.  __________________________________________________________________________Discussion:  The data and graphs illustrate the effect of “offset” on the capacitance as well as on the stored energy of the capacitor.  The graphs clearly indicate that as the dielectric moves closer to the capacitor the capacitance increases and the stored energy decreases.  Also, as the dielectric moves away from the capacitor, the capacitance decreases and the stored energy increases.  This can be further explored by using arrow analysis on the equation:U=Q^2/2CFirst, I determined that the basic equation would apply to the dielectric at its starting point (outside of the capacitor).  Next, I wanted to find out how the equation would change if the dielectric moved in toward the capacitor.  The charge does not change and will be constant.  The capacitance will increase as the dielectric is moved in between the charged plates of the capacitor.U=?(??Q^2  )/(2C?)Therefore, U?If Q is constant and C increases in value, then U will decrease.  The result is a lowered stored or potential energy state.  This further proves that the dielectric will be drawn into the space between the charged plates.  According to Rex and Wolfson (2010), “Inserting a material with dielectric constant k doesn’t alter the charge, but increases the capacitance by a factor of k.  The new capacitance is C=kC0, so the energy becomesU=Q^2/2C=Q^2/(2kC_0 )or U = U0/k.  Thus the store energy decreases by a factor of k.”  My arrow analysis falls in line with this statement.__________________________________________________________________________Conclusion:    In regards to a dielectric in a capacitor, the capacitances and the stored energy of said capacitor will have opposite responses to its movement.  When the dielectric moves towards the middle of the capacitor the capacitance will increase as the stored energy decreases.  This fact is essential to the experiment because the dielectric will be drawn to the area of least potential energy.  Therefore, a dielectric free to move will be pulled into the capacitor.  In addition, the equation used  for stored energy also applies to work, consequently, this means that for the dielectric to be pushed out more work is required.__________________________________________________________________________?References:Hyperphysics (n.d.). Energy stored on a capacitor. Retrieved from http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.htmlRex, A. & Wolfson, R. (2010). Essential College Physics. Glenview, IL: Pearson Education, Inc.