Lab 2 - Capacitance

# When the lcr meter is connected to the foil capacitor

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When the LCR meter is connected to the foil capacitor is the capacitence contributed by the wire leads additive? If so why? In the following section the contribution from the LCR meter will be an important factor to consider. 2

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Finally, your TA will show the charging of the foil capacitor using a high voltage power supply. What is the crackling you hear? What’s happening to the foil as charge flows into the conduc- tors? Approximate how much energy is being stored in your capacitor. Neglecting fringing, what is the strength of the electric field storing this energy? 5 Capacitor Basics To better understand capacitors and how they function we will construct one and measure it’s capacitance. By changing the types of materials located between the capacitor plates we can see how the permittivity of various dielectrics affects capacitance. Dielectrics are insulating materials that can be polarized by incident electric fields. This polarization occurs on the microscopic level and can result in a change in the strength of the electric field (either opposing or supporting the externally applied field). Changing the strength of the E-field will result in a change in the capacitance of the entire device by way of changing the potential difference between the plates. With the calipers start by measuring the area of the capacitor plates you were provided as well as the thickness of the dielectric medium you will be using. Neatly record these values in your lab notebook to be used later. Slide a piece of the clear acrylic between the plates of the capacitor. Press the two plates together so that they are flush with it. Using the LCR Meter measure the capacitance of your set-up. NOTE: You will need to take into account the capacitance of the wire leads and subtract this value from the total to get that of the set-up. To do so disconnect the LCR meter from the plates without disturbing the wires and record the resulting value. Now slide the acrylic slab out while maintaining the plate separation (be sure not to touch the two plates together). Again measure the capacitance. How do these two values of capacitance compare? Which one is larger; is this what you expected? It is a good approximation to assume the relative permittivity of air is 1 (actual value of 1.00059), thus as a dielectric it behaves essentially like vacuum, for our purposes. Therefore assuming the capacitor without the acrylic slab has a relative permittivity of one ( ε r = 1) then, based on your Capacitance measurements and the fact that the geometry stayed the same, calculate the relative permittivity of the acrylic slab (hint: Use Eq. 3 and your second capacitance measurement to solve for the term ε 0 A d . This should be constant for both measurements). When charge is placed on the plates of the capacitor what is created between them? How is this related to the voltage difference between the plates? How much energy would it take to move a single charge from one plate to the other?
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