Preparation: A Click on the "Dielectric" tab at the top. Make sure all the check boxes on the right side of the screen are checked. Choose "Glass" as the dielectric material (from the drop down menu on the right). K glass- 4.7 A Adjust the plate separation to its minimum value of 5 [mm], and the plate area to its maximum value of 400 [mm^2] by clicking and dragging the double-sided arrows. A Place the tip of the red probe of the Voltmeter on the top plate, and the tip of the black probe on the lower plate. A Make sure the battery is connected and adjust the battery voltage slider UP to +1.5 [V]. The Voltmeter should read "+1.5 V" Part A: 1) Calculate the capacitance (C) of the capacitor, and the charge (Q) held separated on the plates. Compare your calculations to the meters in the simulator. (Note that the minus signs in the exponents on the meters may be hard to read.) Why is the voltage across the capacitor plates (c) (as measured by the Voltmeter) the same as the battery voltage? 2) Disconnect the battery. Calculate the electric potential energy stored in the electric field between the capacitor plates (Uc) . Compare your answer to the meter at the top of the simulation. Why is there still energy stored even though the battery is disconnected? 3) If we slide the piece of glass (DIELECTRIC) all the way between the plates, will the energy stored by the capacitor increase, decrease, or stay the same? Do not plug in numbers here. But explain your educated guess (you may use equations in variables to do so). Please answer before using the simulation. ***You will NOT lose points for the wrong answer. You will lose points if you do not try to explain your answer. 4) Now (with the battery still disconnected) slide the glass all the way between the plates. Write down the new values of the capacitance (C), charge held separated (Q), voltage across the capacitor (c ) , and stored energy (Uc') . Calculate these values and compare your answers. 5) Explain why some values changed and some did not. If the stored energy changed, what accounts for that change given the Law of Conservation of Energy? Pull the glass out and put it back in to see that the process is "reversible".Part C: 1) Last step. If we were to disconnect the battery and THEN remove the glass, would the capacitance (C"), charge held separated (Q"), voltage across the capacitor (c ) , and stored potential energy ( U .) increase, decrease, or stay the same compared to their corresponding double-primed values. Explain. 2) Now disconnect the battery. THEN remove the glass completely. Calculate the capacitance (C"), charge held separated (Q"), voltage across the capacitor ( ) , and stored potential energy ( U ) and compare them to the meter readings in the simulator. Given the change in stored potential energy, did we "break" the Law of Conservation of Energy? (You can once again slide the glass in and out to see that the process is "reversible".)Part B: 1) Leave the glass between the plates. BEFORE reconnecting the battery, write down whether you think the capacitance (C'), charge held separated (Q'), voltage across the capacitor ( Vc ' ') , and stored potential energy (U c.') will increase, decrease, or stay the same after the battery is reconnected and ES EQ is reestablished. Explain your reasoning. Again, do not correct your answers after reconnecting the battery. I want you to compare your intuition to what you calculate. 2) Now reconnect the battery. Calculate the capacitance (C'), charge held separated (Q'), voltage across the capacitor ( Vc '') , and stored potential energy ( Uc'') and compare your answers to the meter readings. 3) Explain why the electric potential energy stored between the plates changed (or did not) after reconnecting the battery