A homemade capacitor is to be made from two 0.30 m wide x 0.40 m long sheets of

Question

A homemade capacitor is to be made from two 0.30 m wide x 0.40 m long sheets of aluminum foil, separated by a 0.030 mm thick sheet of paper. (a) Draw a diagram showing the capacitor in a charged state, including a positive plate, a negative plate, the gap distance d, and a qualitative representation of the electric field between the plates (exaggerate the gap distance so you have room to draw the field). (b) Find the capacitance (you will need to consult Table 20-1). (c) How much charge is stored on this capacitor if it is connected to opposite terminals of a 12 V car battery? (d) What is the magnitude of the electric field between the plates? (e) If the negative plate is taken to be at 0 V, what is the potential at a point inside the paper, 0.010 mm from the surface of the positive plate? (f) If a second, otherwise identical capacitor is made with a 0.015 mm thick sheet of mylar instead of the paper, what voltage is required to place the same amount of charge on the second capacitor as that found in part (c)? (g) How much energy is stored in each capacitor when they both store the amount of charge found in part (c)?

***I only need help with F and G***

Explanation / Answer

F)

relative permittivity, r (for paper) = 3.85

relative permittivity, r (for mylar) = 3.10 (I took these values from wikipedia)

now capacitance is given by,

C = orA/d

now Cp = oA x 3.85/(0.03 x 10-3)

and Cm = oA x 3.10/(0.015 x 10-3)

Qp = CpVp = Qm = CmVm

=> Vm = (oA x 3.85/(0.03 x 10-3)) x 12/(oA x 3.10/(0.015 x 10-3))

=> Vm = 3.85 x 12/(2 x 3.10) = 7.4516 V

G) Ep = 0.5(CpVp2) = 0.5 x (oA x 3.85/(0.03 x 10-3)) x 122 = 9240000oA

now oA = 8.85 x 10-12 x 0.3 x 0.4 = 1.062 x 10-12

=> Ep = 9240000 x 1.062 x 10-12 = 9.81288 x 10-6 J

Em = 0.5(CmVm2) = 0.5 x (oA x 3.10/(0.015 x 10-3)) x 7.45162 = 5737722oA

now oA = 8.85 x 10-12 x 0.3 x 0.4 = 1.062 x 10-12

=> Em = 5737722 x 1.062 x 10-12 = 6.09346 x 10-6 J

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