16.0.083 05 men 1 mm 05me connected to anything) originally has a potential diff

Question

16.0.083 05 men 1 mm 05me connected to anything) originally has a potential difference of 1010 volits with an air gap of 2 mm. Then a plastic slab 1 mm thick, with dielectric constant 3.3, is inserted into the middle of the air gap as shown in the figure. As shown in the diagram, the capacitor, location 2 is at the left edge of the plastic siab, location 3 is at the right edge of the siab, and location 4 is at the right plate of the capacitor. All of these locations are near the center of the capacitor. Calculate the following potential differences V What assumptions or approximations did you make in this calculation? V1-V4= Assume the electric field inside the plastic slab is zero G Assume the plastic slab did not polarize Assume the positive plate of the capact or does not contribute to the electric field in the gap between the plastic and the negative plate D Assume the radius of the plates was much larger than the distance between the plates O Assume that the field due to the polarized plastic slab was negligible in the air gaps Additional Materials section 169

Explanation / Answer

Electric field is given by

E=V/d =1010/2 =505 V/mm

a)

V1-V2=505*0.5 =252.5 Volts

b)

V2-V3=(505/3.3)*1=153.03 Volts

c)

V3-V4=505*0.4=252.5 Volts

d)

V1-V4=252.5+153.03+252.5 =658.03 Volts

e)

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