A capacitor is made by cutting a metal sheet into two Triskaidecagons with side

Question

A capacitor is made by cutting a metal sheet into two Triskaidecagons with side length 1.2cm and placing a material in between them. a) If the material between the plates is air how far do the plates need to be separated so that the capacitance is 19 mu F? b) A new Material is placed in between the plates with a thickness of 1 mm. The Capacitance drops to 0.14 nF. What is a possible candidate for this new material? c) 14 V of potential is then applied to the plates. How much charge builds up on each plate?

Explanation / Answer

Given

Side length of Triskaidecagon a = 12 x 10-2 m

Initial capacitance of the capacitor C = 19 x 10-6 F

Solution

Area of Triskaidecagon

A = 13a2/4tan(/13)

A = 13 x (12 x 10-2)2/4 tan(13.846o)

A = 0.1899 m2

Capacitance

C = oA/d

d = oA/C

d = 8.854 x 10-12 x 0.1899 / 19 x 10-6

d = 0.0884934 x 10-6 m

d = 8.85 x 10-8 m

Distance between the plates is 8.85 x 10-8 m

Now the new capacitance

C’ = roA/d’

r = C’d’/oA

r = 0.14 x 10-9 x 1 x 10-3 / 8.854 x 10-12 x 0.1899

r = 0.08326

Form the value of relative permittivity we know that the material inserted is TiO2

Assuming we still keep TiO2 in place

C = q/V

q = CV

q = 0.14 x 10-9 x 14

q = 1.96 x 10-9 C

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