A capapcitor consist of two parallel plates but one of them can move relative to

Question

A capapcitor consist of two parallel plates but one of them can move relative to the another. Air Fills the space between the plates and the capacitance is 32pF when the capacitance between the pates is d=0.500 cm

A) Calculate the Area of the capacitor

B) A battery supplying a potential difference V=9.00v in connected to the plates. Determine the surface charge density of each plate.

C) Determine the Surface charge density if d is changed to 0.250

D) With d=0.500cm, the battery is disconnected from the platyes. The plates are then moved so that d=0.250cm. Calculate the pontential difference "Delta V" between the plates

Explanation / Answer

sigma = Q / A

The charge on the capacitor: Q = C V = 32 x 10-12 * 9 = 2.88 x 10-10 C

a) The are of the plates A = Cd / (epsilon0) = (32 x 10-12 * 0.5 x 10-2) / (8.85 x 10-12)

= 0.018 m2

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b) sigma = Q / A = ( 2.88 x 10-10) / 0.018

= 1.59 x 10-8 C/m2  

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if d is changed to 0.250 cm

The charge on the capacitor is doubled, and sigma is doubled

C' = 64 pF

sigma = 2 * 1.59 * 10-8

sigma = 3.2 x 10-8 C/m2

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