A capacitor consists of two parallel plates, but one of them can move relative t

Question

A capacitor consists of two parallel plates, but one of them can move relative to the other as shown in the figure. Air fills the space between the plates, and the capacitance is 32.0 pF when the separation between plates is d = 0.400 cm. a) A battery supplying a potential difference V = 9.00 V is connected to the plates. What is the charge distribution, sigma, on the left plate? What are the capacitance, C', and the charge distribution, sigma', when d is changed to 0.20 cm? b) With d = 0.400 cm, the battery is disconnected from the plates. The plates are then moved so that d = 0.200 cm. What is the potential difference V', between the plates?

Explanation / Answer

(a) ans

Given that

capacitance C=32pF

distance d=0.4 cm

voltage v=9 v

now we find the charge distribution

electric field E=V/d=9/0.4*10^-2=2250 N/c

charge distribution =2250*8.85*10^-12=19.9*10^-9 C/m^2

now we find the charge distribution and capacitance at distance d=0.2 cm

capacitance C'=d2/d1*C=0.2/0.4*32*10^-12=16 pF

electric field E=9/0.2*10^-2=4500 N/c

charge distribution =4500*8.85*10^-12=39.83*10^-9 C/m^2

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(b) ans

votage V'=d2/d1*v=0.2*9/0.4=4.5 v

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