Question

Two sheets of aluminum foil have the same area, a separation of 1mm, a capacitance of 10pF, and are charged to 12V. Thereafter, the source is being disconnected. (a) The separation is now decreased by 0.1mm with the charge held constant. What is the new capacitance? (b) By how much does the potential difference between the plates change as a result? (c) How might you build a microphone based on this principle? (d) What is the amount of work you must do in order to move the plates between the two cases? (e) What is the energy density of the electric field between the plates in each case?

Answer 1

(A) C = e0 A / d

d is changed so C d = constant

(10 pF) (1 mm) = C' (0.1mm)

C' = 100 pF  

(B) charge on capacitor will not change as it is disconnecetd.

Q = C V = C' V'

(10)(12) = (100)V'

V' = 1.2 Volt

(d) U = C V^2 / 2

Ui = (10 x 10^-12) (12^2)/2 = 720 x 10^-12 J

Uf = (100 x 10^-12)(1.2^2)/2 = 72 x 10^-12 J

Work done = Uf - Ui = - 648 x 10^-12 J

(e) E = V/d = (12 V) / (1 x 10^-3 m)

E = 12 x 10^3 V/m

Energy density = e0 E^2 / 2

= (8.854 x 10^-12)(12 x 10^3)^2 /2

= 637.5 x 10^-6 J/m^3

Energy density is same before and after.