24.3 A parallel-plate air capacitor with a capacitance of 250 pF has a charge of
ID: 1546943 • Letter: 2
Question
24.3 A parallel-plate air capacitor with a capacitance of 250 pF has a charge of magnitude 0.149 C on each plate. The plates have a separation of 0.254 mm . Part A What is the potential difference between the plates? Part B What is the area of each plate? Use 8.854×1012 F/m for the permittivity of free space. Part C What is the electric field magnitude between the plates? Part D What is the surface-charge density on each plate? 24.28 A parallel-plate vacuum capacitor has 6.00 J of energy stored in it. The separation between the plates is 3.80 mm . If the separation is decreased to 1.90 mm , Part A what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed? Part B What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed?
Explanation / Answer
(a)we have V =q/C, so v(potential difference)=596 volt
(b)c=(8.85*10^-12)*A/d where d=0.254mm, so A(area)=7.175*10^-3 .
(c)v=Ed so E=v/d, E(electric field)=2.35*10^6 v/m
(d)surface charge density=q/A=2.076*10^-5
2. 1.Final energy = Uf = initial energy *d2/d1
Final energy = Uf =6.0 *1.90/3.80
(A) Final energy = Uf = 3J
B) If the capacitor remained connected to the potential source while the separation of the plates was changed, potential remains same.
Final energy = Uf = initial energy *d1/d2
Final energy = Uf =6.0 *3.8/1.9
Final energy = Uf =12 J
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