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Sagot :
Answer:
a) ΔV ’= 1.66 10¹ V= 16.6 V, b) U = 55.64 10⁻¹² J, c) U_f = 1.18 10⁻¹⁰ J
d) W = 6.236 10⁻¹¹ J
Explanation:
Capacitance can be found for a parallel plate capacitor
C = ε₀ [tex]\frac{A}{d}[/tex]
Let's reduce the magnitudes to the SI system
A = 9.30 cm² (1 m / 10² cm) 2 = 9.30 10⁻⁴ m²
c = 4.50 mm (1 m / 1000 mm) = 4.50 10⁻³ m
Co = 8.85 10⁻¹² 9.30 10⁻⁴ /4.50 10⁻³
Co = 1.829 10⁻¹² F
when the plates separate at d = 9.60 10⁻³ m, the capcitance changes to
C = ε₀ \frac{A}{d_1}
C = 8.85 10⁻¹² 9.30 10⁻⁴/9.60 10⁻³
C = 8.57 10⁻¹³ F
a) the potential difference
C =
since the capacitor is not discharged, let's look for the initial charge
Co = \frac{Q}{ \Delta V}
Q = C₀ ΔV
Q = 1.829 10⁻¹² 7.80
Q = 14.2662 10⁻¹² C
when the condensate plates are separated
C = \frac{Q}{ \Delta V' }
ΔV ’= Q / C
ΔV ’= 14.266 10⁻¹² / 8.57 10⁻¹³
ΔV ’= 1.66 10¹ V= 16.6 V
b) the stored energy is
U = ½ C ΔV²
for initial separation
U = ½ C₀ ΔV²
U = ½ 1.829 10⁻¹² 7.80²
U = 55.64 10⁻¹² J
c) The energy for end separation;
U_f = ½ C DV’2
U_f = ½ 8.57 10⁻¹³ 16,6²2
U_f = 1.18 10⁻¹⁰ J
d) The work
as there are no losses, the work is equal to the variation of the energy
W = ΔU = U_f -U₀
W = 1.18 10⁻¹⁰ - 55.64 10-12
W = 6.236 10⁻¹¹ J
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