Respuesta :
Answer:
a) Q₀ = Q = 4.27 10² pC, b) C = 1.26 10⁻¹⁰ F, ΔV = 3.375 V, c) ΔE = 56.88 nJ
Explanation:
a) the capacitance of a parallel plate capacitor is
C = ε₀ A / d
let's reduce the magnitudes to the SI system
d = 1.40 cm = 0.0140 m
A = 25.0 cm² (1 m / 100 cm) ² = 25.0 10⁻⁴ m²
C = 8.85 10⁻¹² 25.0 10⁻⁴ / 0.0140
C = 1.58 10⁻¹² F
the capacitance is also
C = Q / ΔV
Q = V ΔV
Q = 270 1.58 10⁻¹²
Q = 4.27 10⁻¹⁰ C
When the battery is removed and the capacitor is inserted into the dielectric, the charge should remain the same,
Q₀ = Q = 4.27 10⁻¹⁰ C
let's reduce to pC
Q₀ = 4.27 10⁻¹⁰ C (10¹² pC / 1C)
Q₀ = Q = 4.27 10² pC
b) After being immersed in the fluid of constant k = 80.0
capacitance is
C = k Co
C = 80 1.58 10⁻¹²
C = 1.26 10⁻¹⁰ F
the voltage difference is
ΔV = ΔV₀ / k
ΔV = 270/80
ΔV = 3.375 V
c) We stop the energy before the dive
E₀ = ½ C₀ ΔV₀²
after the dive
E = ½ (k C₀) (ΔV₀/k)²
E = ½ C₀ ΔV₀² / k
E = E₀ / k
the change in energy is
ΔE = E -E₀
ΔE = E₀ / k - E₀
ΔE = E₀ ( [tex]\frac{1}{k} - 1[/tex] )
we calculate
E₀ = ½ 1.58 10⁻¹² 270²
E₀ = 5.76 10⁻⁸ J
ΔE = 5.76 10⁻⁸ ([tex]\frac{1}{80} -1[/tex])
ΔE = 5.688 10⁻⁸ J
we reduce to nJ
ΔE = 5.688 10⁻⁸ J (10⁹ nJ / 1J)
ΔE = 56.88 nJ
