Respuesta :
Answer:
14 x 10⁻⁶ J
1377 x 10⁻⁶ J
Explanation:
C = Capacitance of the capacitor = 1 x 10⁻⁶ F
ΔV = Original potential difference across the plates = 6.0 Volts
U₀ = Original energy stored in the capacitor
Original energy stored in the capacitor is given as
U₀ = (0.5) C ΔV² eq-1
ΔV' = Potential difference across the plates after increase = 8.0 Volts
U'₀ = New energy stored in the capacitor
New energy stored in the capacitor is given as
U'₀ = (0.5) C ΔV'² eq-2
U = Additional energy stored
Additional energy stored by the capacitor is given as
U = U'₀ - U₀
U = (0.5) C ΔV'² - (0.5) C ΔV²
U = (0.5) (1 x 10⁻⁶) (8)² - (0.5) (1 x 10⁻⁶) (6)²
U = 14 x 10⁻⁶ J
[tex]k_{final}[/tex] = final dielectric constant = 76.5
[tex]k_{initial}[/tex] = initial dielectric constant = 1
Energy stored in the capacitor is directly proportional to the dielectric constant, hence increase in the energy is given as
[tex]U_{inc}=(k_{final} - k_{initial})U_{o}[/tex]
Original energy stored in the capacitor is given as
U₀ = (0.5) C ΔV² = (0.5) (1 x 10⁻⁶) (6)² = 18 x 10⁻⁶ J
[tex]U_{inc}=(k_{final} - k_{initial})U_{o}[/tex]
[tex]U_{inc} = (76.5 - 1)(18\times 10^{-6})[/tex]
[tex]U_{inc} = 1377\times 10^{-6})[/tex]
