Respuesta :
Answer:
Q / Q₀ = 3.4 , ΔQ = 31.14 10⁻⁶ C
Explanation:
For this exercise we will calculate the charge of the capacitor with and without dielectric
the capacitance is given by
C = Q /ΔV
we also use the ratio of capacitances
C = k C₀
Q /ΔV = kQ₀ /ΔV
Q = k Q₀
Q / Q₀ = k
Q / Q₀ = 3.4
If we want a numerical value
C = k Q / DV
Q = C ΔV / k
Q = 2.5 10⁻⁶ 60 / 3.4
Q = 44.12 10⁻⁶ C
whereby the charge without dielectric is
Q₀ = Q / 3.4
Q₀ = 44.12 / 3.4 10⁻⁶
Q₀ = 12.98 10⁻⁶ C
therefore the charge when removing the dielectric is reduced, the amount of charge that flows is
ΔQ = Q - Q₀
ΔQ = (44.12 - 12.98) 10⁻⁶
ΔQ = 31.14 10⁻⁶ C
The amount of charge should be Q / Q₀ = 3.4 , ΔQ = 31.14 10⁻⁶ C
Capacitor:
For this given situation, we will determine the charge of the capacitor with and without dielectric
We know that
the capacitance is given by
C = Q /ΔV
Now here we can applied the ratio of capacitances
So,
C = k C₀
Q /ΔV = kQ₀ /ΔV
Q = k Q₀
Q / Q₀ = k
Q / Q₀ = 3.4
In case of numerical value
C = k Q / DV
Q = C ΔV / k
Q = 2.5 10⁻⁶ 60 / 3.4
Q = 44.12 10⁻⁶ C
whereby the charge without dielectric is
Q₀ = Q / 3.4
Q₀ = 44.12 / 3.4 10⁻⁶
Q₀ = 12.98 10⁻⁶ C
Thus the charge at the time when removing the dielectric is decreased , the amount of charge that flows is
ΔQ = Q - Q₀
ΔQ = (44.12 - 12.98) 10⁻⁶
ΔQ = 31.14 10⁻⁶ C
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