The plates of an air-filled parallel-plate capacitor with a plate area of 18.5 cm2 and a separation of 9.25 mm are charged to a 125-V potential difference. After the plates are disconnected from the source, a porcelain dielectric with κ = 6.5 is inserted between the plates of the capacitor.
(a) What is the charge on the capacitor before and after the dielectric is inserted?
Qi=C Qf=C
(b) What is the capacitance of the capacitor after the dielectric is inserted?
F
(c) What is the potential difference between the plates of the capacitor after the dielectric is inserted?
V
(d) What is the magnitude of the change in the energy stored in the capacitor after the dielectric is inserted?J
J
To answer these questions, we need to use the formulas and equations related to capacitors.
(a) To calculate the charge on the capacitor before and after the dielectric is inserted, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference.
Before inserting the dielectric:
The capacitance of a parallel-plate capacitor is given by the formula C = ε₀A/d, where ε₀ is the vacuum permittivity (8.85 x 10^-12 F/m), A is the cross-sectional area of the plates, and d is the separation between the plates.
Let's convert the given values to SI units:
A = 18.5 cm² = 18.5 x 10^-4 m²
d = 9.25 mm = 9.25 x 10^-3 m
Using these values, we can calculate the capacitance as:
C = (8.85 x 10^-12 F/m) * (18.5 x 10^-4 m²) / (9.25 x 10^-3 m) = 1.77 x 10^-11 F
Now, we can calculate the charge on the capacitor:
Q_before = C_before * V = (1.77 x 10^-11 F) * (125 V) = 2.21 x 10^-9 C
After inserting the dielectric:
When the dielectric is inserted, the capacitance increases by a factor of κ, which is the dielectric constant.
Therefore, the new capacitance, C_after, is given by:
C_after = κ * C_before = 6.5 * (1.77 x 10^-11 F) = 1.15 x 10^-10 F
Now, we can calculate the charge on the capacitor after the dielectric is inserted:
Q_after = C_after * V = (1.15 x 10^-10 F) * (125 V) = 1.43 x 10^-8 C
So, the charge on the capacitor before inserting the dielectric is 2.21 x 10^-9 C, and after inserting the dielectric, it becomes 1.43 x 10^-8 C.
(b) To calculate the capacitance of the capacitor after the dielectric is inserted, we already found it to be 1.15 x 10^-10 F.
(c) The potential difference between the plates of the capacitor remains the same, which is 125 V, regardless of the presence of the dielectric.
(d) To calculate the magnitude of the change in the energy stored in the capacitor after the dielectric is inserted, we can use the formula for the energy stored in a capacitor: U = (1/2) CV².
The initial energy stored in the capacitor, U_before, is given by:
U_before = (1/2) * C_before * V² = (1/2) * (1.77 x 10^-11 F) * (125 V)²
The final energy stored in the capacitor, U_after, is given by:
U_after = (1/2) * C_after * V² = (1/2) * (1.15 x 10^-10 F) * (125 V)²
The magnitude of the change in energy is then:
ΔU = |U_after - U_before| = |(1/2) * (1.15 x 10^-10 F) * (125 V)² - (1/2) * (1.77 x 10^-11 F) * (125 V)²|
By substituting the values and simplifying the expression, you can calculate the magnitude of the change in energy in joules.
(a) The charge on a capacitor can be calculated using the formula Qi = CV. Given that the potential difference (V) is 125 V, we need to find the initial capacitance (C) to determine the initial charge (Qi).
The capacitance of a parallel-plate capacitor can be calculated using the formula C = ε₀(A/d), where ε₀ is the vacuum permittivity, A is the plate area, and d is the separation distance.
The plate area (A) is given as 18.5 cm², which is equal to 0.0185 m². The separation distance (d) is given as 9.25 mm, which is equal to 0.00925 m. The vacuum permittivity (ε₀) is a constant value of 8.854 x 10⁻¹² F/m.
Using the formula for capacitance, we can calculate the initial capacitance (C) as follows:
C = ε₀(A/d)
C = (8.854 x 10⁻¹² F/m)(0.0185 m²/0.00925 m)
C = 17.708 x 10⁻¹² F
Now, we can calculate the initial charge (Qi) using the formula Qi = CV:
Qi = (17.708 x 10⁻¹² F)(125 V)
Qi = 2.2135 x 10⁻⁹ C
Therefore, the charge on the capacitor before the dielectric is inserted is 2.2135 x 10⁻⁹ C.
After the dielectric is inserted, the charge on the capacitor remains the same. Therefore, the charge on the capacitor after the dielectric is inserted is also 2.2135 x 10⁻⁹ C.
(b) The capacitance of a capacitor with a dielectric can be calculated using the formula C' = κC, where C' is the capacitance with the dielectric, κ is the dielectric constant, and C is the initial capacitance.
Using the given dielectric constant (κ) of 6.5 and the initial capacitance (C) of 17.708 x 10⁻¹² F, we can calculate the capacitance of the capacitor after the dielectric is inserted (C') as follows:
C' = κC
C' = (6.5)(17.708 x 10⁻¹² F)
C' = 114.77 x 10⁻¹² F
C' = 1.1477 x 10⁻¹⁰ F
Therefore, the capacitance of the capacitor after the dielectric is inserted is 1.1477 x 10⁻¹⁰ F.
(c) The potential difference between the plates of a capacitor remains the same when a dielectric is inserted. Therefore, the potential difference between the plates after the dielectric is inserted is still 125 V.
(d) The change in energy stored in a capacitor with a dielectric can be calculated using the formula ΔU = (1/2)Q²(1/C - 1/C'), where ΔU is the change in energy, Q is the charge on the capacitor, C is the initial capacitance, and C' is the capacitance with the dielectric.
Using the charge on the capacitor (Q) of 2.2135 x 10⁻⁹ C, the initial capacitance (C) of 17.708 x 10⁻¹² F, and the capacitance with the dielectric (C') of 1.1477 x 10⁻¹⁰ F, we can calculate the change in energy (ΔU) as follows:
ΔU = (1/2)Q²(1/C - 1/C')
ΔU = (1/2)(2.2135 x 10⁻⁹ C)²(1/(17.708 x 10⁻¹² F) - 1/(1.1477 x 10⁻¹⁰ F))
ΔU = 7.854 x 10⁻⁷ J
Therefore, the magnitude of the change in the energy stored in the capacitor after the dielectric is inserted is 7.854 x 10⁻⁷ J.