a parallel plate capacitor has an area of 5cm^2 and a capacitance of 3.5PF. the capacitor is connected to a 12v battery .after the capacitor is completely charge up the battery is removed.
a)what must be the separation of the plates and the charge density on the plate
b)what is the magnitude of the electric field between the plate and how much energy is stored in the electric field between the plates
c)if a sheet of mylar is placed between the plates how will the voltage across the plate and the charge density on the plate change?
d)if the sheet of mylar has an area of 3cm^2 so that it doesn't cover the whole area of the plate what would be the effective capacitance of this arrangement
d = epsilon A/C
E = V/d (E field)
E = 1/2CV^2 (Energy)
Don't know the kappa for mylar but
C = kappa epsilon A/d
if the mylar is removed and the plates are moved 1mm further from each other what will be the voltage across the plate and how much work will it take to move the plate apart
a) To find the separation of the plates, we can use the formula for capacitance of a parallel plate capacitor:
C = ε₀ * (A / d)
Where C is the capacitance, ε₀ is the vacuum permittivity (constant), A is the area of the plates, and d is the separation of the plates.
Rearranging the formula, we can solve for d:
d = A / (C * ε₀)
Plugging in the values, we have:
A = 5 cm² = 5 * 10^(-4) m²
C = 3.5 pF = 3.5 * 10^(-12) F
ε₀ = 8.85 * 10^(-12) F/m (vacuum permittivity)
Now we can calculate the separation:
d = (5 * 10^(-4) m²) / (3.5 * 10^(-12) F * 8.85 * 10^(-12) F/m) = 0.16 m = 1.6 cm
Therefore, the separation of the plates is 1.6 cm.
To calculate the charge density (σ) on the plates, we can use the formula:
σ = Q / A
Where σ is the charge density, Q is the charge on the plates, and A is the area of the plates.
We know the charge on the plates is equal to the voltage times the capacitance:
Q = C * V = (3.5 * 10^(-12) F) * (12 V)
Plugging in the values, we have:
Q = 4.2 * 10^(-11) C
And now we can calculate the charge density:
σ = (4.2 * 10^(-11) C) / (5 * 10^(-4) m²) = 8.4 * 10^(-7) C/m²
Therefore, the charge density on the plates is 8.4 * 10^(-7) C/m².
b) The magnitude of the electric field (E) between the plates can be calculated using the formula:
E = V / d
Where E is the electric field, V is the voltage, and d is the separation of the plates.
Plugging in the values, we have:
E = (12 V) / (0.016 m) = 750 V/m
Therefore, the magnitude of the electric field between the plates is 750 V/m.
The energy stored in the electric field (U) between the plates can be calculated using the formula:
U = (1/2) * C * V²
Plugging in the values, we have:
U = (1/2) * (3.5 * 10^(-12) F) * (12 V)^2 = 2.52 * 10^(-10) J
Therefore, the energy stored in the electric field between the plates is 2.52 * 10^(-10) J.
c) If a sheet of mylar is placed between the plates, the voltage across the plates will remain the same because the voltage only depends on the charge and the capacitance. However, the charge density on the plate will decrease because the mylar will reduce the effective area available for charge storage, leading to a lower charge density on the plates.
d) If the mylar sheet has an area of 3 cm² and doesn't cover the whole area of the plate, we need to consider the combined capacitance of the mylar and the air gap between the mylar and the plate. The effective capacitance (C_eff) can be calculated by adding the individual capacitances of the mylar (C_mylar) and the air gap (C_air):
C_eff = C_mylar + C_air
Using the formula for capacitance, we can calculate the individual capacitances:
C_mylar = ε₀ * (A_mylar / d_mylar)
C_air = ε₀ * (A_air / d_air)
Where ε₀ is the vacuum permittivity, A_mylar is the area of the mylar sheet, d_mylar is the thickness of the mylar sheet (assumed to be negligible), A_air is the remaining area of the plate, and d_air is the separation between the mylar and the plate.
Plugging in the values, we have:
C_mylar = (8.85 * 10^(-12) F/m) * (3 * 10^(-4) m² / d_mylar)
C_air = (8.85 * 10^(-12) F/m) * ((5 * 10^(-4) m² - 3 * 10^(-4) m²) / d_air)
Since we don't have specific values for d_mylar and d_air, we can't calculate the exact individual capacitances. However, the effective capacitance will be lower than 3.5 pF due to the reduced area and increased separation caused by the mylar sheet.