A parallel plate capacitor has a rectangular plates with length 5cm and width 3cm. They are separated by a distance 12m and hooked up to a 20V battery
A.) what is the area of each plate in m^2
B.) assuming nothing else is present on the circuit what is the potential difference between the plates?
C.) what is the capacitance of this parallel plate capacitor?
D.)what is the total charge on the positively charged plate of this capacitor?
E.) what is the surface charge density of this capacitor?
F.) what is the electric field outside the capacitor?
G.) what is the area of the sides of the Gaussian surface
H.)what is the total charge enclosed by the Gaussian surface?
I.) find the total electric flux going through the Gaussian surface
J.) using Guass's law, what is the electric field at some point inside the capacitor?
A.) To find the area of each plate in square meters, we need to convert the given length and width from centimeters to meters.
Length = 5 cm = 0.05 m
Width = 3 cm = 0.03 m
The formula for the area of a rectangle is A = length x width.
So, the area of each plate is A = 0.05 m x 0.03 m = 0.0015 m²
B.) The potential difference between the plates is given by the voltage of the battery, which is 20V.
C.) The capacitance of a parallel plate capacitor is given by the formula C = ε₀ * (A / d), where ε₀ is the permittivity of free space, A is the area of each plate, and d is the distance between the plates.
Given values:
Area (A) = 0.0015 m²
Distance (d) = 12 m
The permittivity of free space, ε₀, is approximately 8.854 x 10⁻¹² F/m.
Using the formula, we have:
C = (8.854 x 10⁻¹² F/m) * (0.0015 m² / 12 m)
C ≈ 1.106 x 10⁻¹³ F
D.) The total charge on the positively charged plate of the capacitor is equal to the product of the capacitance and the potential difference.
Q = C * V
Q = (1.106 x 10⁻¹³ F) * (20V)
Q ≈ 2.212 x 10⁻¹² C
E.) The surface charge density of the capacitor is given by the formula σ = Q / A, where σ is the surface charge density, Q is the total charge on the plate, and A is the area of the plate.
Using the values we found,
σ = (2.212 x 10⁻¹² C) / (0.0015 m²)
σ ≈ 1.475 x 10⁻⁹ C/m²
F.) The electric field outside the capacitor is zero. It exists only between the plates of the capacitor.
G.) The area of the sides of a Gaussian surface is equal to the area of the capacitor plates since we choose the Gaussian surface to enclose the entire capacitor.
The area of each plate is 0.0015 m², so the area of the sides of the Gaussian surface is also 0.0015 m².
H.) The total charge enclosed by the Gaussian surface is equal to the total charge on the positively charged plate, which we found earlier to be approximately 2.212 x 10⁻¹² C.
I.) The total electric flux going through the Gaussian surface is given by Gauss's Law, which states that the electric flux Φ = Q / ε₀, where Φ is the electric flux, Q is the total charge enclosed by the surface, and ε₀ is the permittivity of free space.
Using the values we found,
Φ = (2.212 x 10⁻¹² C) / (8.854 x 10⁻¹² F/m)
Φ ≈ 0.25
J.) Gauss's Law can be used to find the electric field at some point inside the capacitor. The electric field inside is given by the formula E = σ / ε₀, where E is the electric field, σ is the surface charge density, and ε₀ is the permittivity of free space.
Using the values we found,
E = (1.475 x 10⁻⁹ C/m²) / (8.854 x 10⁻¹² F/m)
E ≈ 166.65 N/C