Very large capacitors have been considered as a means for storing electrical energy. If we constructed a very large parallel-plate capacitor of plate area 1.0 m2 using paper (k= 3.7) of thickness 1.0 mm as a dielectric, how much electrical energy would it store at a plate voltage of 5 000 V?
0.41J
0.41
To calculate the electrical energy stored in a capacitor, we can use the formula:
E = (1/2) * C * V^2
Where:
E = Electrical Energy stored in the capacitor
C = Capacitance of the capacitor
V = Voltage across the capacitor
First, let's calculate the capacitance of the parallel-plate capacitor using the formula:
C = (ε0 * εr * A) / d
Where:
ε0 = Permittivity of free space (8.85 x 10^-12 F/m)
εr = Relative permittivity or dielectric constant of the material (In this case, k = 3.7)
A = Area of the plates (1.0 m^2)
d = Distance or thickness of the dielectric (1.0 mm = 0.001 m)
C = (8.85 x 10^-12 F/m * 3.7 * 1.0 m^2) / 0.001 m
C = 3.0765 x 10^-8 F
Now, we can calculate the electrical energy stored in the capacitor using the given voltage of 5,000 V:
E = (1/2) * C * V^2
E = (1/2) * 3.0765 x 10^-8 F * (5,000 V)^2
E = 1.923 x 10^-1 J
Therefore, the electrical energy stored in the capacitor would be approximately 0.192 J.
To calculate the electrical energy stored in a capacitor, we use the formula E = 0.5 * C * V^2, where E is the energy stored, C is the capacitance, and V is the voltage.
To find the capacitance, we need to know the formula for the capacitance of a parallel-plate capacitor: C = (ε * A) / d, where C is the capacitance, ε is the permittivity of the dielectric, A is the area of the plates, and d is the separation between the plates.
Given:
Plate area (A) = 1.0 m^2
Dielectric thickness (d) = 1.0 mm = 0.001 m
Relative permittivity of the dielectric (k) = 3.7
Plate voltage (V) = 5,000 V
First, we need to calculate the capacitance of the capacitor using the formula C = (ε * A) / d.
The permittivity (ε) can be determined using the formula ε = k * ε₀, where ε₀ is the permittivity of free space (8.85 x 10^-12 F/m).
ε₀ = 8.85 x 10^-12 F/m
k = 3.7
ε = k * ε₀
ε = 3.7 * 8.85 x 10^-12 F/m
ε = 3.25 x 10^-11 F/m
Now, we can calculate the capacitance (C):
C = (ε * A) / d
C = (3.25 x 10^-11 F/m * 1.0 m^2) / 0.001 m
C = 3.25 x 10^-11 F
Next, we can calculate the energy (E) stored in the capacitor using the formula E = 0.5 * C * V^2:
E = 0.5 * C * V^2
E = 0.5 * 3.25 x 10^-11 F * (5,000 V)^2
E = 0.5 * 3.25 x 10^-11 F * 25,000,000 V^2
Now, let's substitute the values into the formula and calculate the result:
E = 0.5 * 3.25 x 10^-11 F * 25,000,000 V^2
E = 0.5 * 3.25 x 10^-11 F * 625,000,000 V^2
E = 101,562,500 x 10^-11 F
So, the electrical energy stored in the capacitor would be approximately 101,562,500 x 10^-11 F, or 1.015625 J (joules).