How much energy is stored by the electric field between two square plates, 7.9 cm on a side, separated by a 2.5-mm air gap? The charges on the plates are equal and opposite and of magnitude 15 nC .
Answer--> PE=_____ J
PE = potential energy?
I got this wrong twice. [119805.7 J]
ever find out the answer?
To calculate the potential energy stored in the electric field between two square plates, you can use the formula:
PE = (1/2) * C * V^2
Where:
PE is the potential energy
C is the capacitance of the capacitor
V is the voltage difference between the plates
To find the capacitance, you can use the formula:
C = (ε₀ * A) / d
Where:
ε₀ is the permittivity of free space (8.85 x 10^-12 F/m)
A is the area of one plate
d is the distance between the plates
First, let's calculate the capacitance:
A = (side length)^2 = (0.079 m)^2
d = 2.5 mm = 0.0025 m
C = (8.85 x 10^-12 F/m) * (0.079 m^2) / (0.0025 m)
C = 2.817 x 10^-9 F
Next, calculate the voltage difference between the plates:
q = 15 nC = 15 x 10^-9 C
From the definition of capacitance, we have:
C = q / V
V = q / C
V = (15 x 10^-9 C) / (2.817 x 10^-9 F)
V = 5.323 V
Now, substitute the values of C and V into the formula for potential energy:
PE = (1/2) * (2.817 x 10^-9 F) * (5.323 V)^2
PE = 119.81 x 10^-6 J
Therefore, the potential energy stored in the electric field between the two square plates is approximately 119.81 x 10^-6 J.
To determine the potential energy (PE) stored by the electric field between two square plates, we can use the formula:
PE = (1/2) * (Q^2 / C)
Where:
- PE is the potential energy in Joules (J)
- Q is the magnitude of the charge on one of the plates in Coulombs (C)
- C is the capacitance of the capacitor formed by the plates in Farads (F)
First, let's calculate the capacitance (C) using the given information:
C = ε * (A / d)
Where:
- ε is the permittivity of the medium between the plates, which is the permittivity of free space (ε₀) for air (ε₀ = 8.854 x 10^-12 F/m)
- A is the area of the plates in square meters (m²)
- d is the distance between the plates in meters (m)
Given:
- Side length of square plates, a = 7.9 cm = 0.079 m
- Air gap between the plates, d = 2.5 mm = 0.0025 m
Calculating the area of the plates:
A = a² = (0.079 m)^2 = 0.006241 m²
Calculating the capacitance:
C = ε₀ * (A / d) = 8.854 x 10^-12 F/m * (0.006241 m² / 0.0025 m)
C ≈ 2.2136 x 10^-11 F
Now, we can substitute the magnitude of the charge (Q) and the capacitance (C) into the formula for potential energy (PE):
PE = (1/2) * (Q^2 / C) = (1/2) * (15 x 10^-9 C)^2 / (2.2136 x 10^-11 F)
PE ≈ (1/2) * (225 x 10^-18 C^2) / (2.2136 x 10^-11 F)
PE ≈ 20243.6 J
Therefore, the potential energy stored by the electric field between the two square plates is approximately 20243.6 Joules (J).