An electromagnetic wave has an electric field strength of 145 V/m at a point P in space at time t. (a) What is the electric field energy density at P? (b) A parallel-plate capacitor whose plates have area 0.110 m2 has a uniform electric field between its plates with the same energy density as the answer to part a. What is the charge on the capacitor?
To find the electric field energy density at point P, we can use the formula:
Energy density = (1/2) * ε₀ * E²
where ε₀ is the permittivity of free space and E is the electric field strength.
(a) Electric Field Energy Density at Point P:
We are given the electric field strength E = 145 V/m. The permittivity of free space, ε₀, is a constant with a value of approximately 8.85 × 10⁻¹² C²/(N·m²).
Using the formula, we can calculate the energy density:
Energy density = (1/2) * ε₀ * E² = (1/2) * (8.85 × 10⁻¹² C²/(N·m²)) * (145 V/m)² = 5.69 × 10⁻⁶ J/m³
(b) Charge on the Capacitor:
We are given that the electric field energy density between the plates of the capacitor is the same as the energy density at point P.
The energy density between the plates of a parallel-plate capacitor is given by the equation:
Energy density = (1/2) * ε₀ * E²
Given the area of the plates, A = 0.110 m², and the energy density from part (a), we can write:
(1/2) * ε₀ * E² = 5.69 × 10⁻⁶ J/m³
We can rearrange this equation to solve for the electric field strength E:
E² = (2 * 5.69 × 10⁻⁶ J/m³) / ε₀
E² = 1.283 × 10⁻⁵ N/C²
E = √ (1.283 × 10⁻⁵ N/C²)
E = 0.01133 N/C
Now, we can calculate the charge on the capacitor:
The charge on a capacitor is given by the equation:
Q = C * V
Here, C is the capacitance, which is given by the equation:
C = ε₀ * (A/d)
where d is the separation between the plates.
Given the area, A = 0.110 m², the energy density, ε₀ = 8.85 × 10⁻¹² C²/(N·m²), and the electric field strength, E = 0.01133 N/C, we need to find the separation distance d.
Using the equation for energy density:
(1/2) * ε₀ * E² = (1/2) * (ε₀ * (A/d)) * E²
Simplifying the equation:
E² = (A/d) * E²
d = A / E²
Substituting the given values:
d = (0.110 m²) / (0.01133 N/C)²
d = 873.53 m
Now, we can calculate the charge on the capacitor:
Q = C * V
Q = ε₀ * (A/d) * V
Q = (8.85 × 10⁻¹² C²/(N·m²)) * (0.110 m²) * (0.01133 N/C)
Q = 1.06 × 10⁻¹⁰ C
Therefore, the charge on the parallel-plate capacitor is approximately 1.06 × 10⁻¹⁰ C.