To find the energy contained in a cylinder of cross-section area \(10 \, \text{cm}^2\) and length \(50 \, \text{cm}\) along the x-axis, we will follow these steps:
### Step 1: Identify the given parameters
- The electric field is given by:
\[
E = 50 \, \text{N/C} \cdot \sin(\omega(t - \frac{x}{c}))
\]
- The maximum electric field \(E_0 = 50 \, \text{N/C}\).
- Cross-sectional area \(A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 = 1 \times 10^{-3} \, \text{m}^2\).
- Length of the cylinder \(L = 50 \, \text{cm} = 0.5 \, \text{m}\).
### Step 2: Calculate the volume of the cylinder
The volume \(V\) of the cylinder can be calculated using the formula:
\[
V = A \times L
\]
Substituting the values:
\[
V = (1 \times 10^{-3} \, \text{m}^2) \times (0.5 \, \text{m}) = 5 \times 10^{-4} \, \text{m}^3
\]
### Step 3: Calculate the energy density
The energy density \(u\) (energy per unit volume) in an electromagnetic wave is given by:
\[
u = \frac{1}{2} \epsilon_0 E_0^2
\]
Where \(\epsilon_0\) (the permittivity of free space) is approximately:
\[
\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}
\]
Substituting the values:
\[
u = \frac{1}{2} \times (8.85 \times 10^{-12} \, \text{F/m}) \times (50 \, \text{N/C})^2
\]
Calculating \( (50 \, \text{N/C})^2 \):
\[
(50)^2 = 2500 \, \text{N}^2/\text{C}^2
\]
Now substituting back:
\[
u = \frac{1}{2} \times (8.85 \times 10^{-12}) \times 2500
\]
Calculating:
\[
u = \frac{1}{2} \times 2.2125 \times 10^{-8} = 1.10625 \times 10^{-8} \, \text{J/m}^3
\]
### Step 4: Calculate the total energy contained in the cylinder
The total energy \(U\) contained in the cylinder can be calculated using:
\[
U = u \times V
\]
Substituting the values:
\[
U = (1.10625 \times 10^{-8} \, \text{J/m}^3) \times (5 \times 10^{-4} \, \text{m}^3)
\]
Calculating:
\[
U = 5.53125 \times 10^{-12} \, \text{J}
\]
### Final Answer
The energy contained in the cylinder is approximately:
\[
U \approx 5.53 \times 10^{-12} \, \text{J}
\]
