The electric field in a particular space is 𝐸→=((1500.0𝑧+3000.0) N/C)kˆ  with 𝑧 in meters. Consider a cylindrical Gaussian surface of radius 13.0 cm that is coaxial with the 𝑧-axis. One end of the cylinder is at 𝑧=0, the other end of the cylinder is at 𝑧=2.07 m. What is the net charge contained by the cylinder?

We are tasked with finding the net charge enclosed by a cylindrical Gaussian surface given the electric field:

\[
\vec{E} = \left(1500.0z + 3000.0 \, \text{N/C}\right) \hat{k}
\]

The Gaussian surface is a cylinder of radius \( R = 13.0 \, \text{cm} = 0.13 \, \text{m} \), with one end at \( z = 0 \) and the other at \( z = 2.07 \, \text{m} \).

We will use **Gauss's Law**, which states:

\[
\oint_{\text{surface}} \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\epsilon_0}
\]

Where \( q_{\text{enc}} \) is the enclosed charge, and \( \epsilon_0 \) is the permittivity of free space \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2 \).

### Steps:
1. **Electric flux** through the cylinder's ends:
   The electric field points along the \( z \)-axis (axial), so we need to compute the electric flux only through the two circular end caps. The lateral surface does not contribute to the flux because the electric field is perpendicular to the side surface.

   The flux through each end is given by:
   
   \[
   \Phi = \vec{E} \cdot A = E(z) \cdot \pi R^2
   \]
   
   where \( E(z) \) is the electric field at a given \( z \), and \( A = \pi R^2 \) is the area of the circular end caps.

2. **Flux at \( z = 2.07 \, \text{m} \):**
   \[
   E(2.07) = 1500(2.07) + 3000 = 6105 \, \text{N/C}
   \]
   The flux at \( z = 2.07 \) is:
   \[
   \Phi_2 = 6105 \cdot \pi (0.13)^2
   \]

3. **Flux at \( z = 0 \):**
   \[
   E(0) = 1500(0) + 3000 = 3000 \, \text{N/C}
   \]
   The flux at \( z = 0 \) is:
   \[
   \Phi_1 = 3000 \cdot \pi (0.13)^2
   \]

4. **Net flux:**
   \[
   \text{Net flux} = \Phi_2 - \Phi_1
   \]

5. **Find enclosed charge using Gauss's Law:**
   \[
   q_{\text{enc}} = \epsilon_0 \cdot \text{Net flux}
   \]

Let's compute this step by step.The net charge enclosed by the cylindrical Gaussian surface is approximately:

\[
q_{\text{enc}} = 1.46 \times 10^{-9} \, \text{C}
\]

Would you like more details on the process or have any questions?

Here are 5 related questions to consider:
1. How would the result change if the electric field varied differently with \( z \)?
2. What if the cylindrical surface had a different radius? How would that affect the enclosed charge?
3. How can Gauss's Law be applied to non-uniform fields in other geometric configurations?
4. Why is there no flux contribution from the curved surface of the cylinder?
5. How does the concept of electric flux help in determining the charge distribution in different regions?

**Tip:** Always remember that Gauss’s Law is most effective in cases with high symmetry (spherical, cylindrical, or planar).

Learn how to calculate the net charge enclosed by a cylindrical Gaussian surface with a given electric field. This detailed step-by-step solution covers key concepts like Gauss's Law, electric flux, and cylindrical geometry. Suitable for university-level physics students.

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