Find the electric field a distance z above the center of a square loop (side a) carrying uniform line charge λ (See attached image for the figure of the loop).
Find the electric field a distance z above the center of a square loop (side a) carrying uniform line charge λ (See attached image for the figure of the loop).
Related questions
Question
Find the electric field a distance z above the center of a square loop (side a) carrying uniform line charge λ (See attached image for the figure of the loop).
(Hint: Use the result of Example. 2.2, which is also attatched in an image)
Transcribed Image Text:The diagram depicts a right-angled perspective of a plane with a point labeled "P" positioned above it. There's a vertical dashed line indicating the distance between the point "P" and the plane, labeled as "z." On the plane, a line with arrows at both ends represents a dimension labeled "a," indicating a length across the plane.
The elements in the image are as follows:
- **Point P**: Located above the plane, connected by a dashed line indicating vertical distance.
- **Distance z**: The vertical height from the plane to the point "P."
- **Plane**: A flat surface depicted in perspective view.
- **Length a**: The horizontal dimension of the plane shown with bidirectional arrows.
This image might be used to represent concepts in geometry or physics, such as finding the distance from a point to a plane.
![**Example 2.2.** Find the electric field a distance \( z \) above the midpoint of a straight line segment of length \( 2L \) that carries a uniform line charge \( \lambda \) (Fig. 2.6).
**FIGURE 2.6 Explanation:**
The diagram shows a line segment on the x-axis from \(-L\) to \(L\). The point \(P\) is located a vertical distance \(z\) above the midpoint of this line. From point \(P\) to the line segment, the vector \(\mathbf{r}\) is shown, directed along the positive z-axis. A differential element \(dx\) is illustrated on the line, with a vector pointing along the x-axis.
**Solution**
The simplest method is to chop the line into symmetrically placed pairs (at \(\pm x\)), quote the result of Ex. 2.1 (with \(d/2 \to x\), \(q \to \lambda \, dx\)), and integrate (\(x : 0 \to L\)). But here’s a more general approach:\(^3\)
\[
\mathbf{r} = z \, \mathbf{\hat{z}}, \quad \mathbf{r}' = x \, \mathbf{\hat{x}}, \quad d\mathbf{l}' = dx \, \mathbf{\hat{x}};
\]
\[
\mathbf{n} = \mathbf{r} - \mathbf{r'} = z \, \mathbf{\hat{z}} - x \, \mathbf{\hat{x}}, \quad n = \sqrt{z^2 + x^2}, \quad \mathbf{\hat{n}} = \frac{z \, \mathbf{\hat{z}} - x \, \mathbf{\hat{x}}}{\sqrt{z^2 + x^2}}.
\]
\[
\mathbf{E} = \frac{1}{4\pi\varepsilon_0} \int_{-L}^{L} \frac{\lambda}{z^2 + x^2} \frac{z \, \mathbf{\hat{z}} - x \, \mathbf{\hat{x}}}{\sqrt{z^2 + x^2}} \, dx
\]
\[
= \frac{\lambda}{4\pi\varepsilon_0} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5142c08c-3b37-45fa-beed-241deb836fb1%2F46ff311b-bcfa-4465-8d0b-53fec0318fd4%2F78lzisg_processed.png&w=3840&q=75)
Transcribed Image Text:**Example 2.2.** Find the electric field a distance \( z \) above the midpoint of a straight line segment of length \( 2L \) that carries a uniform line charge \( \lambda \) (Fig. 2.6).
**FIGURE 2.6 Explanation:**
The diagram shows a line segment on the x-axis from \(-L\) to \(L\). The point \(P\) is located a vertical distance \(z\) above the midpoint of this line. From point \(P\) to the line segment, the vector \(\mathbf{r}\) is shown, directed along the positive z-axis. A differential element \(dx\) is illustrated on the line, with a vector pointing along the x-axis.
**Solution**
The simplest method is to chop the line into symmetrically placed pairs (at \(\pm x\)), quote the result of Ex. 2.1 (with \(d/2 \to x\), \(q \to \lambda \, dx\)), and integrate (\(x : 0 \to L\)). But here’s a more general approach:\(^3\)
\[
\mathbf{r} = z \, \mathbf{\hat{z}}, \quad \mathbf{r}' = x \, \mathbf{\hat{x}}, \quad d\mathbf{l}' = dx \, \mathbf{\hat{x}};
\]
\[
\mathbf{n} = \mathbf{r} - \mathbf{r'} = z \, \mathbf{\hat{z}} - x \, \mathbf{\hat{x}}, \quad n = \sqrt{z^2 + x^2}, \quad \mathbf{\hat{n}} = \frac{z \, \mathbf{\hat{z}} - x \, \mathbf{\hat{x}}}{\sqrt{z^2 + x^2}}.
\]
\[
\mathbf{E} = \frac{1}{4\pi\varepsilon_0} \int_{-L}^{L} \frac{\lambda}{z^2 + x^2} \frac{z \, \mathbf{\hat{z}} - x \, \mathbf{\hat{x}}}{\sqrt{z^2 + x^2}} \, dx
\]
\[
= \frac{\lambda}{4\pi\varepsilon_0} \
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
