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5) Electric field of a rod again--this time with calculus and a different geometry •Part a: A thin nonconducting rod of length L has charge q uniformly distributed along i as shown in Fig. 22-55 (associated with problem 32). SHOW that the magnitude of the electric field at point P (on the perpendicular bisector of the rod) is given by: 1 2TE ,R JL² +4R² If your integral calculus is rusty, you might want to look at Appendix E in your text. (Again, since I've given you the final form, you must SHOW the steps leading to the result above.) •Part b: Justify the functional form of your expression for E, by considering what it reduces to in the limit of R>>L. Is this expression what you expect? Also find an expression for the opposite limit of L>>R. We will discuss this shortly in the context of Gauss' law, so remember what this is!

5) Electric field of a rod again--this time with calculus and a different geometry •Part a: A thin nonconducting rod of length L has charge q uniformly distributed along i as shown in Fig. 22-55 (associated with problem 32). SHOW that the magnitude of the electric field at point P (on the perpendicular bisector of the rod) is given by: 1 2TE ,R JL² +4R² If your integral calculus is rusty, you might want to look at Appendix E in your text. (Again, since I've given you the final form, you must SHOW the steps leading to the result above.) •Part b: Justify the functional form of your expression for E, by considering what it reduces to in the limit of R>>L. Is this expression what you expect? Also find an expression for the opposite limit of L>>R. We will discuss this shortly in the context of Gauss' law, so remember what this is!

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### 5) Electric Field of a Rod Again—This Time with Calculus and a Different Geometry #### Part a: A thin nonconducting rod of length \( L \) has charge \( q \) uniformly distributed along it as shown in Fig. 22-55 (associated with problem 32). **Show that the magnitude of the electric field at point \( P \) (on the perpendicular bisector of the rod) is given by:** \[ E = \frac{q}{2 \pi \epsilon_0 R \sqrt{L^2 + 4R^2}} \] If your integral calculus is rusty, you might want to look at Appendix E in your text. (Again, since I've given you the final form, you must show the steps leading to the result above.) #### Part b: Justify the functional form of your expression for \( E \), by considering what it reduces to in the limit of \( R \gg L \). Is this expression what you expect? Also, find an expression for the opposite limit of \( L \gg R \). We will discuss this shortly in the context of Gauss' law, so remember what this is! --- ### Detailed Explanation of Concepts: **Figure 22-55:** Refers to a diagram (not included here) that visually represents a thin nonconducting rod of length \( L \) with a uniformly distributed charge \( q \). The diagram likely includes a perpendicular bisector and a point \( P \) where the electric field is being calculated. **Expression for Electric Field:** The formula \[ E = \frac{q}{2 \pi \epsilon_0 R \sqrt{L^2 + 4R^2}} \] involves: - \( q \): total charge on the rod. - \( \epsilon_0 \): permittivity of free space. - \( R \): distance from the rod to point \( P \). - \( L \): length of the rod. **Integral Calculus:** The steps leading to the result likely involve breaking the rod into infinitesimally small elements of charge and using integration to sum the contributions of each of these elements to the electric field at point \( P \). **Limits of \( R \gg L \) and \( L \gg R \):** - For \( R \gg L \): When the distance \( R \) is much greater than the length
Transcribed Image Text:### 5) Electric Field of a Rod Again—This Time with Calculus and a Different Geometry #### Part a: A thin nonconducting rod of length \( L \) has charge \( q \) uniformly distributed along it as shown in Fig. 22-55 (associated with problem 32). **Show that the magnitude of the electric field at point \( P \) (on the perpendicular bisector of the rod) is given by:** \[ E = \frac{q}{2 \pi \epsilon_0 R \sqrt{L^2 + 4R^2}} \] If your integral calculus is rusty, you might want to look at Appendix E in your text. (Again, since I've given you the final form, you must show the steps leading to the result above.) #### Part b: Justify the functional form of your expression for \( E \), by considering what it reduces to in the limit of \( R \gg L \). Is this expression what you expect? Also, find an expression for the opposite limit of \( L \gg R \). We will discuss this shortly in the context of Gauss' law, so remember what this is! --- ### Detailed Explanation of Concepts: **Figure 22-55:** Refers to a diagram (not included here) that visually represents a thin nonconducting rod of length \( L \) with a uniformly distributed charge \( q \). The diagram likely includes a perpendicular bisector and a point \( P \) where the electric field is being calculated. **Expression for Electric Field:** The formula \[ E = \frac{q}{2 \pi \epsilon_0 R \sqrt{L^2 + 4R^2}} \] involves: - \( q \): total charge on the rod. - \( \epsilon_0 \): permittivity of free space. - \( R \): distance from the rod to point \( P \). - \( L \): length of the rod. **Integral Calculus:** The steps leading to the result likely involve breaking the rod into infinitesimally small elements of charge and using integration to sum the contributions of each of these elements to the electric field at point \( P \). **Limits of \( R \gg L \) and \( L \gg R \):** - For \( R \gg L \): When the distance \( R \) is much greater than the length
**Figure 22-55. Problem 32.** This figure depicts a positively charged thin, horizontal rod. The rod is lying along the x-axis and extends a horizontal distance of length \(L\). The positive sign markings along the rod indicate a uniform positive charge distribution. A point \(P\) is located at a distance \(R\) above the center of the rod along the y-axis. The distance from the center of the rod to point \(P\) is denoted as \(R\), and it is perpendicular to the rod. In the x-y coordinate system, the rod lies between \(-\frac{L}{2}\) and \(\frac{L}{2}\) along the x-axis while the point \(P\) lies at \((0, R)\). The goal of the problem is presumably to find some quantity related to the electric field or potential at point \(P\) due to the uniformly charged rod.
Transcribed Image Text:**Figure 22-55. Problem 32.** This figure depicts a positively charged thin, horizontal rod. The rod is lying along the x-axis and extends a horizontal distance of length \(L\). The positive sign markings along the rod indicate a uniform positive charge distribution. A point \(P\) is located at a distance \(R\) above the center of the rod along the y-axis. The distance from the center of the rod to point \(P\) is denoted as \(R\), and it is perpendicular to the rod. In the x-y coordinate system, the rod lies between \(-\frac{L}{2}\) and \(\frac{L}{2}\) along the x-axis while the point \(P\) lies at \((0, R)\). The goal of the problem is presumably to find some quantity related to the electric field or potential at point \(P\) due to the uniformly charged rod.
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