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64. The surface charge density on a long straight metallic pipe is o. What is the electric potential outside and inside the pipe? Assume the pipe has a diameter of 2a. I O + + + + + + + + + + + +

64. The surface charge density on a long straight metallic pipe is o. What is the electric potential outside and inside the pipe? Assume the pipe has a diameter of 2a. I O + + + + + + + + + + + +

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Can you explain how they got to this solution for problem 64? I’m really confused.
**64. The surface charge density on a long straight metallic pipe is \(\sigma\). What is the electric potential outside and inside the pipe? Assume the pipe has a diameter of \(2a\).** **Diagram Explanation:** The diagram illustrates a long, straight, hollow metallic pipe with a surface charge density \(\sigma\). The pipe has a diameter of \(2a\). The arrows in the diagram indicate the direction of potential, extending to infinity (\( \infty \)) both above and below the pipe. - The pipe is cylindrical in shape. - Positive charges (+) are uniformly distributed along the surface of the pipe. - Two vertical lines with arrows point outward from the top and bottom of the pipe, symbolizing the infinite length of the pipe. For educational explanation, the diagram emphasizes the distribution of the surface charge density and the infinite nature of the pipe.
Transcribed Image Text:**64. The surface charge density on a long straight metallic pipe is \(\sigma\). What is the electric potential outside and inside the pipe? Assume the pipe has a diameter of \(2a\).** **Diagram Explanation:** The diagram illustrates a long, straight, hollow metallic pipe with a surface charge density \(\sigma\). The pipe has a diameter of \(2a\). The arrows in the diagram indicate the direction of potential, extending to infinity (\( \infty \)) both above and below the pipe. - The pipe is cylindrical in shape. - Positive charges (+) are uniformly distributed along the surface of the pipe. - Two vertical lines with arrows point outward from the top and bottom of the pipe, symbolizing the infinite length of the pipe. For educational explanation, the diagram emphasizes the distribution of the surface charge density and the infinite nature of the pipe.
--- ### Electrostatic Calculation in a Conducting Pipe **Solution:** The electric field outside the pipe, denoted as \( E_P \), can be expressed using the formula from the previous chapter as: \[ E_P = \frac{\sigma_0}{\epsilon_0} \frac{1}{s} \] where \(\sigma_0\) is the surface charge density and \(\epsilon_0\) is the permittivity of free space. By choosing the zero potential to be at \( R_0 > a \), we can find the potential \( V_P \) as: \[ V_P = -\int_{R_0}^{a} \frac{\sigma_0}{\epsilon_0} \frac{1}{s} ds = -\frac{\sigma_0}{\epsilon_0} \ln \frac{a}{R_0} \] **Additional Problem (65):** Consider two concentric conducting spherical shells carrying charges \( Q \) and \( -Q \) respectively. The inner shell is assumed to have negligible thickness. The problem is to find the potential difference between the shells. **Diagram Description:** The diagram illustrates two concentric spherical shells: - The inner shell, with radius \( b \), carrying a positive charge (+) - The outer shell, depicted with an unspecified radius, carrying a negative charge (-) To find the potential difference between the shells, we will need to apply Gauss's law and the principles of electrostatics to determine the potential at the surface of both shells and subtract one from the other. --- This explanation helps outline the steps in determining the potential in and around conducting surfaces, suitable for students learning electrostatics in an introductory physics course.
Transcribed Image Text:--- ### Electrostatic Calculation in a Conducting Pipe **Solution:** The electric field outside the pipe, denoted as \( E_P \), can be expressed using the formula from the previous chapter as: \[ E_P = \frac{\sigma_0}{\epsilon_0} \frac{1}{s} \] where \(\sigma_0\) is the surface charge density and \(\epsilon_0\) is the permittivity of free space. By choosing the zero potential to be at \( R_0 > a \), we can find the potential \( V_P \) as: \[ V_P = -\int_{R_0}^{a} \frac{\sigma_0}{\epsilon_0} \frac{1}{s} ds = -\frac{\sigma_0}{\epsilon_0} \ln \frac{a}{R_0} \] **Additional Problem (65):** Consider two concentric conducting spherical shells carrying charges \( Q \) and \( -Q \) respectively. The inner shell is assumed to have negligible thickness. The problem is to find the potential difference between the shells. **Diagram Description:** The diagram illustrates two concentric spherical shells: - The inner shell, with radius \( b \), carrying a positive charge (+) - The outer shell, depicted with an unspecified radius, carrying a negative charge (-) To find the potential difference between the shells, we will need to apply Gauss's law and the principles of electrostatics to determine the potential at the surface of both shells and subtract one from the other. --- This explanation helps outline the steps in determining the potential in and around conducting surfaces, suitable for students learning electrostatics in an introductory physics course.
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