Part 2: Potential of Long Conducting Cylindrical Shell We will now study the electric field of a system consisting of an infinitely long solid conducting cylinder (1 cm radius) held at a positive potential Vo = 10 V that is centered within a uniformly long conducting cylindrical shell (10 cm radius) held at ground (our zero of potential). You can picture the painted conductive paper as being a perpendicular slice of this infinitely long system. Preliminary calculations: 2a) Using Gauss's Law, (§ Ē·dà = Qenc/to) show that the electric field outside an infinitely long charged conducting cylinder is given by X Ē = -↑ 2πeor (3) where is the charge per length, r is the distance from the axis and is the radial unit vector. Make sure you draw a Gaussian surface and show all your calculations, even the trivial ones. Note that there should be two different cylinders in your drawing - the real conducting cylinder and the Gaussian surface. 26) Show that with the electric field derived above (which is the case for points between the cylinders) and a path of integration (dī) that is radially outward, the potential difference between a point at radius ro and a point at radius r (AV = — fr E dl) can be expressed as . AV = V(r) - V(ro) X 27E0 r To (4)
Part 2: Potential of Long Conducting Cylindrical Shell We will now study the electric field of a system consisting of an infinitely long solid conducting cylinder (1 cm radius) held at a positive potential Vo = 10 V that is centered within a uniformly long conducting cylindrical shell (10 cm radius) held at ground (our zero of potential). You can picture the painted conductive paper as being a perpendicular slice of this infinitely long system. Preliminary calculations: 2a) Using Gauss's Law, (§ Ē·dà = Qenc/to) show that the electric field outside an infinitely long charged conducting cylinder is given by X Ē = -↑ 2πeor (3) where is the charge per length, r is the distance from the axis and is the radial unit vector. Make sure you draw a Gaussian surface and show all your calculations, even the trivial ones. Note that there should be two different cylinders in your drawing - the real conducting cylinder and the Gaussian surface. 26) Show that with the electric field derived above (which is the case for points between the cylinders) and a path of integration (dī) that is radially outward, the potential difference between a point at radius ro and a point at radius r (AV = — fr E dl) can be expressed as . AV = V(r) - V(ro) X 27E0 r To (4)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Part 2: Potential of Long Conducting Cylindrical Shell**
We will now study the electric field of a system consisting of an infinitely long solid conducting cylinder (1 cm radius) held at a positive potential \( V_0 = 10 \, \text{V} \) that is centered within a uniformly long conducting cylindrical shell (10 cm radius) held at ground (our zero of potential). You can picture the painted conductive paper as being a perpendicular slice of this infinitely long system.
*Preliminary calculations:*
2a) Using Gauss’s Law, \(( \oint \vec{E} \cdot d\vec{A} = Q_{\text{enc}}/\epsilon_0 )\) show that the electric field outside an infinitely long charged conducting cylinder is given by
\[
\vec{E} = \frac{\lambda}{2 \pi \epsilon_0 r} \hat{r}
\]
where \( \lambda \) is the charge per length, \( r \) is the distance from the axis and \( \hat{r} \) is the radial unit vector. Make sure you draw a Gaussian surface and show all your calculations, even the trivial ones. Note that there should be two different cylinders in your drawing - the real conducting cylinder and the Gaussian surface.
2b) *Show* that with the electric field derived above (which is the case for points between the cylinders) and a path of integration (\( d\vec{l} \)) that is radially outward, the potential difference between a point at radius \( r_0 \) and a point at radius \( r \) \(( \Delta V = -\int_{r_0}^{r} \vec{E} \cdot d\vec{l} )\) can be expressed as
\[
\Delta V = V(r) - V(r_0) = -\frac{\lambda}{2 \pi \epsilon_0} \ln \left( \frac{r}{r_0} \right)
\]
**Explanation of Equations:**
- **Equation 1**: The expression for the electric field \( \vec{E} \) in terms of charge per unit length \( \lambda \) is derived using Gauss’s Law. It shows that the electric field decreases with distance \( r \) from the axis of the cylinder.
- **Equation 2**: The](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Facfb296d-c42b-4e28-876a-1baf5375f9c9%2F25d57d93-ddba-467c-aacc-cca27072b1b3%2F652v9tl_processed.png&w=3840&q=75)
Transcribed Image Text:**Part 2: Potential of Long Conducting Cylindrical Shell**
We will now study the electric field of a system consisting of an infinitely long solid conducting cylinder (1 cm radius) held at a positive potential \( V_0 = 10 \, \text{V} \) that is centered within a uniformly long conducting cylindrical shell (10 cm radius) held at ground (our zero of potential). You can picture the painted conductive paper as being a perpendicular slice of this infinitely long system.
*Preliminary calculations:*
2a) Using Gauss’s Law, \(( \oint \vec{E} \cdot d\vec{A} = Q_{\text{enc}}/\epsilon_0 )\) show that the electric field outside an infinitely long charged conducting cylinder is given by
\[
\vec{E} = \frac{\lambda}{2 \pi \epsilon_0 r} \hat{r}
\]
where \( \lambda \) is the charge per length, \( r \) is the distance from the axis and \( \hat{r} \) is the radial unit vector. Make sure you draw a Gaussian surface and show all your calculations, even the trivial ones. Note that there should be two different cylinders in your drawing - the real conducting cylinder and the Gaussian surface.
2b) *Show* that with the electric field derived above (which is the case for points between the cylinders) and a path of integration (\( d\vec{l} \)) that is radially outward, the potential difference between a point at radius \( r_0 \) and a point at radius \( r \) \(( \Delta V = -\int_{r_0}^{r} \vec{E} \cdot d\vec{l} )\) can be expressed as
\[
\Delta V = V(r) - V(r_0) = -\frac{\lambda}{2 \pi \epsilon_0} \ln \left( \frac{r}{r_0} \right)
\]
**Explanation of Equations:**
- **Equation 1**: The expression for the electric field \( \vec{E} \) in terms of charge per unit length \( \lambda \) is derived using Gauss’s Law. It shows that the electric field decreases with distance \( r \) from the axis of the cylinder.
- **Equation 2**: The
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