Problem 38 - University Physics with Modern Physics

Question

22\. 38. A very long conducting tube (hollow cylinder) has inner radius \(a\) and outer radius \(b\) . It carries charge per unit length \(+\alpha\) , where \(\alpha\) is a positive constant with units of \(\mathrm{C} / \mathrm{m}\) . A line of charge lies along the axis of the tube. The line of charge has charge per unit length \(+\alpha\) (a) Calculate the electric field in terms of \(\alpha\) and the distance \(r\) from the axis of the tube for \((i) rb\) . Show your results in a graph of \(E\) as a function of \(r\) . (b) What is the charge per unit length on (i) the inner surface of the tube and (ii) the outer surface of the tube?

Step-by-Step Solution

Answer

(a) E = 0 for r< a; E = \(\frac{\alpha}{2\pi\varepsilon_{0}r}\) for a < r < b; E = \(\frac{\alpha}{\pi\varepsilon_{0}r}\) for r > b. (b) Inner surface: -\(\alpha\); Outer surface: +\(\alpha\).

Step 1: Gauss's Law for Electric Field

To find the electric field created by a symmetrical charge distribution, we use Gauss's Law, which is \[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_{0}} \] where \(Q_{\text{enc}}\) is the charge enclosed by the Gaussian surface. The electric field \(\mathbf{E}\) depends on the distance \(r\) from the line.

Step 2: Case (i) - Electric Field for r < a

For \(r < a\), the Gaussian surface is a cylinder of radius \(r\) which does not enclose any charge since we are inside the tube. Thus, \[ E \times 2 \pi r L = 0 \rightarrow E = 0 \]The electric field is zero inside the hollow part of the tube.

Step 3: Case (ii) - Electric Field for a < r < b

For \(a < r < b\), the Gaussian cylinder encloses the linear charge along the axis, which has charge per unit length \(+\alpha\). Thus, \[ E \times 2 \pi r L = \frac{\alpha L}{\varepsilon_{0}} \rightarrow E = \frac{\alpha}{2 \pi \varepsilon_{0} r} \]The electric field at a distance \(r\) from the axis increases as \(\frac{1}{r}\) within this region.

Step 4: Case (iii) - Electric Field for r > b

For \(r > b\), the charge enclosed by the Gaussian surface includes the charge from both the axis and the tube. Per unit length, this totals to \(+2\alpha\) (line and the inner surface of the tube). Hence,\[E \times 2 \pi r L = \frac{2\alpha L}{\varepsilon_{0}} \rightarrow E = \frac{2\alpha}{2 \pi \varepsilon_{0} r} = \frac{\alpha}{\pi \varepsilon_{0} r} \]The electric field decreases as \(\frac{1}{r}\) for points outside the tube.

Step 5: Graphing Electric Field vs. Distance

The graph shows \(E = 0\) for \(r < a\), increases as \(\frac{1}{r}\) for \(a < r < b\), and decreases as \(\frac{1}{r}\) for \(r > b\). The electric field is discontinuous at \(r = a\) and \(r = b\), showing a change in the charge environment.

Step 6: Charge on Surfaces - Inner Surface

By symmetry, the inner surface of the tube must balance the line charge's field within \(a < r < b\). Hence, the charge per unit length on the inner surface is \[-\alpha\].

Step 7: Charge on Surfaces - Outer Surface

The net charge on the tube is \(+\alpha\). With \(-\alpha\) on the inner surface, the outer surface must have \[ +2\alpha - \alpha = +\alpha \]Thus, the charge per unit length on the outer surface is \[ +\alpha \].

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that relates the electric field in a region to the electric charge in that region. It states that the total electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. Mathematically, it is expressed as:\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_{0}} \]This means that if you know the charge enclosed in a surface, you can find the electric field by applying Gauss's Law. It's especially useful for problems with high symmetry, like spheres or cylinders, where the electric field can be considered constant over the surface.

Electric Field

The electric field is a vector field that represents the force experienced by a unit positive charge at any point in space. It is an essential concept when discussing electromagnetism. Depending on the location inside or outside the charged object, the electric field can vary significantly.For the problem given:- **Inside the tube, when \( r < a \):** There is no charge enclosed within the Gaussian surface, so the electric field is zero. This means any point within the hollow section of the cylindrical tube doesn’t experience any electric forces.- **In the region \( a < r < b \):** The electric field originates from the line charge along the axis. The electric field here can be calculated using the formula \( E = \frac{\alpha}{2 \pi \varepsilon_{0} r} \). It decreases with increasing distance, following an inverse relation to \( r \).- **Outside the tube, \( r > b \):** Both the axial line charge and the inner surface charge contribute here. The electric field's formula changes to \( E = \frac{\alpha}{\pi \varepsilon_{0} r} \), and it decreases with distance, similar to a larger single linear charge.

Charge Distribution

Charge distribution is crucial for understanding how electric fields behave in and around charged objects. In this exercise, the distribution of charge on the surfaces of the cylindrical tube and along its axis affects the electric field at various points.- **Inner surface of the tube:** Since the line of charge at the axis attracts charges on the closest surface, there is an induced charge on the inner surface that is exactly \. vanishing the interior electric field: - This charge per unit length is \( -\alpha \), neutralizing the axial charge effect inside the tube.- **Outer surface of the tube:** Any remaining charge after balancing on the inner surface needs to appear here. - The outer surface ends up having a charge per unit length of \( +\alpha \), as it complements the overall positive charge per unit length of the entire tube, calculated considering the charges present on the inner surface.This charge distribution creates the observed electric fields both within and outside the cylinder.