Problem 61 - University Physics with Modern Physics

Question

Coaxial Cylinders. A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius \(b\). The positive charge per unit length on the inner cylinder is \(\lambda\), and there is an equal negative charge per unit length on the outer cylinder. (a) Calculate the potential \(V(r)\) for (i) \(r < a\); (ii) \(a < r < b\); (iii) \(r > b\). (\(Hint:\) The net potential is the sum of the potentials due to the individual conductors.) Take \(V = 0\) at \(r = b\). (b) Show that the potential of the inner cylinder with respect to the outer is $$V^{ab} = \frac{\lambda} {2\pi\epsilon_0} ln \frac{b} {a}$$ (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the cylinders has magnitude $$E(r) = \frac{V_{ab}} {ln(b/a)} \frac{1} {r}$$ (d) What is the potential difference between the two cylinders if the outer cylinder has no net charge?

Step-by-Step Solution

Answer

The potential differences are calculated using integration of the electric fields with respect to radial distances. The potential difference between the cylinders is \(\frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{b}{a}\right)\). Electric fields depend only on radial distance.

Step 1: Identifying Regions for Potential Calculation

To find the potential, we consider the regions: (i) inside the inner cylinder (\( r < a \)); (ii) between the cylinders (\( a < r < b \)); (iii) outside both cylinders (\( r > b \)). For each region, apply Gauss's law to find the electric field and integrate to find the potential.

Step 2: Calculate Electric Field and Potential for r < a

For \( r < a \), since we are inside a conductor, the electric field \( E(r) = 0 \). Therefore, the potential \( V(r) \) is constant. Since \( V = 0 \) at \( r = b \), and no field inside the conductor, \( V(r) = \) constant.

Step 3: Determine Electric Field and Potential for a < r < b

For \( a < r < b \), use a cylindrical Gaussian surface of radius \( r \). From Gauss's law: \( E(r) \cdot 2 \pi r L = \frac{\lambda L}{\epsilon_0} \) gives \( E(r) = \frac{\lambda}{2 \pi \epsilon_0 r} \). Integrating \( E(r) \) from \( b \) to \( r \):\[ V(r) = -\int_b^r \frac{\lambda}{2\pi \epsilon_0 r} \ dr = \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{b}{r}\right) \]

Step 4: Electric Field and Potential for r > b

For \( r > b \), the net charge enclosed is zero, leading to \( E(r) = 0 \). Thus, the potential \( V(r) \) remains constant and equals the potential at \( r = b \), which is zero.

Step 5: Calculate Potential Difference V_ab

To find \( V^{ab} \), use the potential difference between \( r = a \) and \( r = b \):\[ V^{ab} = V(a) - V(b) = \frac{\lambda}{2\pi \epsilon_0} \ln\left(\frac{b}{a}\right) \]

Step 6: Calculate Electric Field E(r) Using V_ab

From the expression for \( V^{ab} \), derive the electric field \( E(r) \) for \( a < r < b \):\[ E(r) = \frac{dV}{dr} = \frac{\lambda}{2\pi\epsilon_0} \frac{1}{r} \]Relate to \( V^{ab} \):\[ E(r) = \frac{V_{ab}}{\ln(b/a)} \frac{1}{r} \]

Step 7: Understanding Potential Difference with No Net Charge

If the outer cylinder has no net charge, the symmetry and equal charge setup still leads to the same potential difference as calculated because potential difference is independent of the path, only depending on the endpoints. Thus, from previous results, the potential difference between cylinders remains \(\frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{b}{a}\right)\).

Key Concepts

Electric Potential

Electric potential is a fundamental concept in electromagnetism. It represents the potential energy per unit charge at a point in space due to electric charges. In simpler terms, it tells us how much work is needed to move a charge to a particular point in the electric field. The unit of electric potential is the volt (V).

For coaxial cylinders, we examine the potential at various points relative to the charged surfaces. The potential depends on the integrated effect of the electric field generated by the charges. The hint of considering the net potential as the sum of potentials from the conductors helps us simplify the calculations. By defining the potential at a distance as zero, typically at the outer surface, we set a reference point for measuring potential differences around the system.

Electric Field

The electric field is a vector field surrounding electric charges. It describes the force that a charge would experience at any point in space. The strength of the field is measured in newtons per coulomb (N/C).

Using Gauss's Law, the electric field between and around the coaxial cylinders can be calculated. The electric field in different regions depends on the charge enclosed. For the region between the cylinders, the field is derived using a cylindrical Gaussian surface, giving the expression: \( E(r) = \frac{\lambda}{2\pi \epsilon_0 r} \).By integrating this expression, we find the electric potential as seen in the previous section. Outside the outer cylinder, since it encloses equal and opposite charges, the net electric field is zero.

Coaxial Cylinders

Coaxial cylinders are a common configuration in electrostatics, consisting of two cylindrical conductors one inside the other. This setup is intriguing because it encases charge distributions that define the behavior of electric fields and potentials in a unique way.

In the given problem, an inner cylinder with a positive charge per unit length \( \lambda \)is surrounded by an outer cylinder with an equal negative charge per length. This symmetry simplifies the electric field and potential calculations. Insulating supports ensure that the cylinders are electrically isolated. Hence, the system nicely illustrates how charges distribute to maintain equilibrium, leading to radial fields and calculable potential differences.

Potential Difference

The potential difference, or voltage, between two points is a measure of the work done in moving a unit charge from one point to another in a field. It's an essential concept for understanding electrical circuits and energy transfer.

In the coaxial cylinder problem, the potential difference between the two cylinders is compactly described by the formula:\[ V^{ab} = \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{b}{a}\right) \]This expression results from the integration of the electric field from radius \( b \) to \( a \), capturing the energy change per unit charge. This potential difference remains the same whether or not the outer cylinder has a net charge, due to the principle that potential difference depends purely on the endpoints, not the path taken.