Problem 33 - University Physics with Modern Physics

Question

A very long insulating cylindrical shell of radius 6.00 \(\mathrm{cm}\) carries charge of linear density 8.50\(\mu \mathrm{C} / \mathrm{m}\) spread uniformly over its outer surface. What would a voltmeter read if it were connected between (a) the surface of the cylinder and a point 4.00 \(\mathrm{cm}\) above the surface, and (b) the surface and a point 1.00 \(\mathrm{cm}\) from the central axis of the cylinder?

Step-by-Step Solution

Answer

(a) 3.93 x 10^4 V, (b) 0 V.

Step 1: Understand the Problem

We are given a cylindrical shell with a radius of 6.00 cm, and it carries a linear charge density of 8.50 \( \mu \mathrm{C/m} \). We need to find the potential difference between given points using Gauss's Law.

Step 2: Calculate Electric Field

For an infinitely long charged cylindrical shell, the electric field \( E \) outside the cylinder at a distance \( r \) from the central axis is given by:\[ E = \frac{\lambda}{2\pi\varepsilon_0 r} \]where \( \lambda = 8.50 \times 10^{-6} \ \mathrm{C/m} \), and \( \varepsilon_0 \) is the permittivity of free space \( 8.85 \times 10^{-12} \ \mathrm{C^2/N \cdot m^2} \).

Step 3: Potential Difference from Surface to 4 cm Above

Since this point lies outside the cylinder, use the formula of potential difference \( V = - \int E \cdot dr \) from the surface to 4 cm above.\[ V = -\int_{0.06}^{0.10} \frac{\lambda}{2\pi\varepsilon_0 r} dr \]Calculating the integral:\[ V = - \left[ \frac{\lambda}{2\pi\varepsilon_0} \ln r \right]_{0.06}^{0.10} = \frac{8.50 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12}} \ln \left( \frac{0.10}{0.06} \right) \]Solve for \( V \).

Step 4: Solve for Potential Difference in Part (a)

Calculating the value:\[ V = \frac{8.50 \times 10^{-6} \cdot 0.510}{2\pi \times 8.85 \times 10^{-12}} = \frac{8.50 \times 0.510}{1.11} \times 10^6 ≈ 3.93 \times 10^4 \ \mathrm{V} \]

Step 5: Potential Inside a Conductor

Inside a charged conductor, the electric field is zero. Thus, for any point inside the conductor (from the surface to 1 cm from the center), the potential difference is zero.

Step 6: Short Answer Conclusions

For part (a), the potential difference is \( 3.93 \times 10^4 \ \mathrm{V} \) between the surface and a point 4.00 cm above. For part (b), it is 0 V as the point is inside the conductor.

Key Concepts

Electric Field Calculation

To understand the electric field calculation using Gauss's Law, we first imagine a cylindrical shell with a uniformly distributed charge across its outer surface. To calculate the electric field at a distance from the shell, we utilize Gauss's Law. This law states that the electric field \[ E = \frac{\lambda}{2\pi\varepsilon_0 r} \] at a distance \( r \) from the axis of the cylinder depends on the linear charge density \( \lambda \) and the permittivity of free space \( \varepsilon_0 \).

• \( \lambda \) is the charge per unit length, given as \( 8.50 \times 10^{-6} \, \mathrm{C/m} \).
• \( \varepsilon_0 \) is a constant \( 8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2} \).
• \( r \) is the distance from the central axis.

By applying these values in the equation, the electric field can be specifically found for any point around the cylinder. Such calculations are crucial in determining the potential difference in following steps.

Potential Difference

Potential difference is a measure of electrical potential energy per unit charge in an electric field. In our problem, we measure the potential difference between two points: one at the surface and another at a certain height or within the cylindrical shell.For a point exterior to the cylindrical shell, we determine the potential difference by evaluating:\[ V = - \int_{r_1}^{r_2} E \, dr \]This integral calculates how the electric potential changes from \( r_1 \) to \( r_2 \). For the given exercise, - the integral is evaluated from the surface at \( 0.06 \, \mathrm{m} \) to a point \( 0.10 \, \mathrm{m} \) from the axis above the surface.For a situation where the point is inside the shell's limits, as in the given problem's part (b), there is no electric field, hence no change in potential:- Inside the conductor, the potential remains constant, making the potential difference zero.Knowing how to evaluate these integrals helps determine voltages due to charge distributions.

Cylindrical Charge Distribution

Understanding cylindrical charge distribution is essential when solving problems dealing with cylindrical geometries like our cylindrical shell. In the exercise, we are dealing with a very long insulating cylindrical shell that has a constant linear charge density.- **Linear charge density**, represented as \( \lambda \), is the amount of charge per unit length. In our situation, it is given as \( 8.50 \mu \mathrm{C/m} \). - It assumes the cylinder is infinitely long, simplifying the math by making edge effects negligible. This lets us use symmetry to analyze the electric field only in a radial component.The characteristics of such a distribution are significant when assessing potential differences and electric field strengths at various points concerning the cylinder. In realistic scenarios, appreciating the behavior due to distributed charges along cylindrical objects helps in designing and understanding systems where geometrical symmetry is pivotal.