A very long insulating cylinder of charge of radius 2.502.50 cm carries a uniform linear density of 15.015.0 nC/m. If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 175175 V?

Ch 23: Electric Potential

Young & Freedman Calc - University Physics 15th Edition

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Young & Freedman Calc 15th Edition

Ch 23: Electric Potential

Problem 34

Chapter 23, Problem 34

A very long insulating cylinder of charge of radius $2.50$ cm carries a uniform linear density of $15.0$ nC/m. If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads $175$ V?

Verified step by step guidance

Start by understanding that the problem involves an insulating cylinder with a uniform linear charge density. The linear charge density $ \lambda $ is given as 15.0 nC/m, which is $ 15.0 \times 10^{-9} $ C/m.

Use Gauss's Law to find the electric field $ E $ outside the cylinder. For a cylindrical surface coaxial with the charged cylinder, the electric field is given by $ E = \frac{\lambda}{2\pi\varepsilon_0 r} $, where $ r $ is the radial distance from the axis of the cylinder and $ \varepsilon_0 $ is the permittivity of free space $ 8.85 \times 10^{-12} $ C²/(N·m²).

Recognize that the voltmeter measures the potential difference $ \Delta V $ between two points. The potential difference between the surface of the cylinder and a point at distance $ r $ from the surface is given by $ \Delta V = \int_{R}^{r} E \, dr $, where $ R $ is the radius of the cylinder (2.50 cm or 0.025 m).

Substitute the expression for $ E $ into the integral to find $ \Delta V = \frac{\lambda}{2\pi\varepsilon_0} \int_{R}^{r} \frac{1}{r} \, dr $. This integral evaluates to $ \Delta V = \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r}{R}\right) $.

Set $ \Delta V = 175 $ V and solve for $ r $ using the equation $ 175 = \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r}{R}\right) $. Rearrange to find $ r = R \cdot e^{\frac{2\pi\varepsilon_0 \cdot 175}{\lambda}} $. Calculate $ r $ to find the distance from the surface where the other probe must be placed.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field of a Charged Cylinder

The electric field around a long, uniformly charged insulating cylinder can be determined using Gauss's Law. For points outside the cylinder, the electric field behaves as if all the charge were concentrated along the axis, decreasing with distance from the surface. This field is crucial for calculating potential differences.

Electric Potential Difference

The electric potential difference between two points in an electric field is the work done in moving a unit charge from one point to the other. For a radial electric field, this involves integrating the electric field over the distance between the two points, which helps determine the voltmeter reading.

Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. For a cylindrical symmetry, it simplifies the calculation of the electric field by considering a Gaussian surface co-axial with the charged cylinder, allowing us to find the field and potential difference efficiently.

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Related Practice

Textbook Question

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by $2.20$ cm. If the surface charge density for each plate has magnitude $47.0$ nC/m², what is the magnitude of $E$ in the region between the plates?

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Textbook Question

A thin spherical shell with radius $R sub 1 equals 3.00$ cm is concentric with a larger thin spherical shell with radius $R sub 2 equals 5.00$ cm. Both shells are made of insulating material. The smaller shell has charge $q sub 1 equals positive 6.00$ nC distributed uniformly over its surface, and the larger shell has charge $q_{2} = - 9.00$ nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. What is the electric potential due to the two shells at the following distance from their common center: (i) $r equals 0$ ; (ii) $r equals 4.00$ cm; (iii) $r equals 6.00$ cm?

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Textbook Question

An infinitely long line of charge has linear charge density $5.00 \times 10^{- 12}$ C/m. A proton (mass $1.67 \times 10^{- 27}$ kg, charge $+ 1.60 \times 10^{- 19}$ C) is $18.0$ cm from the line and moving directly toward the line at $3.50 times 10 cubed$ m/s. Calculate the proton's initial kinetic energy.

Textbook Question

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by $2.20$ cm. What is the potential difference between the two plates?

Textbook Question

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by $2.20$ cm. The surface charge density for each plate has magnitude $47.0$ nC/m^2. If the separation between the plates is doubled while the surface charge density is kept constant at the given value, what happens to the magnitude of the electric field and to the potential difference?

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Textbook Question

An infinitely long line of charge has linear charge density $5.00$\times$10^{-12}$ C/m. A proton (mass $1.67$\times$10^{-27}$ kg, charge $+1.60$\times$10^{-19}$ C) is $18.0$ cm from the line and moving directly toward the line at $3.50$\times$10^3$ m/s. How close does the proton get to the line of charge?

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