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Ch 23: Electric Potential
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoYoung & Freedman Calc 14th EditionCh 23: Electric PotentialProblema 32
Capitolo 23, Problema 32

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?

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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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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.
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Electric Field due to a Point Charge

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.
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Electric Potential

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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Gauss' Law
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Two large, parallel conducting plates carrying op­posite charges of equal magnitude are separated by 2.202.20 cm. The surface charge density for each plate has magnitude 47.047.0 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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An infinitely long line of charge has linear charge den­sity 5.00×10−125.00\(\times\)10^{-12} C/m. A proton (mass 1.67×10−271.67\(\times\)10^{-27} kg, charge +1.60×10−19+1.60\(\times\)10^{-19} C) is 18.018.0 cm from the line and moving directly toward the line at 3.50×1033.50\(\times\)10^3 m/s. How close does the proton get to the line of charge?

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