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Chapter 23: Problem 34

A very long insulating cylinder of charge of radius \(2.50 \mathrm{~cm}\) carries a uniform linear density of \(15.0 \mathrm{nC} / \mathrm{m}\). 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 \(175 \mathrm{~V} ?\)

Short Answer

Expert verified
The distance from the surface of the charging cylinder to the other probe needed for the voltmeter to read 175V, or \(r2\), can be found by performing the calculations laid out in the steps. It is important to ensure all values are represented in proper SI units before substituting into the formulas.

Step by step solution

01

Identify the Formula for Electric Field of a Long Uniformly Charged Cylinder

The electric field, \(E\), of a long uniformly charged cylinder inside can be expressed as \(E = \frac{位}{2蟺蔚_0r}\), where 位 is the linear charge density, \(蔚_0\) is the permittivity of free space and \(r\) is the radial distance from the center of the cylinder. However, this formula is only valid inside the charged cylinder; outside the charged cylinder, the electric field is given by \(E = \frac{位}{2蟺蔚_0r}\). Since we're concerned with the surface and beyond, we should use the latter formula.
02

Calculate Electric Field at the Surface

Substitute 位 = \(15.0 脳 10^{-9} C/m\) and r = \(2.50 脳 10^{-2} m\) into the equation. The permittivity of free space \(蔚_0\) is a known constant, \(蔚_0 = 8.85 脳 10^{-12} C^2/N路m^2\). Therefore, we can calculate the electric field at the surface, \(E_s\), as \(E_s = \frac{位}{2蟺蔚_0r}\)
03

Identify Formula for Electric Potential Difference

The electric potential difference (\(V\)) between two points is the work done per unit charge in moving a positive test charge from one point to another. The formula to calculate the electric potential difference in the electric field of a long straight uniformly charged line from a point at distance r1 to a point at distance r2 is given by \(V = \frac{位}{2蟺蔚_0} ln(\frac{r2}{r1})\)
04

Solve for r2

Set \(V = 175V\), \(r1 = 2.50cm\) and Solve the equation for \(r2\)
05

Final Calculation

Use the values of \(V\), \(位\), \(蔚_0\), and \(r1\) in the formula identified in step 3 to solve for \(r2\), the distance from the surface of the cylinder to the other probe needed for the voltmeter to read 175V

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Potential Difference
Electric potential difference is a fundamental concept in the study of electric fields and charged objects. It represents the amount of work needed to move a unit charge from one point to another in an electric field.
When dealing with a charged cylinder, calculating the electric potential difference involves understanding how the electric field emanates from the charged surface.
In the case of a long, uniformly charged cylinder, the potential difference can be calculated using the formula:
  • \(V = \frac{位}{2蟺蔚_0} \ln(\frac{r_2}{r_1})\)
Here, \(\lambda\) is the linear charge density, \(\epsilon_0\) is the permittivity of free space, and \(r_1\) and \(r_2\) are distances from the center of the cylinder to two different points in space.
Understanding this concept allows one to determine how potential energy changes within the electric field surrounding the charged cylinder.
Linear Charge Density
Linear charge density is a measure of how much electric charge is distributed along a line, such as the axis of a cylinder. It is defined as the charge per unit length, which makes it particularly useful when dealing with long, cylindrical objects with uniform charge distribution.
For the given exercise, the cylinder's linear charge density is \(\lambda = 15.0 \, \text{nC/m} \). This value indicates that for every meter along the axis of the cylinder, there is a charge of 15 nanocoulombs.
In practical terms, understanding linear charge density is key when calculating the electric field and potential difference because it affects how these quantities change as you move away from the surface of the cylinder.
Permittivity of Free Space
Permittivity of free space, also often referred to by the symbol \(\epsilon_0\), is a very important constant in electromagnetism. It describes the ability of a vacuum to permit electric field lines. The permittivity of free space is approximately \(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\).
This constant features prominently in the calculations of electric fields and potential differences, especially when dealing with long cylinders. The permittivity helps dictate how the electric field from a charge extends through space.
When calculating the electric field around the charged cylinder, \(\epsilon_0\) appears in the equation:
  • \(E = \frac{位}{2蟺蔚_0r}\)
Its presence ensures that the electromagnetic forces are expressed accurately under the physical laws governing electric fields.

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Most popular questions from this chapter

A solid conducting sphere of radius \(5.00 \mathrm{~cm}\) carries a net charge. To find the value of the charge, you measure the potential difference \(V_{A B}=V_{A}-V_{B}\) between point \(A,\) which is \(8.00 \mathrm{~cm}\) from the center of the sphere, and point \(B\), which is a distance \(r\) from the center of the sphere. You repeat these measurements for several values of \(r>8.00 \mathrm{~cm} .\) When you plot your data as \(V_{A B}\) versus \(1 / r,\) the values lie close to a straight line with slope \(-18.0 \mathrm{~V} \cdot \mathrm{m}\). What does your data give for the net charge on the sphere? Is the net charge positive or negative?

A thin spherical shell with radius \(R_{1}=3.00 \mathrm{~cm}\) is concentric with a larger thin spherical shell with radius \(R_{2}=5.00 \mathrm{~cm}\). Both shells are made of insulating material. The smaller shell has charge \(q_{1}=+6.00 \mathrm{nC}\) distributed uniformly over its surface, and the larger shell has charge \(q_{2}=-9.00 \mathrm{nC}\) distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the fol- (ii) \(r=4.00 \mathrm{~cm}\) lowing distance from their common center: (i) \(r=0\) (iii) \(r=6.00 \mathrm{~cm} ?\) (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

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) \(rb .\) (Hint: The net potcntial 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 \(V_{a b}=\frac{\lambda}{2 \pi \epsilon_{0}} \ln \frac{b}{a}\) (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_{a b}}{\ln (b / a)} \frac{1}{r} (d) What is the potential difference between the two cylinders if the outer cylinder has no net charge?

A very large plastic sheet carries a uniform charge density of \(-6.00 \mathrm{nC} / \mathrm{m}^{2}\) on one face. (a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? (b) Find the spacing between equipotential surfaces that differ from each other by \(1.00 \mathrm{~V}\). What type of surfaces are these?

Two point charges of equal magnitude \(Q\) are held a distance \(d\) apart. Consider only points on the line passing through both charges: take \(V=0\) at infinity. (a) If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential is zero (is the electric field zero at these points?), and (ii) the electric field is zero (is the potential zero at these points?). (b) Repeat part (a) for two point charges having opposite signs.
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