Question

the figure shows a section of a long, thin-walled metal tube of radius r = 9.90 cm, with a charge per unit length 2.26 nc/m. what is the magnitude e of the electric field at r = 15.0 cm? n/c

Explanation:

Step1: Recall Gauss's Law for Cylindrical Symmetry

For a long, thin - walled metal tube (a cylindrical conductor), when we consider a point outside the tube (\(r > R\), where \(R\) is the radius of the tube), the electric field is given by the formula for the electric field due to an infinite line of charge (since the tube can be approximated as an infinite line of charge for points outside it). The formula from Gauss's law for the electric field \(E\) at a distance \(r\) from an infinite line of charge with linear charge density \(\lambda\) is \(E=\frac{\lambda}{2\pi\epsilon_0r}\), where \(\epsilon_0 = 8.85\times 10^{- 12}\ C^{2}/(N\cdot m^{2})\) is the permittivity of free space.

Step2: Identify the given values

We are given:

• Linear charge density \(\lambda=2.26\ nC/m = 2.26\times 10^{-9}\ C/m\)
• Distance \(r = 15.0\ cm=0.15\ m\)
• \(\epsilon_0 = 8.85\times 10^{-12}\ C^{2}/(N\cdot m^{2})\)

Step3: Substitute the values into the formula

Substitute \(\lambda = 2.26\times 10^{-9}\ C/m\), \(r = 0.15\ m\) and \(\epsilon_0=8.85\times 10^{-12}\ C^{2}/(N\cdot m^{2})\) into the formula \(E=\frac{\lambda}{2\pi\epsilon_0r}\)

First, calculate the denominator: \(2\pi\epsilon_0r=2\times\pi\times8.85\times 10^{-12}\times0.15\)

\(2\times\pi\times8.85\times 10^{-12}\times0.15\approx2\times3.1416\times8.85\times 10^{-12}\times0.15\)

\(2\times3.1416\times8.85\times 10^{-12}\times0.15\approx8.32\times 10^{-12}\)

Then, calculate the numerator: \(\lambda = 2.26\times 10^{-9}\ C/m\)

Now, \(E=\frac{2.26\times 10^{-9}}{8.32\times 10^{-12}}\)

\(E=\frac{2.26\times 10^{3}}{8.32}\approx\frac{2260}{8.32}\approx271.6\ N/C\) (more accurately, using a calculator for \(\frac{2.26\times 10^{-9}}{2\pi\times8.85\times 10^{-12}\times0.15}\))

\(2\pi\times8.85\times 10^{-12}\times0.15=2\times3.14159\times8.85\times 10^{-12}\times0.15 = 8.3203\times 10^{-12}\)

\(E=\frac{2.26\times 10^{-9}}{8.3203\times 10^{-12}}=\frac{2.26\times 10^{3}}{8.3203}\approx271.6\ N/C\approx272\ N/C\) (or more precisely, let's do the calculation step by step)

\(E=\frac{\lambda}{2\pi\epsilon_0r}=\frac{2.26\times 10^{-9}\ C/m}{2\times\pi\times8.85\times 10^{-12}\ C^{2}/(N\cdot m^{2})\times0.15\ m}\)

\(=\frac{2.26\times 10^{-9}}{2\times\pi\times8.85\times 10^{-12}\times0.15}\ N/C\)

\(=\frac{2.26\times 10^{3}}{2\times\pi\times8.85\times 0.15}\ N/C\)

\(2\times\pi\times8.85\times 0.15\approx2\times3.1416\times8.85\times 0.15\approx8.32\)

\(\frac{2.26\times 10^{3}}{8.32}\approx271.6\ N/C\approx272\ N/C\)

Answer:

\(\approx 272\ N/C\) (or more precisely, if we calculate with more decimal places, \(\frac{2.26\times 10^{-9}}{2\times\pi\times8.85\times 10^{-12}\times0.15}\approx271.6\ N/C\))