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Chapter 16: Problem 75

In lab tests it was found that rats can detect electric fields of about $5.0 \mathrm{kN} / \mathrm{C}\( or more. If a point charge of \)1.0 \mu \mathrm{C}$ is sitting in a maze, how close must the rat come to the charge in order to detect it?

Short Answer

Expert verified
Answer: The rat must be at least \(2.35\times10^{-2}\,\mathrm{m}\) or \(23.5\,\mathrm{mm}\) close to the point charge in order to detect it.

Step by step solution

01

Write down the given values

We are given: - Threshold electric field E = \(5.0\,\mathrm{kN/C}\) or \(5.0\times10^3\,\mathrm{N/C}\) - Point charge Q = \(1.0\,\mathrm{\mu C}\) or \(1.0\times10^{-6}\,\mathrm{C}\)
02

Use the electric field formula

We know that the electric field due to a point charge is given by the formula: \(E = k\frac{Q}{r^2}\) where \(k = 8.99\times10^9\,\mathrm{N\cdot m^2/C^2}\) (electrostatic constant) and \(r\) is the distance from the charge.
03

Solve for r

We want to solve for \(r\). First, write the equation with the given values of \(E\) and \(Q\): \(5.0\times10^3\,\mathrm{N/C} = (8.99\times10^9\,\mathrm{N\cdot m^2/C^2})\frac{1.0\times10^{-6}\,\mathrm{C}}{r^2}\). Now, we need to solve for \(r\). First, divide both sides by \(kQ\): \(\frac{5.0\times10^3\,\mathrm{N/C}}{1.0\times10^{-6}\,\mathrm{C}} = \frac{(8.99\times10^9\,\mathrm{N\cdot m^2/C^2})}{r^2}\). Next, isolate \(r^2\) by multiplying both sides by \(\frac{1}{(8.99\times10^9\,\mathrm{N\cdot m^2/C^2})}\): \(r^2 = \frac{5.0\times10^3\,\mathrm{N/C}}{(8.99\times10^9\,\mathrm{N\cdot m^2/C^2})(1.0\times10^{-6}\,\mathrm{C})}\).
04

Calculate r

Now, we can calculate the value of \(r\) by taking the square root of both sides: \(r = \sqrt{\frac{5.0\times10^3\,\mathrm{N/C}}{(8.99\times10^9\,\mathrm{N\cdot m^2/C^2})(1.0\times10^{-6}\,\mathrm{C})}}\) \(r = 2.35\times10^{-2}\,\mathrm{m}\) So, the rat must be at least \(2.35\times10^{-2}\,\mathrm{m}\) or \(23.5\,\mathrm{mm}\) close to the point charge in order to detect it.

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