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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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Most popular questions from this chapter

(a) Use Gauss's law to prove that the electric field outside any spherically symmetric charge distribution is the same as if all of the charge were concentrated into a point charge. (b) Now use Gauss's law to prove that the electric field inside a spherically symmetric charge distribution is zero if none of the charge is at a distance from the center less than that of the point where we determine the field.
The electric field across a cellular membrane is $1.0 \times 10^{7} \mathrm{N} / \mathrm{C}$ directed into the cell. (a) If a pore opens, which way do sodium ions (Na") flow- into the cell or out of the cell? (b) What is the magnitude of the electric force on the sodium ion? The charge on the sodium ion is \(+e\)
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A conducting sphere that carries a total charge of \(+6 \mu \mathrm{C}\) is placed at the center of a conducting spherical shell that also carries a total charge of \(+6 \mu \mathrm{C}\). The conductors are in electrostatic equilibrium. (a) Determine the charge on the inner surface of the shell. (b) Determine the total charge on the outer surface of the shell.
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