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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

Consider two protons (charge \(+e\) ), separated by a distance of $2.0 \times 10^{-15} \mathrm{m}$ (as in a typical atomic nucleus). The electric force between these protons is equal in magnitude to the gravitational force on an object of what mass near Earth's surface?
Two point charges are located on the \(x\) -axis: a charge of \(+6.0 \mathrm{nC}\) at \(x=0\) and an unknown charge \(q\) at \(x=\) \(0.50 \mathrm{m} .\) No other charges are nearby. If the electric field is zero at the point $x=1.0 \mathrm{m},\( what is \)q ?$
Point charges are arranged on the vertices of a square with sides of $2.50 \mathrm{cm} .$ Starting at the upper left corner and going clockwise, we have charge A with a charge of \(0.200 \mu \mathrm{C}, \mathrm{B}\) with a charge of \(-0.150 \mu \mathrm{C}, \mathrm{C}\) with a charge of \(0.300 \mu \mathrm{C},\) and \(\mathrm{D}\) with a mass of \(2.00 \mathrm{g},\) but with an unknown charge. Charges \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are fixed in place, and \(\mathrm{D}\) is free to move. Particle D's instantaneous acceleration at point \(D\) is \(248 \mathrm{m} / \mathrm{s}^{2}\) in a direction \(30.8^{\circ}\) below the negative \(x\) -axis. What is the charge on D?
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