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Chapter 23: Problem 40

Electrical Sensitivity of Sharks. Certain sharks can detect an electric field as weak as 1.0\(\mu \mathrm{V} / \mathrm{m} .\) To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.5-V AA battery across these plates, how far apart would the plates have to be?

Short Answer

Expert verified
The plates must be 1,500 kilometers apart.

Step by step solution

01

Understand the Electric Field Formula

The electric field \(E\) between two parallel plates is given by the formula \(E = \frac{V}{d}\), where \(V\) is the voltage across the plates and \(d\) is the distance between the plates.
02

Rearrange the Formula to Solve for Distance

We want to find the distance \(d\), so we rearrange the formula to solve for \(d\). To do this, multiply both sides of the equation by \(d\) and then divide both sides by \(E\): \(d = \frac{V}{E}\).
03

Substitute the Given Values

Substitute the given values into the formula. Here, \(V = 1.5 \, \text{V}\) and \(E = 1.0 \, \mu \text{V/m} = 1.0 \times 10^{-6} \, \text{V/m}\). Thus, \(d = \frac{1.5}{1.0 \times 10^{-6}}\).
04

Calculate the Distance

Now calculate \(d\) using the substituted values: \(d = \frac{1.5}{1.0 \times 10^{-6}} = 1.5 \times 10^{6} = 1,500,000 \, \text{meters}\).
05

Interpret the Result

The result, \(1,500,000 \, \text{meters}\), means that in order to produce an electric field of 1.0 \(\mu \text{V/m}\) with a 1.5V battery, the plates would have to be 1,500,000 meters, or 1,500 kilometers, apart.

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Key Concepts

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

Voltage
Voltage is an essential concept in understanding electric fields. It represents the electric potential difference between two points. Imagine it like the pressure that pushes electric charges through a circuit. In this context, voltage is the force that makes an electric field possible between two plates.
  • Measured in volts (V), it tells us how strong an electric field can be for a given distance.
  • The greater the voltage, the stronger the electric field when other factors are constant.
Understanding voltage involves realizing how this potential difference can drive the motion of electric charges. In the problem, the voltage of the AA battery is used to determine how far apart plates must be to achieve a weak electric field. With a voltage of 1.5V, we see how even a small battery can influence electric fields over large distances.
Distance Between Plates
The distance between plates in a parallel plate setup is crucial for calculating the electric field formed between them. This distance, denoted by \(d\), is what determines how diluted or concentrated an electric field is given a fixed voltage.
  • When plates are closer together, the electric field is stronger and more concentrated.
  • If the plates are farther apart, the same voltage results in a weaker electric field.
In this problem, by rearranging the formula \(E = \frac{V}{d}\), we can solve for the distance \(d\). Substituting the known values provides insight into how spacing affects field strength. For the specified conditions, calculating \(d\) gives us a colossal distance of 1,500 kilometers, illustrating the extent needed to sustain such a weak electric field with a 1.5V battery.
Electric Field Sensitivity
Electric field sensitivity is a measure of how weak an electric field can be yet still detectable by an organism or an instrument. The concept highlights the remarkable sensitivity some creatures, like certain sharks, possess. These sharks can detect electric fields as weak as 1.0\(\mu \mathrm{V} / \mathrm{m}\), showcasing their highly advanced biological electroreception.
  • 1.0\(\mu \mathrm{V} / \mathrm{m}\) is a microvolt per meter, indicating a very subtle and weak electric field.
  • This sensitivity surpasses many artificial detection devices we're familiar with.
Understanding this level of sensitivity illustrates why calculations like those in the problem are relevant. They allow us to compare human engineering to natural systems and appreciate the precision and sensitivity present in nature.
Physics Problem Solving
Physics problem-solving often involves translating real-world scenarios into mathematical forms. The solution to this problem required understanding and applying the formula for an electric field between two plates.
  • Identifying the given values: voltage and desired field strength.
  • Rearranging the formula to solve for the unknown: distance between plates.
  • Substituting values and performing calculations to find a sensible result.
What makes this exercise intriguing is that it merges conceptual knowledge with mathematical application. By solving it, students not only practice rearranging equations but also gain insights into how electric fields operate in diverse contexts. This reinforces the connection between physics formulas and real-world applications, deepening comprehension through practical examples.

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Most popular questions from this chapter

A point charge \(q_{1}=+2.40 \mu \mathrm{C}\) is held stationary at the origin. A second point charge \(q_{2}=-4.30 \mu \mathrm{C}\) moves from the point \(x=0.150 \mathrm{m}, \quad y=0\) to the point \(x=0.250 \mathrm{m},\) \(y=0.250 \mathrm{m} .\) How much work is done by the electric force on \(q_{2} ?\)

Two plastic spheres, each carrying charge uniformly distributed throughout its interior, are initially placed in contact and then released. One sphere is 60.0 \(\mathrm{cm}\) in diameter, has mass \(50.0 \mathrm{g},\) and contains \(-10.0 \mu \mathrm{C}\) of charge. The other sphere is 40.0 \(\mathrm{cm}\) in diameter, has mass \(150.0 \mathrm{g},\) and contains \(-30.0 \mu \mathrm{c}\) of charge. Find the maximum acceleration and the maximum speed achieved by each sphere (relative to the fixed point of their initial location in space), assuming that no other forces are acting on them. (Hint: The uniformly distributed charges behave as though they were concentrated at the centers of the two spheres.)

A positive charge \(+q\) is located at the point \(x=0\) \(y=-a,\) and a negative charge \(-q\) is located at the point \(x=0,\) \(y=+a .(\) a) Derive an expression for the potential \(V\) at points on the \(y\) -axis as a function of the coordinate \(y .\) Take \(V\) to be zero at an infinite distance from the charges. (b) Graph \(V\) at points on the \(y\) -axis as a function of \(y\) over the range from \(y=-4 a\) to \(y=+4 a\) (c) Show that for \(y>a\) , the potential at a point on the positive \(y\) -axis is given by \(V=-\left(1 / 4 \pi \epsilon_{0}\right) 2 q a / y^{2}\) . (d) What are the answers to parts (a) and (c) if the two charges are interchanged so that \(+q\) is at \(y=+a\) and \(-q\) is at \(y=-a ?\)

A very long insulating cylindrical shell of radius 6.00 \(\mathrm{cm}\) carries charge of linear density 8.50\(\mu \mathrm{C} / \mathrm{m}\) spread uniformly over its outer surface. What would a voltmeter read if it were connected between (a) the surface of the cylinder and a point 4.00 \(\mathrm{cm}\) above the surface, and (b) the surface and a point 1.00 \(\mathrm{cm}\) from the central axis of the cylinder?

A point charge \(q_{1}\) is held stationary at the origin. A second charge \(q_{2}\) is placed at point \(a,\) and the electric potential energy of the pair of charges is \(+5.4 \times 10^{-8} \mathrm{J} .\) When the second charge is moved to point \(b,\) the electric force on the charge does \(-1.9 \times 10^{-8} \mathrm{J}\) of work. What is the electric potential energy of the pair of charges when the second charge is at point \(b\) ?
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