/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Electric eels and electric fish ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Electric eels and electric fish generate large potential differences that are used to stun enemies and prey. These potentials are produced by cells that each can generate \(0.10 \mathrm{~V}\). We can plausibly model such cells as charged capacitors. (a) How should these cells be connected (in series or in parallel) to produce a total potential of more than \(0.10 \mathrm{~V} ?\) (b) Using the connection in part (a), how many cells must be connected together to produce the \(500 \mathrm{~V}\) surge of the electric eel?

Short Answer

Expert verified
The cells should be connected in series to generate a voltage greater than 0.10 V. To produce 500 V, 5000 cells need to be connected together in series.

Step by step solution

01

Identify the type of connection to increase voltage

To increase the total voltage, the cells (capacitors) should be connected in series. In a series connection, the total voltage is the sum of the voltages each cell can generate. Therefore, the cells must be connected in series to produce a total potential of more than 0.10 V.
02

Calculate the number of cells needed to produce 500 V

Each cell connected in series can generate 0.10 V. So, to generate a total voltage of 500 V, we divide the desired total voltage by the voltage per cell: \( \frac{500 V}{0.10 V/cell} = 5000 \) cells. Thus, 5000 cells must be connected in series to produce the 500 V surge of the electric eel.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Capacitors in Series
When dealing with electronic components, understanding how they are connected is crucial for predicting their behavior in a circuit. Capacitors, for instance, can be connected in various ways, each altering their collective properties. One such method of connection is 'in series.'

In series connection, the capacitors are arranged end to end, so that the charge on each capacitor is the same. To visualize this, imagine a lineup of people holding balloons; each balloon is the 'capacitor.' Now, if one person represents a unit charge, they can only pass through each balloon in sequence, paralleling the flow of charge in series capacitors.

Due to this configuration, the overall capacitance (the capacitors' ability to store charge) actually decreases, but the electric potential difference, commonly known as voltage, adds up. This is why when you want to increase the total voltage produced by a group of capacitors, such as bioelectric cells in an electric eel, you would connect them in series. This is exactly the principle leveraged by nature's own electrical engineers, our electrifying friends, the eels.
Electric Eel Bioelectricity
The electric eel showcases one of the most fascinating adaptations in the animal kingdom: its ability to generate significant amounts of bioelectricity. The cells responsible for this remarkable feat are known as electrocytes. These special cells generate electricity much like a capacitor in a circuit.

How does it work? Within the electric eel's body, thousands of electrocytes are lined up, akin to capacitors in series. Each can produce a meager 0.10 V, seemingly insignificant alone, yet formidable collectively. The cumulative effect of these cells allows for a defensive and hunting mechanism that can produce a 500 V surge to stun the eel’s unfortunate adversaries.

Thinking of these cells as a ‘battery’ composed of many small capacitors, we can appreciate how biology mirrors the engineering concepts we use to design electrical circuits. It's a natural example of how the series arrangement of capacitive elements leads to a higher electric potential difference.
Voltage Calculation
Calculating voltage is a fundamental task in understanding and designing electrical circuits. Voltage, defined as the electric potential difference between two points, is a measure indicative of the potential energy per unit charge that's available for doing work, like moving a current in a circuit.

When performing a voltage calculation for capacitors in series, one must simply sum the voltage contributions of each individual capacitor. Here's the math in simple terms: if you have a string of lights and each light adds a certain sparkle, to know how much twinkle the entire string provides, you'd add up the sparkle from each individual light.

Therefore, with each eel’s bioelectric cell producing 0.10 V, a straightforward division tells us how many cells we need to stack together to achieve a given high voltage. To get the desired 500 V, for example, divide the total needed voltage by the voltage each cell contributes — which neatly tells us that 5000 capacitors (or electrocytes) would be necessary to produce such an electrifying result.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A parallel-plate air capacitor of capacitance \(245 \mathrm{pF}\) has a charge of magnitude \(0.148 \mu \mathrm{C}\) on each plate. The plates are \(0.328 \mathrm{~mm}\) apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electric-field magnitude between the plates? (d) What is the surface charge density on each plate?

Two air-filled parallel-plate capacitors with capacitances \(C_{1}\) and \(C_{2}\) are connected in series to a battery that has voltage \(V\) \(C_{1}=3.00 \mu \mathrm{F}\) and \(C_{2}=6.00 \mu \mathrm{F}\). The electric field between the plates of capacitor \(C_{2}\) is \(E_{02}\). While the two capacitors remain connected to the battery, a dielectric with dielectric constant \(K=4\) is inserted between the plates of capacitor \(C_{1}\), completely filling the space between them. After the dielectric is inserted in \(C_{1},\) the electric field between the plates of capacitor \(C_{2}\) is \(E_{2}\). (a) What is the ratio \(E_{2} / E_{02}\) ? When the dielectric is inserted into \(C_{1},\) does the electric field in \(C_{2}\) increase, decrease, or remain the same? (b) Repeat the calculation in part (a) for the two capacitors connected to the battery in parallel.

When lightning strikes a car, the metallic outer shell, which is insulated from the ground by its rubber tires, attains a high voltage. We can estimate how much charge is deposited by roughly modeling the car as a spherical capacitor with the outer radius taken to infinity. (a) Determine the capacitance of a sphere in terms of its radius, either by considering the potential on a sphere relative to infinity as a function of its charge, or by considering a spherical capacitor as the outer shell becomes very large. (These methods provide the same result.) (b) Estimate the radius of a sphere that corresponds to the size of a car. (c) Determine the corresponding capacitance. (d) A typical lighting strike could deliver a \(100 \mathrm{MV}\) potential to a car. Then what net charge would be deposited on the car?

\(\mathrm{A} 12.5 \mu \mathrm{F}\) capacitor is connected to a power supply that keeps a constant potential difference of \(24.0 \mathrm{~V}\) across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of \(1.60 \times 10^{7} \mathrm{~V} / \mathrm{m} .\) The capacitor is to have a capacitance of \(1.25 \times 10^{-9} \mathrm{~F}\) and must be able to withstand a maximum potential difference of \(5500 \mathrm{~V}\). What is the minimum area the plates of the capacitor may have?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.