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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 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 V? (b) Using the connection in part (a), how many cells must be connected together to produce the 500-V surge of the electric eel?

Short Answer

Expert verified
Connect cells in series and use 5000 cells for 500 V.

Step by step solution

01

Understanding the Problem

We need to determine how to connect the cells to create a greater potential difference than 0.10 V with each cell modeled as a capacitor producing 0.10 V. We aim for a 500 V total in Part (b).
02

Type of Connection

To increase the total voltage output above 0.10 V, the cells must be connected in series. In a series connection, the voltages add up cumulatively.
03

Calculate Number of Cells Needed

To find the number of cells needed to reach a total of 500 V, divide the desired voltage by the voltage of one cell: \( N = \frac{500 \text{ V}}{0.10 \text{ V}} \).
04

Performing the Division

Calculate the number of cells: \( N = \frac{500 \text{ V}}{0.10 \text{ V}} = 5000 \text{ cells} \).
05

Conclusion

Therefore, 5000 cells need to be connected in series to achieve the 500 V required for an electric eel's surge.

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Key Concepts

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

Series Connection
In an electric circuit, a series connection is where components are arranged in a single path. Every part is interconnected, forming a loop. When you connect electrical components, such as capacitors, in series, the electric current must flow through each component one after the other. This results in the voltages across each component adding together.
For example:
  • If you have three capacitors in series, each contributing 0.10 V, the total voltage would be 0.10 V + 0.10 V + 0.10 V = 0.30 V.
  • Series connections are particularly useful when you need to increase the voltage supplied by components like batteries or capacitors.
In our case, connecting the cells in a series allows us to achieve a higher potential difference than each cell can provide individually. Therefore, to achieve the high voltages electric eels generate, cells must be connected in series.
Voltage Calculation
Voltage calculation in a series connection involves adding the voltages of each component. Understanding this principle is key when determining the total voltage achievable with several cells or units.
The formula is simple: - In a series connection, the total voltage, Vtotal, is the sum of the individual voltages: \[ V_{total} = V_1 + V_2 + ext{...} + V_n \] - For our exercise, the calculation involves determining how many 0.10 V cells we need to sum to reach 500 V.
Following the formula, you divide the target voltage by the voltage of one cell:\[ N = \frac{V_{target}}{V_{cell}} = \frac{500\text{ V}}{0.10\text{ V}} = 5000 \text{ cells} \] Therefore, we need 5000 cells to reach the desired total of 500 V, showing how effective a series connection is for achieving high voltage levels.
Capacitor Model
A capacitor is an electrical component that can store and release electrical energy. In our exercise, cells that make up the electric eels are modeled as capacitors. Each capacitor provides a specific voltage, in this case, 0.10 V.
Some key points about capacitors:
  • They store energy in an electric field created between a pair of conductive plates.
  • In this context, capacitors are used to model the potential difference each cell in an electric eel can generate.
  • Capacitors are crucial in circuits where storage and release of energy at certain voltages is required.
When you connect capacitors in series, the total voltage across them becomes the sum of the voltages across each capacitor. Therefore, modeling the eel's cells as capacitors helps us understand and visualize how these cells work to produce such high voltages.
Electric Potential
Electric potential, often referred to as voltage, represents the potential energy per unit charge at a point in an electric field. It's the driving force that pushes charge through a circuit, allowing electrical devices to function.

Some essential aspects include:
  • Electric potential is measured in volts (V).
  • In a circuit, the electric potential determines the amount of work needed to move a charge between two points.
  • The cells in electrical eels generate electric potential to produce powerful surges.
In the context of electric eels, the cells produce a small voltage, 0.10 V, but when connected in a series, they can accumulate to the high electric potentials observed, like the 500 V surge mentioned.
This high electric potential is crucial for the eel's ability to stun its prey, illustrating the practical application of these circuits in nature.

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Most popular questions from this chapter

A cylindrical air capacitor of length 15.0 m stores 3.20 \(\times\) 10\(^{-9}\) J of energy when the potential difference between the two conductors is 4.00 V. (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.

Cell membranes (the walled enclosure around a cell) are typically about 7.5 nm thick. They are partially permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside and outside faces of such a membrane, and these charges prevent additional charges from passing through the cell wall. We can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organic material to give the membrane a dielectric constant of about 10. (See \(\textbf{Fig. P24.48}\).) (a) What is the capacitance per square centimeter of such a cell wall? (b) In its normal resting state, a cell has a potential difference of 85 mV across its membrane. What is the electric field inside this membrane?

A 10.0- \(\mu\)F parallel-plate capacitor with circular plates is connected to a 12.0-V battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the 12.0-V battery after the radius of each plate was doubled without changing their separation?

A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few sheets of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 mm. (a) If the sheets of paper measure 22 \(\times\)28 cm and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of 12.0 mm, instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 nF of capacitance? (c) Suppose she goes high-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Teflon than of posterboard? Explain.

Three capacitors having capacitances of 8.4, 8.4, and 4.2 \(\mu\)F are connected in series across a 36-V potential difference. (a) What is the charge on the 4.2-\(\mu\)F capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?

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