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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 capacitor consists of a solid inner conducting core with radius 0.250 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 cm. The capacitance is 36.7 pF. (a) Calculate the inner radius of the hollow tube. (b) When the capacitor is charged to 125 V, \(\textbf{what is the charge per unit length}\) \(\lambda\) on the capacitor?

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \({1 \over 675}\) s with an average light power output of 2.70 \(\times\) 10\(^5\) W. (a) If the conversion of electrical energy to light is 95% efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value calculated in part (a). What is the capacitance?

A 5.00\(\mu\)F parallel-plate capacitor is connected to a 12.0-V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. (a) A voltmeter is connected across the two plates without discharging them. What does it read? (b) What would the voltmeter read if (i) the plate separation were doubled; (ii) the radius of each plate were doubled but their separation was unchanged?

Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm\) 0.50 \(\times\) 10\(^{-3}\) C/m\(^2\), the cell wall is 5.0 nm thick, and the cell- wall material is air. (a) Find the magnitude of \(\vec{E}\) S in the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) A typical cell in the human body has a volume of 10\(^{-16}\) m\(^3\). Estimate the total electric-field energy stored in the wall of a cell of this size. (\(Hint\): Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of 5.4. Repeat parts (a) and (b) in this case.

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 12.5 cm, and the outer sphere has radius 14.8 cm. A potential difference of 120 V is applied to the capacitor. (a) What is the energy density at \(r\) = 12.6 cm, just outside the inner sphere? (b) What is the energy density at \(r\) = 14.7 cm, just inside the outer sphere? (c) For a parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges of the plates. Is this also true for a spherical capacitor?

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