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BIO Electric Eels. 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
Connect cells in series; 5000 cells are needed to achieve 500 V.

Step by step solution

01

Understanding Electric Potential

The electric potential generated by a cell is 0.10 V. To achieve a potential greater than this, a specific connection of cells is required.
02

Determine Connection Type for Increased Potential

To achieve a higher voltage than an individual cell, the cells should be connected in series. In a series connection, the total potential difference is the sum of the potential differences of each cell.
03

Calculate Number of Cells Needed for 500 V

We need the total voltage to be 500 V. With cells connected in series, the voltage adds up linearly. Hence, if each cell provides 0.10 V, dividing the required total potential by the potential per cell gives the number of cells needed. So, we have:\[\text{Number of cells} = \frac{500\ \text{V}}{0.10\ \text{V/cell}} = 5000.\]

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Key Concepts

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

Electric Potential
Electric potential is a crucial concept in understanding how electric eels generate shocks. It is the amount of electric energy per unit charge at a point in space. The electric potential difference between two points, measured in volts, determines how electric charges move; from higher potential to lower potential areas.
In electric eels, special cells work like small batteries. If each cell generates 0.10 V, it means that the electric potential at one end of the cell is 0.10 volts higher than at the other end. This slight difference sums up to create the eel's stunning discharge. By organizing these cells cleverly, electric eels can generate large voltage surges to protect themselves and capture prey.
Series Connection
To increase electric potential, multiple cells need to be connected in series. In a series connection, the voltage of each cell adds up to the total voltage. This is akin to connecting several batteries end to end, making the total potential difference the sum of each individual battery.
  • If a single cell has a potential of 0.10 V, and you connect two in series, the combined potential becomes 0.20 V.
  • This process continues as more cells are added in series, each time increasing the total voltage.
  • Such a configuration does not increase electric current capacity but increases the voltage.
Electric eels benefit from this setup because they can effectively stack the small voltages from numerous cells to generate significant electric shocks, enough to achieve their protective and hunting purposes.
Capacitors
Capacitors are devices that store electric charge and energy. In the case of electric eels, cells behaving like capacitors quickly store and discharge large energy amounts. This ability is critical since it allows eels to deliver a sudden and powerful voltage surge to incapacitate prey or defend themselves.
In electrical terms, a capacitor consists of two conductors separated by an insulator. When connected to a battery or other charge source, it accumulates charge, creating an electric field and storing energy in that field. When needed, capacitors rapidly discharge the stored energy. This makes them perfect for electric eels, requiring a brief but potent burst of electricity at will.
Understanding capacitors helps explain why electric eels need multiple cells connected in series to produce higher voltages, effectively allowing them to operate as natural, efficient energy-storing batteries.

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Most popular questions from this chapter

An air capacitor is made from two flat parallel plates 1.50 \(\mathrm{mm}\) apart. The magnitude of charge on each plate is 0.0180\(\mu \mathrm{C}\) when the potential difference is 200 \(\mathrm{V}\) . (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of \(3.0 \times 10^{6} \mathrm{V} / \mathrm{m} .\) ) (d) When the charge is \(0.0180 \mu \mathrm{C},\) what total energy is stored?

A parallel-plate air capacitor of capacitance 245 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?

A 10.0 -\muF 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?

You are working on an electronics project requiring a variety of capacitors, but you have only a large supply of 100 -nF capacitors available. Show how you can connect these capacitors to produce each of the following equivalent capacitances: (a) 50 \(\mathrm{nF}\) ; (b) \(450 \mathrm{nF} ;\) (c) \(25 \mathrm{nF} ;\) (d) 75 \(\mathrm{nF}\) .

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

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