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Chapter 17: Problem 36

It is believed that a large electric fish known as Torpedo occidentalis uses electricity to shock its victims. A typical fish can deliver a potential difference of \(0.20 \mathrm{kV}\) for a duration of \(1.5 \mathrm{ms}\). This pulse delivers charge at a rate of \(18 \mathrm{C} / \mathrm{s} .\) (a) What is the rate at which work is done by the electric organs during a pulse? (b) What is the total amount of work done during one pulse?

Short Answer

Expert verified
Question: An electric fish can deliver an electric shock with a potential difference of 0.20 kV and a rate of 18 Coulombs per second. The pulse lasts for 1.5 ms. Calculate (a) the rate at which work is done by the electric organs during a pulse and (b) the total amount of work done during one pulse. Answer: (a) The rate at which work is done during a pulse is 3600 watts. (b) The total amount of work done during one pulse is 5.4 joules.

Step by step solution

01

Determine the electric current during the pulse

To find the electric current, we will use the formula for current: \(I = \frac{Q}{t}\). The given rate is 18 Coulombs per second, and we are given a time duration of \(1.5 ms\). Rearrange the formula to solve for charge \(Q\): \(Q = I \cdot t\). \(I = \frac{18 \mathrm{C}}{1 \mathrm{s}}\) \(t = 1.5 \times 10^{-3} \mathrm{s}\) Now, substitute the given values into the equation: \(Q = (18 \mathrm{C}/\mathrm{s}) \times (1.5 \times 10^{-3} \mathrm{s})\) \(Q = 0.027 \mathrm{C}\)
02

Calculate the rate at which work is done (power)

To find the power, we can use the formula: \(P = V \cdot I\). We are given the potential difference (\(V = 0.20 \,\mathrm{kV}\)), and we have found the current \(I\) during the pulse. Now, we need to convert the potential difference to volts and plug in the values. \(V = 0.20 \,\mathrm{kV} \times 10^{3} \,\mathrm{V/kV} = 200 \,\mathrm{V}\) \(I = \frac{Q}{t} = \frac{0.027 \,\mathrm{C}}{1.5\times10^{-3} \,\mathrm{s}} = 18 \,\mathrm{A}\) \(P = V \cdot I\) \(P = (200 \,\mathrm{V}) \times (18 \,\mathrm{A})\) \(P = 3600\, \mathrm{W}\) The rate at which work is done during a pulse is 3600 watts.
03

Calculate the total amount of work done during one pulse

To find the total work done, we will use the formula: \(W = P \cdot t\). We have found the power in step 2 and we have the time duration from the given information. \(W = (3600 \,\mathrm{W}) \times (1.5 \times 10^{-3} \,\mathrm{s})\) \(W = 5.4 \,\mathrm{J}\) The total amount of work done during one pulse is 5.4 joules.

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Most popular questions from this chapter

A spherical conductor of radius \(R\) carries a total charge Q. (a) Show that the magnitude of the electric field just outside the sphere is \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the charge per unit area on the conductor's surface. (b) Construct an argument to show why the electric field at a point \(P\) just outside any conductor in electrostatic equilibrium has magnitude \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the local surface charge density. [Hint: Consider a tiny area of an arbitrary conductor and compare it to an area of the same size on a spherical conductor with the same charge density. Think about the number of field lines starting or ending on the two areas.]
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