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Chapter 17: Problem 86

The bottom of a thundercloud is at a potential of $-1.00 \times 10^{8} \mathrm{V}\( with respect to Earth's surface. If a charge of \)-25.0 \mathrm{C}$ is transferred to the Earth during a lightning strike, find the electric potential energy released. (Assume that the system acts like a capacitor-as charge flows, the potential difference decreases to zero.)

Short Answer

Expert verified
Answer: The electric potential energy released during the lightning strike is 1.25 x 10^9 J.

Step by step solution

01

Identify the relevant formula for the energy stored in a capacitor

We can use the formula for the energy stored in a capacitor to calculate the electric potential energy released: \(U = \frac{1}{2}QV\) Where: \(U\) is the electric potential energy, \(Q\) is the charge transferred during the lightning strike, and \(V\) is the initial potential difference between the thundercloud and the Earth.
02

Plug the given values into the formula

We are given \(Q = -25.0\mathrm{C}\) and \(V = -1.00 \times 10^{8}\mathrm{V}\). We can plug these values into the formula: \(U = \frac{1}{2}(-25.0\mathrm{C})(-1.00 \times 10^{8}\mathrm{V})\)
03

Calculate the electric potential energy released

Simplify and solve for \(U\): \(U = \frac{1}{2}(25.0\mathrm{C})(1.00 \times 10^{8}\mathrm{V})\) \(U = 12.5\mathrm{C} \times 10^{8}\mathrm{V}\) \(U = 1.25 \times 10^{9}\mathrm{J}\)
04

Write the final answer

The electric potential energy released during the lightning strike is \(1.25 \times 10^{9}\mathrm{J}\).

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Most popular questions from this chapter

A parallel plate capacitor is connected to a \(12-\mathrm{V}\) battery. While the battery remains connected, the plates are pushed together so the spacing is decreased. What is the effect on (a) the potential difference between the plates? (b) the electric field between the plates? (c) the magnitude of charge on the plates?
A parallel plate capacitor has a charge of \(5.5 \times 10^{-7} \mathrm{C}\) on one plate and \(-5.5 \times 10^{-7} \mathrm{C}\) on the other. The distance between the plates is increased by \(50 \%\) while the charge on each plate stays the same. What happens to the energy stored in the capacitor?
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.]
A 2.0 - \(\mu\) F capacitor is connected to a \(9.0-\mathrm{V}\) battery. What is the magnitude of the charge on each plate?
A van de Graaff generator has a metal sphere of radius \(15 \mathrm{cm} .\) To what potential can it be charged before the electric field at its surface exceeds \(3.0 \times 10^{6} \mathrm{N} / \mathrm{C}\) (which is sufficient to break down dry air and initiate a spark)?
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