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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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