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Chapter 22: Problem 23

A point charge, \(q=4.00 \cdot 10^{-9} \mathrm{C},\) is placed on the \(x\) -axis at the origin. What is the electric field produced at \(x=25.0 \mathrm{~cm} ?\)

Short Answer

Expert verified
Answer: The electric field produced by the point charge at a distance of 25.0 cm is 361.9 N/C.

Step by step solution

01

Identify given variables

We are given the following information: - Point charge, \(q = 4.00 \cdot 10^{-9} \mathrm{C}\) - Distance from the point charge, \(x = 25.0 \mathrm{cm}\), which we need to convert to meters: \(x = 0.25 \mathrm{m}\) Our goal is to find the electric field, \(E\), at this distance.
02

Use Coulomb's Law to find the electric field

Coulomb's Law for the electric field produced by a point charge is given by the formula: \(E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}\) Where: - \(E\) is the electric field - \(\epsilon_0\) is the vacuum permittivity (electric constant), with a value of \(8.85 \cdot 10^{-12} \mathrm{C^2/N} \cdot \mathrm{m^2}\), - \(q\) is the charge of the point charge - \(r\) is the distance from the point charge to the point where we want to find the electric field. In this exercise, we will use \(x\) instead of \(r\) for distance.
03

Substitute the given values and calculate the electric field

We will now plug in the given values to find the electric field at the desired \(x\): \(E = \frac{1}{4\pi(8.85 \cdot 10^{-12}\,\mathrm{C^2/N} \cdot \mathrm{m^2})} \cdot \frac{4.00 \cdot 10^{-9}\,\mathrm{C}}{(0.25\, \mathrm{m})^2}\) Calculate the electric field: \(E = 361.9\,\mathrm{N/C}\)
04

Write the final answer

The electric field produced by the point charge at \(x = 25.0\,\mathrm{cm}\) is \(361.9\,\mathrm{N/C}\).

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Most popular questions from this chapter

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