/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Research suggests that the elect... [FREE SOLUTION] | 91Ó°ÊÓ

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Research suggests that the electric fields in some thunderstorm clouds can be on the order of \(10.0 \mathrm{kN} / \mathrm{C}\). Calculate the magnitude of the electric force acting on a particle with two excess electrons in the presence of a \(10.0-\mathrm{kN} / \mathrm{C}\) field.

Short Answer

Expert verified
Answer: The magnitude of the electric force acting on the particle is approximately 3.2 × 10^{-15} N.

Step by step solution

01

Identify the given values

In this problem, we are given: - Electric field strength E: 10.0 kN/C - Number of excess electrons: 2
02

Convert the electric field strength to standard units

The electric field strength is given in kN/C. To convert it into the standard unit of N/C, we will multiply by 1000: E = 10.0 kN/C × 1000 = 10,000 N/C
03

Determine the charge of the particle

Since the particle has two excess electrons, we will calculate the charge of the particle by multiplying the charge of one electron by the number of excess electrons. The charge of one electron (e) is approximately -1.6 × 10^{-19} C: Charge (q) = Number of excess electrons × Charge of one electron = 2 × (-1.6 × 10^{-19} C)
04

Calculate the electric force

The electric force (F) acting on the particle is the product of the electric field strength (E) and the charge of the particle (q): F = E × q Substitute the values of E and q from the previous steps: F = (10,000 N/C) × (2 × -1.6 × 10^{-19} C) Now, perform the calculation to obtain the electric force: F ≈ -3.2 × 10^{-15} N The negative sign indicates that the force is in the opposite direction of the electric field. In conclusion, the magnitude of the electric force acting on the particle with two excess electrons in the presence of a 10.0 kN/C electric field is approximately 3.2 × 10^{-15} N.

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Most popular questions from this chapter

The electric flux through a spherical Gaussian surface of radius \(R\) centered on a charge \(Q\) is \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right) .\) What is the electric flux through a cubic Gaussian surface of side \(R\) centered on the same charge \(Q ?\) a) less than \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\) b) more than \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\) c) equal to \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\) d) cannot be determined from the information given

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