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Chapter 22: Problem 38

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

A solid nonconducting sphere of radius \(a\) has a total charge \(+Q\) uniformly distributed throughout its volume. The surface of the sphere is coated with a very thin (negligible thickness) conducting layer of gold. A total charge of \(-2 Q\) is placed on this conducting layer. Use Gauss's Law to do the following. a) Find the electric field \(E(r)\) for \(ra\) (outside the coated sphere, beyond the sphere and the gold layer). c) Sketch the graph of \(E(r)\) versus \(r\). Comment on the continuity or discontinuity of the electric field, and relate this to the surface charge distribution on the gold layer.

A charge of \(+2 q\) is placed at the center of an uncharged conducting shell. What will be the charges on the inner and outer surfaces of the shell, respectively? a) \(-2 q,+2 q\) b) \(-q,+q\) c) \(-2 q,-2 q\) d) \(-2 q,+4 q\)

A thin glass rod is bent into a semicircle of radius \(R\). A charge \(+Q\) is uniformly distributed along the upper half, and a charge \(-Q\) is uniformly distributed along the lower half as shown in the figure. Find the magnitude and direction of the electric field \(\vec{E}\) (in component form) at point \(P\), the center of the semicircle.

Two parallel, infinite, nonconducting plates are \(10.0 \mathrm{~cm}\) apart and have charge distributions of \(+1.00 \mu \mathrm{C} / \mathrm{m}^{2}\) and \(-1.00 \mu \mathrm{C} / \mathrm{m}^{2} .\) What is the force on an electron in the space between the plates? What is the force on an electron located outside the two plates near the surface of one of the two plates?

Two uniformly charged insulating rods are bent in a semicircular shape with radius \(r=10.0 \mathrm{~cm} .\) If they are positioned so they form a circle but do not touch and have opposite charges of \(+1.00 \mu \mathrm{C}\) and \(-1.00 \mu \mathrm{C}\) find the magnitude and direction of the electric field at the center of the composite circular charge configuration.
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