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Chapter 22: Problem 41

An electron is observed traveling at a speed of \(27.5 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) parallel to an electric field of magnitude \(11,400 \mathrm{~N} / \mathrm{C}\) How far will the electron travel before coming to a stop?

Short Answer

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Question: An electron with an initial speed of 27.5 million meters per second travels parallel to a uniform electric field of 11,400 N/C. Find the distance the electron travels before coming to a stop. Answer: The electron will travel approximately 0.000189 meters (189 micrometers) before coming to a stop.

Step by step solution

01

Find the force experienced by the electron

We are given the magnitude of the electric field, \(E=11,400 \mathrm{~N}/\mathrm{C}\), and the charge of an electron \(q=-1.6\times 10^{-19} \mathrm{~C}\). Let's calculate the force using the formula \(F=qE\): \(F=(-1.6\times 10^{-19}\mathrm{~C})(11,400\mathrm{~N}/\mathrm{C})=-1.824\times10^{-16} \mathrm{~N}\).
02

Calculate the acceleration

To find acceleration, we can use Newton's second law of motion, which states that \(F = ma\), where \(m\) is the mass of the electron, which we know is \(9.11 \times 10^{-31} \mathrm{~kg}\). So, the acceleration is: \(a = \frac{F}{m} =\frac{-1.824\times10^{-16}\mathrm{~N}}{9.11\times 10^{-31}\mathrm{~kg}} =-2\times10^{14} \mathrm{~m/s^2}\) (Note: The negative sign indicates acceleration in the opposite direction of motion)
03

Calculate the distance traveled before coming to a stop

To find the distance traveled before the electron comes to a stop, we can use the following equation of motion: \(v^2 = u^2 + 2as\). Since the final velocity is 0, the equation becomes: \(0^2 = (27.5 \cdot 10^{6} \mathrm{~m/s})^2 + 2(-2\times10^{14} \mathrm{~m/s^2})(s)\). Solving for \(s\): \(s = \frac{(27.5 \cdot 10^{6} \mathrm{~m/s})^2}{2\times 2 \times 10^{14} \mathrm{~m/s^2}} = \frac{7.56 \times 10^{13} \mathrm{~m^2/s^2}}{4 \times 10^{14} \mathrm{~m/s^2}} = 0.000189\mathrm{~m}\). So, the electron will travel approximately 0.000189 meters(189 micrometers) before coming to a stop.

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Most popular questions from this chapter

At which of the following locations is the electric field the strongest? a) a point \(1 \mathrm{~m}\) from a \(1 \mathrm{C}\) point charge b) a point \(1 \mathrm{~m}\) (perpendicular distance) from the center of a \(1-\mathrm{m}\) -long wire with \(1 \mathrm{C}\) of charge distributed on it c) a point \(1 \mathrm{~m}\) (perpendicular distance) from the center of a \(1-\mathrm{m}^{2}\) sheet of charge with \(1 \mathrm{C}\) of charge distributed on it d) a point \(1 \mathrm{~m}\) from the surface of a charged spherical shell of charge \(1 \mathrm{C}\) with a radius of \(1 \mathrm{~m}\) e) a point \(1 \mathrm{~m}\) from the surface of a charged spherical shell of charge \(1 \mathrm{C}\) with a radius of \(0.5 \mathrm{~m}\)

A solid, nonconducting sphere of radius \(a\) has total charge \(Q\) and a uniform charge distribution. Using Gauss's Law, determine the electric field (as a vector) in the regions \(ra\) in terms of \(Q\).

There is a uniform charge distribution of \(\lambda=\) \(8.00 \cdot 10^{-8} \mathrm{C} / \mathrm{m}\) along a thin wire of length \(L=6.00 \mathrm{~cm}\) The wire is then curved into a semicircle that is centered about the origin, so the radius of the semicircle is \(R=L / \pi .\) Find the magnitude of the electric field at the center of the semicircle.

A dipole is completely enclosed by a spherical surface. Describe how the total electric flux through this surface varies with the strength of the dipole.

Electric dipole moments of molecules are often measured in debyes \((\mathrm{D}),\) where \(1 \mathrm{D}=3.34 \cdot 10^{-30} \mathrm{C} \mathrm{m} .\) For instance, the dipole moment of hydrogen chloride gas molecules is \(1.05 \mathrm{D}\). Calculate the maximum torque such a molecule can experience in the presence of an electric field of magnitude \(160.0 \mathrm{~N} / \mathrm{C}\).
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