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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

Expert verified
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 point charge, \(+Q\), is located on the \(x\) -axis at \(x=a\), and a second point charge, \(-Q\), is located on the \(x\) -axis at \(x=-a\). A Gaussian surface with radius \(r=2 a\) is centered at the origin. The flux through this Gaussian surface is a) zero. b) greater than zero. c) less than zero. d) none of the above.

Consider a hollow spherical conductor with total charge \(+5 e\). The outer and inner radii are \(a\) and \(b\), respectively. (a) Calculate the charge on the sphere's inner and outer surfaces if a charge of \(-3 e\) is placed at the center of the sphere. (b) What is the total net charge of the sphere?

A charge per unit length \(+\lambda\) is uniformly distributed along the positive \(y\) -axis from \(y=0\) to \(y=+a\). A charge per unit length \(-\lambda\) is uniformly distributed along the negative \(y\) axis from \(y=0\) to \(y=-a\). Write an expression for the electric field (magnitude and direction) at a point on the \(x\) -axis a distance \(x\) from the origin.

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.
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