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Chapter 16: Problem 48

An electron is placed in a uniform electric field of strength $232 \mathrm{N} / \mathrm{C}$. If the electron is at rest at the origin of a coordinate system at \(t=0\) and the electric field is in the positive \(x\) -direction, what are the \(x\) - and \(y\) coordinates of the electron at \(t=2.30 \mathrm{ns} ?\)

Short Answer

Expert verified
Answer: At t = 2.30 ns, the electron is at the coordinates (-1.08 × 10^-5 m, 0).

Step by step solution

01

Calculate the force acting on the electron

To calculate the force acting on the electron, we can use the formula: $$F = qE$$ where $$F$$ is the force, $$q$$ is the charge of the electron, and $$E$$ is the strength of the electric field. The charge of an electron is $$q = -1.6 \times 10^{-19} \mathrm{C}$$. Given the electric field $$E = 232 \mathrm{N/C}$$, the force acting on the electron is: $$F = -1.6 \times 10^{-19} \mathrm{C} \times 232 \mathrm{N/C} = -3.712 \times 10^{-17} \mathrm{N}$$ The negative force value indicates that the force acts in the negative x-direction, which is opposite to the positive x-direction of the electric field.
02

Calculate the acceleration of the electron

To calculate the acceleration of the electron, we can use the formula: $$a = \frac{F}{m}$$ where $$a$$ is the acceleration, $$F$$ is the force, and $$m$$ is the mass of the electron. The mass of an electron is $$m = 9.1 \times 10^{-31} \mathrm{kg}$$. Using the force calculated in Step 1, the acceleration of the electron is: $$a = \frac{-3.712 \times 10^{-17} \mathrm{N}}{9.1 \times 10^{-31} \mathrm{kg}} = -4.08 \times 10^{13} \mathrm{m/s^2}$$
03

Calculate the x-coordinate of the electron at t = 2.30 ns

We know that the electron is at rest at the origin when $$t=0$$. Thus, we can use the equation of motion: $$x = x_0 + v_0t + \frac{1}{2}at^2$$ where $$x$$ is the position at time $$t$$, $$x_0$$ is the initial position, $$v_0$$ is the initial velocity, and $$a$$ is the acceleration. The initial position and initial velocity are both zero, so the equation simplifies to: $$x = \frac{1}{2}at^2$$ Given the acceleration from Step 2 and the time $$t = 2.30\,\text{ns} = 2.30 \times 10^{-9}\,\text{s}$$, we can calculate the x-coordinate: $$x = \frac{1}{2} \times (-4.08 \times 10^{13}\,\text{m/s}^2) \times (2.30 \times 10^{-9}\,\text{s})^2 = -1.08 \times 10^{-5} \mathrm{m}$$
04

Determine the y-coordinate of the electron

Since the electric field is only in the x-direction and there is no force acting in the y-direction, the electron will not have any motion in the y-direction. Therefore, the y-coordinate remains unchanged from its initial position, which is at the origin: $$y = 0$$
05

Write down the final coordinates of the electron at t = 2.30 ns

Finally, we can write down the x- and y-coordinates of the electron at $$t = 2.30\,\text{ns}$$: $$x = -1.08 \times 10^{-5} \mathrm{m}$$ $$y = 0$$ So at $$t = 2.30\,\text{ns}$$, the electron is at the coordinates \((-1.08 \times 10^{-5}\,\text{m}, 0)\).

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Most popular questions from this chapter

Two tiny objects with equal charges of \(7.00 \mu \mathrm{C}\) are placed at the two lower corners of a square with sides of \(0.300 \mathrm{m},\) as shown. Find the electric field at point \(C\) the center of the square.
A point charge \(q_{1}=+5.0 \mu \mathrm{C}\) is fixed in place at \(x=0\) and a point charge \(q_{2}=-3.0 \mu \mathrm{C}\) is fixed at \(x=\) $-20.0 \mathrm{cm} .\( Where can we place a point charge \)q_{3}=-8.0 \mu \mathrm{C}$ so that the net electric force on \(q_{1}\) due to \(q_{2}\) and \(q_{3}\) is zero?
An electron traveling horizontally from west to east enters a region where a uniform electric field is directed upward. What is the direction of the electric force exerted on the electron once it has entered the field?
In this problem, you can show from Coulomb's law that the constant of proportionality in Gauss's law must be \(1 / \epsilon_{0} .\) Imagine a sphere with its center at a point charge q. (a) Write an expression for the electric flux in terms of the field strength \(E\) and the radius \(r\) of the sphere. [Hint: The field strength \(E\) is the same everywhere on the sphere and the field lines cross the sphere perpendicular to its surface.] (b) Use Gauss's law in the form \(\Phi_{\mathrm{E}}=c q(\text { where } c\) is the constant of proportionality) and the electric field strength given by Coulomb's law to show that $c=1 / \epsilon_{0}$
A positively charged rod is brought near two uncharged conducting spheres of the same size that are initially touching each other (diagram a). The spheres are moved apart and then the charged rod is removed (diagram b). (a) What is the sign of the net charge on sphere 1 in diagram b? (b) In comparison with the charge on sphere \(1,\) how much and what sign of charge is on sphere \(2 ?\)
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