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Chapter 17: Problem 41

An electron is accelerated from rest through a potential difference $\Delta V\(. If the electron reaches a speed of \)7.26 \times 10^{6} \mathrm{m} / \mathrm{s},$ what is the potential difference? Be sure to include the correct sign. (Does the electron move through an increase or a decrease in potential?)

Short Answer

Expert verified
Answer: The electron moved through a potential difference of approximately -1174 V, and it went through a decrease in potential.

Step by step solution

01

Write down the known values

We know the following values: - The final speed of the electron, \(v_f = 7.26 \times 10^{6} \mathrm{m} / \mathrm{s}\) - The initial speed of the electron, \(v_i = 0 \mathrm{m} / \mathrm{s}\) (since it starts from rest) - The mass of the electron, \(m_e = 9.11 \times 10^{-31} \mathrm{kg}\) - The elementary charge (charge of an electron), \(e = -1.6 \times 10^{-19} \mathrm{C}\)
02

Use the work-energy principle

Since the electron is accelerated by an electric field, we can use the work-energy principle, which states that the work done (\(W\)) on an object equals the change in kinetic energy: $$W = \Delta KE$$ Since the work is done by the electric field, we can relate work to the potential difference: $$W = q \Delta V$$ Where \(q\) is the charge of the electron and \(\Delta V\) is the potential difference.
03

Calculate the change in kinetic energy

The change in kinetic energy can be calculated using the formula: $$\Delta KE = \frac{1}{2}m(v_f^2 - v_i^2)$$ Plugging in the known values, we get: $$\Delta KE = \frac{1}{2}(9.11 \times 10^{-31} \mathrm{kg})(7.26 \times 10^{6} \mathrm{m} / \mathrm{s})^2$$ $$\Delta KE = 1.879 \times 10^{-17} \mathrm{J}$$
04

Calculate the potential difference

Now that we have the change in kinetic energy, we can calculate the potential difference using the equation: $$\Delta V = \frac{W}{q}$$ Replacing \(W\) with \(\Delta KE\), we have: $$\Delta V = \frac{1.879 \times 10^{-17} \mathrm{J}}{-1.6 \times 10^{-19} \mathrm{C}}$$ $$\Delta V \approx -1174 \mathrm{V}$$
05

Determine the sign of the potential difference

Since the potential difference is negative, it means that the electron went through a decrease in potential. This is because the electric field accelerated the negatively charged electron towards the higher potential side, causing a decrease in potential as it gained kinetic energy.

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Most popular questions from this chapter

Two metal spheres are separated by a distance of \(1.0 \mathrm{cm}\) and a power supply maintains a constant potential difference of \(900 \mathrm{V}\) between them. The spheres are brought closer to one another until a spark flies between them. If the dielectric strength of dry air is $3.0 \times 10^{6} \mathrm{V} / \mathrm{m},$ what is the distance between the spheres at this time?
Find the electric potential energy for the following array of charges: charge \(q_{1}=+4.0 \mu \mathrm{C}\) is located at \((x, y)=(0.0,0.0) \mathrm{m} ;\) charge \(q_{2}=+3.0 \mu \mathrm{C}\) is located at \((4.0,3.0) \mathrm{m} ;\) and charge \(q_{3}=-1.0 \mu \mathrm{C}\) is located at (0.0,3.0) \(\mathrm{m}\)
A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\) The parallel plates are separated by \(0.40 \mathrm{mm}\) of bakelite. What is the capacitance of this capacitor?
Capacitors are used in many applications where you need to supply a short burst of energy. A \(100.0-\mu \mathrm{F}\) capacitor in an electronic flash lamp supplies an average power of \(10.0 \mathrm{kW}\) to the lamp for $2.0 \mathrm{ms}$. (a) To what potential difference must the capacitor initially be charged? (b) What is its initial charge?
An electron (charge \(-e\) ) is projected horizontally into the space between two oppositely charged parallel plates. The electric field between the plates is \(500.0 \mathrm{N} / \mathrm{C}\) upward. If the vertical deflection of the electron as it leaves the plates has magnitude \(3.0 \mathrm{mm},\) how much has its kinetic energy increased due to the electric field? [Hint: First find the potential difference through which the electron moves.]
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