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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

A spherical conductor of radius \(R\) carries a total charge Q. (a) Show that the magnitude of the electric field just outside the sphere is \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the charge per unit area on the conductor's surface. (b) Construct an argument to show why the electric field at a point \(P\) just outside any conductor in electrostatic equilibrium has magnitude \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the local surface charge density. [Hint: Consider a tiny area of an arbitrary conductor and compare it to an area of the same size on a spherical conductor with the same charge density. Think about the number of field lines starting or ending on the two areas.]
Draw some electric field lines and a few equipotential surfaces outside a negatively charged hollow conducting sphere. What shape are the equipotential surfaces?
The bottom of a thundercloud is at a potential of $-1.00 \times 10^{8} \mathrm{V}\( with respect to Earth's surface. If a charge of \)-25.0 \mathrm{C}$ is transferred to the Earth during a lightning strike, find the electric potential energy released. (Assume that the system acts like a capacitor-as charge flows, the potential difference decreases to zero.)
Before a lightning strike can occur, the breakdown limit for damp air must be reached. If this occurs for an electric field of $3.33 \times 10^{5} \mathrm{V} / \mathrm{m},$ what is the maximum possible height above the Earth for the bottom of a thundercloud, which is at a potential $1.00 \times 10^{8} \mathrm{V}$ below Earth's surface potential, if there is to be a lightning strike?
(a) Calculate the capacitance per unit length of an axon of radius $5.0 \mu \mathrm{m} \text { (see Fig. } 17.14) .$ The membrane acts as an insulator between the conducting fluids inside and outside the neuron. The membrane is 6.0 nm thick and has a dielectric constant of \(7.0 .\) (Note: The membrane is thin compared with the radius of the axon, so the axon can be treated as a parallel plate capacitor.) (b) In its resting state (no signal being transmitted), the potential of the fluid inside is about \(85 \mathrm{mV}\) lower than the outside. Therefore, there must be small net charges \(\pm Q\) on either side of the membrane. Which side has positive charge? What is the magnitude of the charge density on the surfaces of the membrane?
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