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Chapter 17: Problem 106

An alpha particle (helium nucleus, charge \(+2 e\) ) starts from rest and travels a distance of \(1.0 \mathrm{cm}\) under the influence of a uniform electric field of magnitude \(10.0 \mathrm{kV} / \mathrm{m}\) What is the final kinetic energy of the alpha particle?

Short Answer

Expert verified
Answer: The final kinetic energy of the alpha particle is 3.2 × 10^{-17} J.

Step by step solution

01

Find the force on the alpha particle due to the electric field

The force on a charged particle in an electric field is given by the equation: F = qE where F is the force, q is the charge of the particle and E is the electric field strength. Since the charge of an alpha particle is +2e and the electric field strength is 10.0 kV/m, we have: F = (2e)(10.0 kV/m) = (2e)(10,000 N/C)
02

Find the work done by the force to move the alpha particle

The work done by a constant force on an object is given by the equation: W = Fd where W is the work done, F is the force and d is the distance moved. Using the force obtained in step 1 and the given distance of 1.0 cm, we calculate the work done: W = (2e)(10,000 N/C)(0.01 m) = (2e)(100 J/C)
03

Use the work-energy theorem to find the final kinetic energy

The work-energy theorem states that the work done on an object is equal to the change in kinetic energy: W = ΔK Since the alpha particle starts from rest, its initial kinetic energy is zero. Therefore, the work done on the alpha particle is equal to its final kinetic energy: ΔK = (2e)(100 J/C)
04

Calculate the final kinetic energy of the alpha particle

Now we will plug in the value of the fundamental charge e to find the final kinetic energy of the alpha particle: ΔK = (2)(1.6 × 10^{-19} C)(100 J/C) = 3.2 × 10^{-17} J So, the final kinetic energy of the alpha particle is 3.2 × 10^{-17} J.

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Most popular questions from this chapter

The electric potential at a distance of \(20.0 \mathrm{cm}\) from a point charge is \(+1.0 \mathrm{kV}\) (assuming \(V=0\) at infinity). (a) Is the point charge positive or negative? (b) At what distance is the potential $+2.0 \mathrm{kV} ?$
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?
Two metal spheres have charges of equal magnitude, $3.2 \times 10^{-14} \mathrm{C},$ but opposite sign. If the potential difference between the two spheres is \(4.0 \mathrm{mV},\) what is the capacitance? [Hint: The "plates" are not parallel, but the definition of capacitance holds.]
A 4.00 - \(\mu \mathrm{F}\) air gap capacitor is connected to a \(100.0-\mathrm{V}\) battery until the capacitor is fully charged. The battery is removed and then a dielectric of dielectric constant 6.0 is inserted between the plates without allowing any charge to leak off the plates. (a) Find the energy stored in the capacitor before and after the dielectric is inserted. [Hint: First find the new capacitance and potential difference.] (b) Does an external agent have to do positive work to insert the dielectric or to remove the dielectric? Explain.
The potential difference across a cell membrane is \(-90 \mathrm{mV} .\) If the membrane's thickness is \(10 \mathrm{nm},\) what is the magnitude of the electric field in the membrane? Assume the field is uniform.
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