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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 脳 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 = 螖K 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: 螖K = (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: 螖K = (2)(1.6 脳 10^{-19} C)(100 J/C) = 3.2 脳 10^{-17} J So, the final kinetic energy of the alpha particle is 3.2 脳 10^{-17} J.

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Most popular questions from this chapter

A parallel plate capacitor is connected to a battery. The space between the plates is filled with air. The electric field strength between the plates is \(20.0 \mathrm{V} / \mathrm{m} .\) Then, with the battery still connected, a slab of dielectric \((\kappa=4.0)\) is inserted between the plates. The thickness of the dielectric is half the distance between the plates. Find the electric field inside the dielectric.
A capacitor can be made from two sheets of aluminum foil separated by a sheet of waxed paper. If the sheets of aluminum are \(0.30 \mathrm{m}\) by $0.40 \mathrm{m}$ and the waxed paper, of slightly larger dimensions, is of thickness \(0.030 \mathrm{mm}\) and dielectric constant \(\kappa=2.5,\) what is the capacitance of this capacitor?

A parallel plate capacitor is attached to a battery that supplies a constant voltage. While the battery remains attached to the capacitor, the distance between the parallel plates increases by \(25 \% .\) What happens to the energy stored in the capacitor?
A 200.0 - \(\mu\) F capacitor is placed across a \(12.0-\mathrm{V}\) battery. When a switch is thrown, the battery is removed from the capacitor and the capacitor is connected across a heater that is immersed in $1.00 \mathrm{cm}^{3}$ of water. Assuming that all the energy from the capacitor is delivered to the water, what is the temperature change of the water?
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