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Chapter 22: Problem 2

One proton is accelerated from rest by a uniform electric field, another proton by a nonuniform electric field. If they move through the same potential difference, how do their final speeds compare?

Short Answer

Expert verified
The final speeds of the protons will be the same, irrespective of whether they were accelerated by a uniform or a non-uniform electric field, as long as they move through the same potential difference.

Step by step solution

01

Understanding The Physics Concepts

Understand that the electric potential difference, often simply called potential difference or voltage, is work done per unit charge. In essence, it is the energy supplied to a unit positive charge to move it from one point to another. In this case, regardless of whether the electric field is uniform or not, if the protons move through the same potential difference, it means they have been supplied the same amount of energy. Kinetic energy, on the other hand, is the energy that a particle possesses due to its motion, calculated as (1/2)mv^2, where m is the mass and v is speed.
02

Relating Potential Difference to the Final Speed

Relate the potential difference to the kinetic energy the protons acquire. The energy provided to the proton by the potential difference will equal the final kinetic energy of the protons, assuming they start from rest. If the protons move through the same potential difference, they will have the same kinetic energy.
03

Comparing the Final Speeds

Understand that since both protons will have the same kinetic energy, and kinetic energy is defined as (1/2)mv^2, and the mass m of the protons is the same, it follows that their speeds v, must be equal too. So, despite the electric fields being different (uniform and non-uniform), the final speeds of the protons will be the same.

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Most popular questions from this chapter

A thin rod of length \(L\) carries charge \(Q\) distributed uniformly over its length. (a) Show that the potential in the plane that perpendicularly bisects the rod is given by $$ V(r)=\frac{2 k Q}{L} \ln \left[\frac{L}{2 r}+\sqrt{1+\frac{L^{2}}{4 r^{2}}}\right] $$ where \(r\) is the perpendicular distance from the rod center and where the zero of potential is taken at infinity. (b) Show that this expression reduces to an expected result when \(r \gg L\). (Hint: See Appendix A for a series expansion of the logarithm.)

A disk of radius \(a\) carries nonuniform surface charge density \(\sigma=\sigma_{0}(r / a),\) where \(\sigma_{0}\) is a constant. (a) Find the potential at an arbitrary point \(x\) on the disk axis, where \(x=0\) is the disk center. (b) Use the result of (a) to find the electric field on the disk axis, and (c) show that the field reduces to an expected form for \(x \gg a\).

"Cherry picker" trucks for working on power lines often carry electrocution hazard signs. Explain how this hazard arises and why it might be more of a danger to someone on the ground than to a worker on the truck.

Proton-beam therapy can be preferable to X rays for cancer treatment (although much more expensive) because protons deliver most of their energy to the tumor, with less damage to healthy tissue. A cyclotron used to accelerate protons for cancer treatment repeatedly passes the protons through a 15 -kV potential difference. (a) How many passes are needed to bring the protons' kinetic energy to \(1.2 \times 10^{-11} \mathrm{J} ?\) (b) What's that energy in eV?

Can equipotential surfaces intersect? Explain.
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