Velocity from electric potential

• tony873004
In summary, the final speed of a proton accelerated through a potential difference of 0.100 MV is approximately 1.39 x 10^8 m/s. This value is half the speed of light and may require the use of relativistic equations. Additionally, the energy gained by an electron accelerated through 1 volt is known as an electron volt (eV) and is equal to 1.602 x 10^-19 joules.
tony873004
Gold Member

Homework Statement

Can someone double check that this is right? I only multiplied charge of a proton by given potential difference to make the units work, without actually knowing if this is the right way to do this problem.

A proton is accelerated from rest through a potential difference of 0.100 MV. What is the final speed of the proton?

The Attempt at a Solution

$$\begin{array}{l} E_k = \frac{1}{2}mv^2 \,\,\,\, \Rightarrow \,\,\,\,v = \sqrt {\frac{{2E_k }}{m}} = \sqrt {\frac{{2 \cdot 100\,{\rm{MV}} \cdot 10^6 {\rm{V/MV}} \cdot 1.602 \cdot 10^{ - 19} {\rm{C}}}}{{1.67 \times 10^{ - 27} {\rm{ kg}}}}} \\ \\ = \sqrt {\frac{{2 \cdot 100\, \cdot 10^6 {\rm{V}} \cdot 1.602 \cdot 10^{ - 19} {\rm{C}}}}{{1.67 \times 10^{ - 27} {\rm{ kg}}}}} = \sqrt {\frac{{2 \cdot 100\, \cdot 10^6 \frac{{{\rm{kg}} \cdot m^2 }}{{\frac{C}{s}s^3 }} \cdot 1.602 \cdot 10^{ - 19} {\rm{C}}}}{{1.67 \times 10^{ - 27} {\rm{ kg}}}}} \\ = \sqrt {\frac{{2 \cdot 100\, \cdot 10^6 \frac{{{\rm{ \times m}}^2 }}{{\frac{{}}{{}} \cdot {\rm{s}}^2 }} \cdot 1.602 \cdot 10^{ - 19} }}{{1.67 \times 10^{ - 27} {\rm{ }}}}} =1.39x10^{8}m/s \\ \end{array} \$$

The final speed is rather high. This is one problem with E&M physics as opposed to mechanics. If a ball was tossed in the air and landed with this speed, I could use my intuition to tell me I did something wrong.

This velocity is half the speed of light. Wouldn't relativity come into play in these types of problems?

to get the exactly correct answer you would use the relativistic expressions, but as long as $$mc^2$$ is a lot bigger than the potential energy difference (V) you can get away with using the nonrelativistic formula. That is, you should use the relativistic
$$v=\frac{pc^2}{V+mc^2}$$
and
$$p^2 c^2 = V^2 + 2Vmc^2$$

... but if $$mc^2 >> V$$ then the first equation reduces to
$$v\approx p/m$$
and the second reduces to
$$p^2\approx 2mV$$

so...

Ever hear of the unit of energy called an electron volt?(eV?)it's the energy gained by an electron accelerated through 1 volt, so it's 1.602x10^-19 joules. So something with the charge of an electron(a proton!)through .1 MV should give .1 MeV, right?

Neato

Anyways, the point of that is yes you were right to do so. However you have an error while calculating it, it's not 1 MV, it's .1 MV don't forget

You found it for, well, one HUNDRED megavolts, accelerating a little proton through that potential difference would be pretty hefty indeed

Your v is in the region of the speed of light. If that were correct, you should be using the relativistic form, as olgranpappy said. On the other hand, it looks like your calculation is using 100MV for the potential, but the problem you quoted says 0.100MV.

Thanks for catching that. I hate losing points for dumb mistakes like that. Thanks for the info on the eV. I'll look into that.

1. What is velocity from electric potential?

Velocity from electric potential is a measure of the speed at which charged particles move in an electric field. It is directly related to the strength of the electric field and the charge of the particle.

2. How is velocity from electric potential calculated?

The velocity from electric potential can be calculated using the equation v = √(2qV/m), where v is the velocity, q is the charge of the particle, V is the electric potential, and m is the mass of the particle.

3. What is the difference between velocity from electric potential and velocity from electric field?

Velocity from electric potential and velocity from electric field are related, but not the same. Velocity from electric potential takes into account the kinetic energy of the charged particle, while velocity from electric field only considers the force exerted on the particle by the electric field.

4. How does electric potential affect velocity?

The electric potential directly affects the velocity of charged particles. A higher electric potential will result in a higher velocity, while a lower potential will result in a lower velocity.

5. Can velocity from electric potential be negative?

Yes, velocity from electric potential can be negative. This typically occurs when the particle has a negative charge and is moving in the opposite direction of the electric field. Negative velocity from electric potential does not necessarily mean the particle is moving in the opposite direction of the electric field, as the direction of the electric field can also be negative.

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