In my high school physics class, we were calculating charges being accelerated through electric fields. The questions usually asked for the capacitance, work to move the charge, resistance of charge, number of electrons in the charge, acceleration of the charge, time of acceleration of the charge, etc, but they would never ask for the final velocity of the charge (hmm suspicious).

So on a random problem, I tried using the same kinematics equations that we used for solving for $a$ and $t$ to solve for final velocity of the charge, and I was somewhat surprised to get a value faster than the speed of light. I know our high school physics class is pretty watered down, so I suspect there's probably something the curriculum doesn't teach on purpose. I think perhaps there's some relativistic effect that comes into play as the speed of electron approaches the speed of light so I can't just use an equation that models classical mechanics, but its just my hypothesis. What is the explanation for this, and how would the velocity of the charge be correctly calculated?

Edit: My teacher told us to use these equations for our calculations

$F=kq_1q_2/d^2$ for electric force

$F=ma$ for acceleration

$d = v_ot + \frac12 at^2$ for time

The mass of the charge is simply derived using the constants for the charge and mass of an electron. Here is one example of a homework problem:

A -4.00C charge is 2.50cm south of a +8.00C charge. Answer all of the following questions, which pertain to the -4.00C charge, during the time it takes it to move 1.00 mm as a result.

  1. Find the electric force (& its direction since it’s a vector) on the -4.00 C charge.
  2. Find the number of electrons on/in the -4.00 C charge.
  3. Find the mass of the -4.00 C charge, assuming it is composed entirely of electrons.
  4. Find acceleration of the -4.00C charge.
  5. Find the time of travel during its movement.

The answer key says acceleration = $2.02\dot{} 10^{25}m/s^2$ and time = $9.94 \dot{} 10^{-15}s$. I assumed it would be okay to use $v = v_0 + at$ since its in the same family of equations my teacher told us to use. This results in a velocity of $~2.0\dot{}10^{11} m/s$.

  • 3
    $\begingroup$ You are doing something wrong, then, depending on what potential you are accelerating across. Remember, the rest energy for an electron is 511keV, so by 100kV you will need to be including relativistic effects on the velocity. (And, if you accelerate across a potential of 100kV, the electron will have a final energy of 100keV, the only problem is what the velocity is). $\endgroup$ – Jon Custer Mar 31 '17 at 19:59
  • $\begingroup$ Yes, please include an example. $\endgroup$ – sammy gerbil Apr 1 '17 at 0:18
  • $\begingroup$ Wow, those are some big charges. See Can High Charges like 1 milli coulomb be acheived $\endgroup$ – M. Enns Nov 14 '17 at 1:51

At high speeds, relativistic effects come to play. You cannot solve this kind of questions using classical mechanics; you need to learn relativistic mechanics. Classical mechanics has no speed limits; the speed limit arises due to relativity. Your method of solving the problem does not account for the relativistic effects. The answer to your question would involve explaining special relativity which is an entire theory. Therefore, the answer to your question is too broad.

The equations you have mentioned work for objects traveling at speeds much lower than the speed of light.

As the velocity of the object approaches the speed of light, its mass increases. The mass is given by:

$$m = m_0\gamma =m_0\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

where $c$ is the speed of light and $m_0$ is the rest mass. As the velocity of the object approachs the speed of light, its mass approaches infinity. The amount of acceleration a force would produce decreases. This way you can never go past the speed of light.

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There seem to be a few flaws with your calculation, particularly in the use of the mass f the charge). Electrons have charges of magnitude $e \approx 10^{-19}$ C but it would not be correct to argue that the mass you use is $n \times m_e$, where $n = \frac{4}{e} \approx 4 \times 10^{19}$. Yes, this number of electrons have to be added, but you couldn't keep those electrons close to each other because of the electrical repulsion (admittedly, though, you were told to use this estimate).

Another problem is that the force depends upon the distance between the charges, which varies. So, besides the need to use relativistic dynamics, you also need to take into account the fact that the force varies with the distance between the charges.

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