Why use conservation of energy in particle velocity problems

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1. Jun 18, 2015

Suppose we have an electric field, and push a proton or electron through it (from rest). We're accelerating it through an electric potential difference of some kind. Why do we apply conservation of energy to the particle-field system to find the speed of the particle?

Example:
Calculate the speed of a proton that is accelerated from rest through a ΔV of 150V.

Applying conservation of energy (and using kinetic energy?),

$$K_i + Ui = K_f + U_f$$

becomes the following conservation of energy equation:

$$0 + qV_i = \frac{1}{2}m_pv_p^2 + 0$$

Solving for the initial potential energy, we get $$qV_i = 2.4 \times 10^{-17}J$$

Plug that back into the equation and solve for final velocity:

$$v_f = \sqrt{\frac{2qV_i}{m_p}} = \sqrt{\frac{2(2.4 \times 10^{-17} J)}{1.67 \times 10^{-27}kg }} = 1.695 \times 10^5 \frac{m}{s}$$

1. Why do we use this to find velocity? Simply because it contains the variables we need in order to calculate it?
2. Why does U_i = qV_i ?
3. Why does U_f = 0 ?
4. Likewise when we accelerate an electron through the same potential difference, why do we set both initial kinetic and potential energies to 0 ?

Last edited: Jun 18, 2015
2. Jun 18, 2015

ShayanJ

Here we have a system consisted of two parts: a charged particle, an EM field. Because in the problem, there is no mechanism that extracts energy from this system, we know that whatever happens, the total energy should be constant.
The total energy has two parts: The potential energy which belongs to the system as a whole and the kinetic energy associated to the charged particle. So this some is a constant.
Now the other piece of the answer is that, the origin of the potential energy i.e. the point where you associate zero potential energy to, is arbitrary.
So we put the origin at the (unknown) position the particle is going to be. But actually it doesn't matter. You may assume that point has a potential of e.g. 12421V and then the starting point has a potential of (12421+150)V. But then because 12421 is the same in both sides, it just gets cancelled so the result is the same no matter what origin you choose.
The initial kinetic energy is set to zero simply because the problem said you accelerate the proton from rest! It wasn't in motion, so there was no kinetic energy!

3. Jun 18, 2015

So ΔK + ΔPE = 0 is just a special case of the total energy of a system? I don't know why I haven't looked at it that way before... now it seems too stupidly obvious. (That's actually how I feel about most of physics)

Sorry if I didn't catch this from your answer, but why is U_i = qV_i ? I didn't think that had anything to do with kinetic energy. (I assume U_i is the initial potential energy and V_i is the initial potential difference)

4. Jun 18, 2015

ShayanJ

I didn't explain that. That's just the definition of potential. At first we have Coloumb's law $F=k\frac{q_s q_t}{r^2}$(s for source and t for test, this distinction is not physical and I'm only making it for later explanations. But one justification maybe that $q_t<<q_s$). Then we find out we can express it as the gradient of another function which we call potential energy(the thing you write in the conservation of energy) $F=-\vec \nabla U$. This definition depends on the charge of the test particle but because its more convienient to work with something depending only on source's properties, we instead work with potential $V=\frac{U}{q_t}$ which gives $U=q_t V$.

5. Jun 18, 2015

Noctisdark

We use conservation of energy because it's the simplest way we can use without solving a very hairy differential equation, If you want a proof that energy is conserved, then I could help, Consider E = T + V let's calculate dE/dt it's d(mv^2/2 + qE(x-x0)),let's assume Electric field to be constant, which really is in your case then dE/dt = m*v*v' + qE*x' and we remind ourselves that mv' = ma = F = qE then thus and hence dE/dt = F*v -F*v = 0 and so energy doesn't change when time does and so it's conserved, Good luck !