# Why use conservation of energy in particle velocity problems

In summary, the reason we use conservation of energy to find the speed of a particle accelerated through an electric potential difference is because it simplifies the problem and allows us to avoid solving a complicated differential equation. This is because the total energy of the system, consisting of the particle and the electric field, is constant and can be expressed as the sum of potential and kinetic energy. We set the initial kinetic energy to zero because the particle was initially at rest, and the initial potential energy is equal to the product of the particle's charge and the initial potential difference. This is because potential energy is defined as the negative gradient of the potential function, and we use the potential instead of potential energy because it is more convenient to work with. Ultimately, the
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 ?

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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 canceled 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!

Shyan said:
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 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)

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)
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 ##.

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 !

## 1. Why is conservation of energy important in particle velocity problems?

Conservation of energy is important in particle velocity problems because it helps us understand and predict the behavior of particles in motion. It states that energy cannot be created or destroyed, only transferred from one form to another. This principle allows us to track the changes in energy as a particle moves and helps us determine its final velocity.

## 2. How does conservation of energy apply to particle velocity problems?

In particle velocity problems, conservation of energy is applied by tracking the changes in kinetic and potential energy of the particle. As the particle moves, its kinetic energy increases while its potential energy decreases. At any given point, the total energy (kinetic energy + potential energy) remains constant, allowing us to use this principle to find the final velocity of the particle.

## 3. Can conservation of energy be used in all particle velocity problems?

Yes, conservation of energy can be used in all particle velocity problems as long as there are no external forces acting on the particle. In the absence of external forces, the total energy of the particle remains constant, making conservation of energy a valid principle to use in these problems.

## 4. What are the benefits of using conservation of energy in particle velocity problems?

The main benefit of using conservation of energy in particle velocity problems is that it allows us to find the final velocity of a particle without having to track its motion over time. This saves us time and effort and also provides a more accurate result. Additionally, it helps us better understand the relationship between energy and motion of particles.

## 5. Are there any limitations to using conservation of energy in particle velocity problems?

One limitation of using conservation of energy in particle velocity problems is that it only applies to systems where there are no external forces acting on the particle. If there are external forces present, such as friction or air resistance, the total energy of the particle will not remain constant and conservation of energy cannot be used. Additionally, this principle assumes that there is no loss of energy due to factors such as heat or sound, which may not always be the case in real-world situations.

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