# Motion of a charged particle -- Changes in KE and PE?

• I
• aaronll
In summary, the potential energy of a charged particle moving through a potential difference (a potential difference is a difference in potential energy) gains kinetic energy, but it also loses potential energy.
aaronll
If a charged particle moves through a potential difference, it gains kinetic energy but does it also lose potential energy?
When I accelerate a particle and then I "free it", what happen to its potential energy if the total energy should be conserved?

Let us see a pendulum. It has maximum kinetic energy at the bottom where potential energy is minimum. It has minimum zero kinetic energy at the top turning points where potential energy is maximum. This example suggests your thought.

vanhees71, Delta2 and aaronll
aaronll said:
If a charged particle moves through a potential difference, it gains kinetic energy but does it also lose potential energy?
Yes, it must. It can only gain KE if it loses PE. That's the nature of a conservative force.

vanhees71 and aaronll
PeroK said:
Yes, it must. It can only gain KE if it loses PE. That's the nature of a conservative force.
Yes I know, maybe my question is very wrong, but if I have this apparatus:

Before: particle have a kinetic energy with v1 and so total energy = kinetic energy.
Enter: particle is due to the force, that is conservative, so E is conserved and gain kinetic energy ( loss potential energy).
flow Out: It has only kinetic energy, but is different from when is inside, because there are no more "potential" energy they have relatively to the position in the field.

I think I'm wrong because potential energy have a meaning only when the particle is in a field so there is potential difference ( this is what interest) and energy is conserved because inside the field do a work and convert that work in kinetic energy, is correct?

thanks

vanhees71 and PeroK
If we take that diagram at face value: there are six postive charges (held in place by an external force) and likewise six negative charges. Then, we have a charged particle moving through the electric field created by those charges. That electric field does not exist only within the plates. It may be an approximation that the force only acts on the particle while it is between the plates, but if you look at the set-up fully, then there must be a potential outside the plates. In particular, a negatively charged particle must have more potential energy before it enters the plates than afterwards, if it moves as shown on your diagram.

vanhees71 and aaronll
PeroK said:
If we take that diagram at face value: there are six postive charges (held in place by an external force) and likewise six negative charges. Then, we have a charged particle moving through the electric field created by those charges. That electric field does not exist only within the plates. It may be an approximation that the force only acts on the particle while it is between the plates, but if you look at the set-up fully, then there must be a potential outside the plates. In particular, a negatively charged particle must have more potential energy before it enters the plates than afterwards, if it moves as shown on your diagram.
Maybe my question was not clear, I try to reformulate with a different situation:
I have a table, 6m tall relatively to the "ground", with a cube on it.
The cube has total energy of 0 ( is at rest and potential energy relatively to the table surface is 0).
So E = 0.

Now, I move the cube beyond the edge of the table, and now it is 6m from the ground ( also before but that is not my doubt) so its energy is the potential energy relatively to the ground, so E > 0.

The value of energy is changed ( value in joules ) but only because I change the "reference", so energy seems to be not conserved, or change value when I change the position of the object but doing no work.
How It is?

I think it's simpler to get the equations right first. Having a "conservative force" means that the force can be written as the gradient of a scalar potential, and everything is not explicitly time-dependent
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
Your capacitor at a static voltage is an example. Another one is the gravitational force between some very light object and other very heavy objects at rest.

From Newton's equation of motion,
$$m \ddot{\vec{x}}=\vec{F}(\vec{x}) = -\vec{\nabla} V(\vec{x}),$$
you can then derive the energy-conservation law by multiplying with ##\dot{\vec{x}}## and integrating between two arbitrary times. On the left-hand side this gives
$$\int_{t_1}^{t_2} \mathrm{d} t m \dot{\vec{x}} \cdot \ddot{\vec{x}} = \int_{t_1}^{t_2} \mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 \right) = \left [\frac{m}{2} \dot{\vec{x}}(t) \right ]_{t=t_1}^{t=t_2} = T(t_2)-T(t_1),$$
where the kinetic energy at time ##t## is defined as
$$T(t)=\frac{m}{2} \dot{\vec{x}}^2(t).$$
On the right-hand side the above procedure gives
$$\int_{t_1}^{t_2} \mathrm{d} t \mathrm{d} t [-\dot{\vec{x}} \cdot \vec{\nabla} V] = -\int_{t_1}^{t_2} \mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} V = - \left \{ V[\vec{x}(t_2)] - V[\vec{x}(t_1)] \right \}.$$
Setting this equal to the previous equation and sorting the quantities at each time you get
$$T(t_2)+V[\vec{x}(t_2)]=T(t_1)+V[\vec{x}(t_1)] \; \rightarrow \; E=T+V=\text{const}$$
along the trajectory.

That energy-conservation law is very useful, because you don't need to know the solution of the equations of motion to calculate the energy, provided the force has a potential. For a given force a potential exists if the line integral of the force between two points is independent on the path connecting these two points, and then the potential is given by the line integrals connecting some fixed point with any other point,
$$V(\vec{x})=-\int_{C_{\vec{x}_0 \rightarrow \vec{x}}} \mathrm{d} \vec{x}' \cdot \vec{F}(\vec{x}').$$

Now if you have the situation that ##V(\vec{x}) \rightarrow 0## for ##r=|\vec{x}| \rightarrow \infty## and the particle enters with some velocity ##\vec{v}## from infinity, it gets scattered by the force described by the potential and then goes to infinity again. Then far away from the region, where the potential is non-zero it has some speed ##\vec{v}'##, but since ##V=0## in the beginning and at the end of the scattering process the energy-conservation law tells you that the speed is the same initially and finally ##|\vec{v}'|=|\vec{v}|##.

aaronll
aaronll said:
The value of energy is changed ( value in joules ) but only because I change the "reference", so energy seems to be not conserved, or change value when I change the position of the object but doing no work.
How It is?
In comparison of before and after, energy ground level should be in common. If you can change energy ground level any time you like, I cannot say anything right for you without knowing your rule.

vanhees71 and aaronll
aaronll said:
The value of energy is changed ( value in joules ) but only because I change the "reference", so energy seems to be not conserved, or change value when I change the position of the object but doing no work.
How It is?
Conserved means a quantity that does not change over time. Energy of a closed system is conserved. Energy, however, is frame dependent. There is no absolute value for the energy of a system. If you change your reference point for PE, then the energy of the system changes from one reference frame to another.

However, the change in energy of a system should be the same in any reference frame. Wherever you choose your reference point for zero PE, the change in PE when a particle moves over time is the same.

Doc Al and vanhees71
aaronll said:
Yes I know, maybe my question is very wrong, but if I have this apparatus:
View attachment 291013
Before: particle have a kinetic energy with v1 and so total energy = kinetic energy.
Enter: particle is due to the force, that is conservative, so E is conserved and gain kinetic energy ( loss potential energy).
flow Out: It has only kinetic energy, but is different from when is inside, because there are no more "potential" energy they have relatively to the position in the field.

I think I'm wrong because potential energy have a meaning only when the particle is in a field so there is potential difference ( this is what interest) and energy is conserved because inside the field do a work and convert that work in kinetic energy, is correct?

thanks
You did not draw the important part. I mean the line ends where the particle is not yet free.

## 1. What is the relationship between the motion of a charged particle and changes in its kinetic and potential energy?

The motion of a charged particle is directly related to changes in its kinetic and potential energy. As the particle moves, it can gain or lose kinetic energy, which is the energy associated with its motion. Additionally, the particle can also gain or lose potential energy, which is the energy associated with its position in an electric or magnetic field.

## 2. How does the speed of a charged particle affect its kinetic energy?

The kinetic energy of a charged particle is directly proportional to its speed. This means that as the speed of the particle increases, so does its kinetic energy. Conversely, as the speed decreases, the kinetic energy also decreases.

## 3. Can the potential energy of a charged particle change without any change in its kinetic energy?

Yes, the potential energy of a charged particle can change without any change in its kinetic energy. This can occur when the particle moves through an electric or magnetic field, which can change its potential energy without affecting its speed.

## 4. How does the direction of motion of a charged particle affect its kinetic and potential energy?

The direction of motion of a charged particle can affect both its kinetic and potential energy. If the particle moves in the same direction as an electric or magnetic field, its potential energy will decrease, while its kinetic energy will increase. On the other hand, if the particle moves in the opposite direction of the field, its potential energy will increase, while its kinetic energy will decrease.

## 5. Can the total energy of a charged particle change during its motion?

Yes, the total energy of a charged particle can change during its motion. This can occur when the particle moves through a changing electric or magnetic field, which can cause the particle to gain or lose energy. Additionally, collisions with other particles can also cause changes in the total energy of a charged particle.

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