Energy change for a charged particle

1. Aug 2, 2017

Vrbic

1. The problem statement, all variables and given/known data
Without introducing any coordinates or basis vectors, show that, when a charged particle interacts with electric and magnetic fields, its energy changes at a rate $$\frac{dE}{dt}=\vec{v}\cdot \vec{E}$$

2. Relevant equations
$E_{kin} + E_{pot}= En =$ const (1)
$E_{pot}=\vec{r}\cdot\vec{E}$ (2)...suppose homogenic electric field and also that magnetic field does not affect magnitude of velocity. $E$ describes electric field

3. The attempt at a solution
First of all I suppose they ask on kinetic energy. Total energy is conserved. Then I put (2) in (1) and differentiate to get:
$$\frac{dE_k}{dt}=-\frac{d}{dt}(\vec{r}\cdot\vec{E}).$$
I hope, a minus sign vanish in opposite definition of vectors \vec{r}. I hope in homogenic electric field is $\vec{E}=$const. than
$$\frac{dE_k}{dt}=\vec{v}\cdot\vec{E}.$$
Is it allright?

2. Aug 2, 2017

TSny

You don't need to assume a uniform E field, and therefore you don't need to assume that -r⋅E is the potential energy.

The rate at which the energy of the particle increases is the rate at which the electromagnetic force does work on the particle. So, consider the electromagnetic force and the amount of work it does when the particle undergoes a small displacement dr.

3. Aug 2, 2017

Vrbic

Aha I see, so $\frac{dE}{dt}=\frac{d}{dt}\int{\vec{E}d\vec{r}}+\frac{d}{dt}\int{(\vec{v}\times\vec{B})d\vec{r}}$, but now $\vec{E}$ is not constant and integration cannot be done so easy. How may I now get a result? Honestly now I'm not sure why is the latter term zero (how prove it).

4. Aug 2, 2017

TSny

I don't understand why you are integrating. To get the rate of change of energy, you just need to consider an infinitesimal displacement dr of the particle during an infinitesimal time dt. If you have a force F that acts on a particle, how much work does F do on the particle when the particle moves through a displacement dr?

5. Aug 2, 2017

NFuller

What do you know about magnetic fields and work? What is the direction of $\vec{v}\times\vec{B}$ and $\vec{r}$?

6. Aug 2, 2017

Vrbic

Ou yes $dE=dW=\vec{E}d\vec{r}$ sorry I have to thing.

7. Aug 2, 2017

Vrbic

Hmm again, more thinging. Direction of $\vec{v}$ is same as $\vec{r}$. Cross product $\vec{v}\times\vec{B}$ results in ortogonal to both of them i.e. also to $\vec{r}$ i.e. dot product is zero.
Thank you both of you.

8. Aug 2, 2017

TSny

Right. The magnetic force doesn't do any work. So, the change in energy is due solely to the work done by the electric force.