# How do I show that the total energy of a particle moving in an electrostatic field is constant?

Basically the Lorentz force says that:

$$F=qE+qv × B$$

I am asked to show that if a particle moves in an electrostatic field

$$E= - ∇V(x,y,z)$$

and ANY magnetic field, then the energy:

$$\frac{1}{2}mv^{2}+qV$$ is a constant of motion.

Does anyone have absolutely any indication as to where to start? I can try taking partial derivatives and plugging in stuff but I don't really have a sense as to where to start on this. Any hints?

To prove that the total energy is a constant of the motion means to prove that its total derivative with respect to time vanishes.

To compute this derivative note that

$$\frac{d}{dt} v^2 \equiv \frac{d}{dt} \vec v \cdot \vec v = 2 \vec v \cdot \frac{d \vec v}{dt}, \tag 1$$

and

$$\frac{d}{dt} V(\vec x(t)) = \frac{d \vec x}{dt} \cdot \nabla V(\vec x(t)). \tag 2$$

Can you conclude from this?

EDIT:

The total energy of the particle at time $t$ is by definition $$U(t) = \frac{1}{2} m v(t)^2 + q V(\vec x(t)) = \frac{1}{2} m \left( \frac{d \vec x(t)}{dt} \right)^2 + q V(\vec x(t)). \tag 3$$ You want to prove that $\frac{d U(t)}{dt} = 0$. To compute this derivative you just have to use (1) and (2), Newton's second law of motion, and that a vector product $\vec a \times \vec b$ is orthogonal to both $\vec a$ and $\vec b$ (i.e. $\vec a \cdot (\vec a \times \vec b) = 0$.