Here's the question.
Consider a particle of mass $m$ that carries a charge $q$. Suppose that the particle is under the influence of both and electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$ so that the particle's trajectory is described by the path $\mathbf{x}(t)$ for $a\le{t}\le{b}$. Then the total force acting on the particle is given by the Lorentz force
$$\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B}),$$
Where $\mathbf{v}$ denotes the velocity of the trajectory.
(a) Using Newton's second law of motion ($\mathbf{F}=m\mathbf{a}$), where $\mathbf{a}$ denotes the acceleration of the particle to show that
$$m\mathbf{a}\cdot\mathbf{v}=q\mathbf{E}\cdot\mathbf{v}$$
So a quick substitution yields
$$m\mathbf{a}\cdot\mathbf{v}=\mathbf{F}\cdot\mathbf{v}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})\cdot\mathbf{v}=(q\mathbf{E}+q(\mathbf{v}\times\mathbf{B}))\cdot\mathbf{v}$$
So i think the next step is
$$=q\mathbf{E}\cdot\mathbf{v}+q(\mathbf{v}\times\mathbf{B})\cdot\mathbf{v}$$
And since the dot product of those last two is $0$, we get are result. But this is where I'm having trouble (new to line integrals)
(b) If the particle moves with a constant speed, use (a) to show that $\mathbf{E}$ does now work along the path of the particle
So
$$\int_C{q\mathbf{E}\cdot\mathbf{v}}=\int_C{m\mathbf{a}\cdot\mathbf{v}}=\int_C{m\frac{d\mathbf{v}}{dt}}\cdot\mathbf{v}$$
But i'm not sure how to procede…
Best Answer
The problem is that not only the force produced by the electric field acts on the particle, if this were so then the particle would accelerate (it would be subject to a non-zero force, this is N. Second law), so there is a force countering that of the electric field ${\bf F} = -q{\bf E}$, only then can ${\bf v}$ be constant. The work done by such force is
$$W= \int_{t_a}^{t_b} {\bf F \cdot v }dt = -q\int_a^b {\bf E}\cdot d{\bf s} $$
The second integral shows that this can be interpreted as work done by (or rather against) the electric field.