I want prove uniform electric field work : I know that $\mathbf{F}(t)=Q\mathbf{E}(t)$

Question

I want prove uniform electric field work : I know that $\mathbf{F}(t)=Q\mathbf{E}(t)$ but is uniform So $\mathbf{F}=Q\mathbf{E}$ Thus we have : $$ \begin{aligned} W_{(t_a -t_b)}&:=\int_{t_a}^{t_b} \Big( \mathbf{F}(t) \cdot \dfrac{d}{dt}\mathbf{r}(t) \Big) \mathrm{d}t \\ &=\int_{t_a}^{t_b} \Big(Q\mathbf{E}(t)\cdot \dfrac{d}{dt}\mathbf{r}(t) \Big) \mathrm{d}t\\ &=Q\int_{t_a}^{t_b} \Big(\Big(E_x(t) \mathbf{i} +E_y(t) \mathbf{j}+E_z(t) \mathbf{k}\Big)\cdot \Big (\dfrac{d}{dt}r_x(t)\mathbf{i}+\dfrac{d}{dt}r_y(t)\mathbf{j}+\dfrac{d}{dt}r_z(t)\mathbf{k}\Big)\Big) \mathrm{d}t \\ &=Q\int_{t_a}^{t_b} \Big(E_x(t) \dfrac{d}{dt}{r}_x(t) +E_y(t) \dfrac{d}{dt}{r}_y(t) +E_z(t)\dfrac{d}{dt}{r}_z(t) \Big) \mathrm{d}t\\ &= Q\int_{t_a}^{t_b} \Big(E_x\dfrac{d}{dt}{r}_x(t) + E_y\dfrac{d}{dt}{r}_y(t) +E_z\dfrac{d}{dt}{r}_z(t) \Big) \mathrm{d}t \\ \end{aligned} $$ Now what ?

Answer 1

You can take the constant ${\bf E}$ out of the integral and write $$W_{(t_a, \,t_b)}=Q{\bf E}\cdot\int_{t_a}^{t_b}{d\over dt}{\bf r}(t)\>dt=Q{\bf E}\cdot\bigl({\bf r}(t_b)-{\bf r}(t_a)\bigr)\ .$$ Note that the function $${\bf r}(t)=\bigl(x(t),y(t),z(t)\bigr)\qquad(t_a\leq t\leq t_b)$$ has no partial derivatives with respect to $x$,$y$, $z$, but just a time derivative $$\dot{\bf r}(t)=\bigl(\dot x(t),\dot y(t),\dot z(t)\bigr)\ .$$