Question

If A particle of charge $q$and mass $m$moves in the uniform fields $\overrightarrow{E}=E_0\widehat{k}\text{and}\overrightarrow{B}=B_0\widehat{k}$. At $t=0$, the particle has velocity ${\overrightarrow{v}}_0=v_0\widehat{i}$. What is the particle's speed at a later time $r$ ?

Short answer

The particle's speed at a later time$v=\sqrt{v_o^2+{\left(\frac{q{E}_{o}t}{m}\right)}^2}$

Step-by-step solution

Given Values 

$\overrightarrow{E}=E_o\widehat{k}$

$\overrightarrow{B}=B_o\widehat{k}$

$\overrightarrow{v}=v_o\widehat{i}$

$t=0$

We use the equations,

${\overrightarrow{F}}_E=q\overrightarrow{E}=q{E}_o\widehat{k}$

$\overrightarrow{F_M}=\overrightarrow{qv}\times \overrightarrow{B}=q{v}_o{B}_o(-j)$

In the x-y plane, for circular movement:

Equation Solving

$\overrightarrow{F_M}=q{v}_o{B}_o\widehat{r}⟺F_E=q{E}_o\widehat{k}$

$\sum \overrightarrow{F}=m\overrightarrow{a}$

$F_{1z}=q{E}_o-m{a}_z$

$F_{1r}=q{v}_o{B}_o=m{a}_{\tau }=\frac{m{v}_o^2}{2}$

$F_{1t}=0$

Final Equation

$\text{we find}\overrightarrow{v}\left(t\right)\text{:}$

$\overrightarrow{v}\left(t\right)=v_o\widehat{r}+\left(0+a_{z}t\right)\widehat{k}$

$\overrightarrow{v}\left(t\right)=v_o\widehat{i}+a_{z}t\widehat{k}$

$v=\sqrt{v_o^2+{\left(a_{z}t\right)}^2}$ $\left(a_z=\frac{q{E}_o}{m}\cdot \right)$

$v=\sqrt{v_o^2+{\left(\frac{q{E}_{o}t}{m}\right)}^2}$