Motion of a Charged Particle in a Constant Electric Field

The motion of a particle with charge $q$ and mass $m$ in a constant electric field $\vec{E}$ is described by,

$z$

$t$

 $$\vec{r}=\left(x_0 +v_{x0}t+\frac{qE_x}{2m}t^2\right)\,\hat{x}+\left(y_0 +v_{y0}t+\frac{qE_y}{2m}t^2\right)\,\hat{y}+\left(z_0 +v_{z0}t+\frac{qE_z}{2m}t^2\right)\,\hat{z},$$ $$\vec{v}=\left(v_{x0}+\frac{qE_x}{m}t\right)\,\hat{x}+\left(v_{y0}+\frac{qE_y}{m}t\right)\,\hat{y}+\left(v_{z0}+\frac{qE_z}{m}t\right)\,\hat{z},$$ $$\vec{a}=\frac{qE_x}{m}\,\hat{x}+\frac{qE_y}{m}\,\hat{y}+\frac{qE_z}{m}\,\hat{z},$$ $$\vec{F}=qE_x\,\hat{x}+qE_y\,\hat{y}+qE_z\,\hat{z}.$$

$z_0=0$ m   $m=1$ kg

$qE_{z}=$ -1 [N]

$v_{z0}=$ 4 [N]

The force on the particle is constant so the motion is parabolic. See the Constant Force = Parabolic Motion app. This problem could also be solved numerically with a Numerical 6th order differential equation solver.

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