Worked exercise — Charged particle as a projectile near a charged plane

Problem

A charge ($q = -2\cdot 10^{-4}\ \text{C}$, $m = 0{,}1\ \text{kg}$) is thrown with $\vv{v}_0 = (0,\ 100)\ \text{m/s}$ near an infinite charged plane ($\sigma = 10^{-6}\ \text{C/m}^2$). Describe the motion.

Solution

Using $\varepsilon_0 = 8{,}85\cdot 10^{-12}\ \text{F/m}$, $g = 9{,}8\ \text{m/s}^2$. The field of an infinite charged plane is uniform, directed horizontally (perpendicular to the plane):

$E = \frac{\sigma}{2\varepsilon_0} = \frac{10^{-6}}{2\cdot 8{,}85\cdot 10^{-12}} \approx 5{,}65\cdot 10^4\ \text{N/C}$

The charge is negative, so the electric force is attractive towards the plane (opposite to $\vv{E}$), with magnitude:

$F_{\text{el}} = |q|\,E = 2\cdot 10^{-4}\cdot 5{,}65\cdot 10^4 = 11{,}3\ \text{N}$

Accelerations. Two constant forces: the weight $\vv{P} = (0,\ -mg)$ (vertical) and the electric force $\vv{F}_{\text{el}} = (-F_{\text{el}},\ 0)$ (horizontal, towards the plane). Hence:

$a_x = -\frac{F_{\text{el}}}{m} = -\frac{11{,}3}{0{,}1} = -113\ \text{m/s}^2\ (\text{verso il piano}),\qquad a_y = -g = -9{,}8\ \text{m/s}^2$

Motion. With initial vertical velocity $\vv{v}_0 = (0,\ 100)$, the motion is parabolic (constant acceleration not parallel to the velocity), like a projectile but with a horizontal “electric gravity” much more intense than the real one. The equations of motion (origin at the launch point):

$\begin{aligned} x(t) &= -\tfrac12\,(113)\,t^2 \\ y(t) &= 100\,t - \tfrac12\,(9{,}8)\,t^2 \end{aligned}$

The particle rises, decelerates vertically under gravity, and meanwhile accelerates horizontally towards the plane, tracing a parabola that carries it to impact the plane.

$\ev{\text{Moto parabolico: } a_x \approx 113\ \text{m/s}^2\ \text{verso il piano},\ a_y = -g}$

Linked atoms

Topics: Elettrostatica Concepts: Campo elettrico · Moto parabolico Objects: Piano infinito carico · Proiettile