Motion in an Electric Field

A particle moves rectilinearly in an electric field E=E0-ax where a is a positive constant and x is the distance from the point where the particle is initially at rest. Let the particle have a specific charge q/m.

Find:

(I) the distance covered by the particle till the moment at which it once again comes to rest, and

(II) acceleration of the particle at this moment.

Solution:

A particle moves rectilinearly in an electric field $$E=E_0-ax$$ where a is a positive constant and x is the distance from the point where the particle is intially at rest.
The particle has a specific charge q/m.
Now,

$$ F=q(E_0-ax)$$
$$or, a = \frac{q(E_o-ax)}{m}$$
At x=0,

$$a=\frac{qE_0}{m}$$
Particle will move in the x direction
$$\frac{vdv}{dx}=a$$

$$v\frac{dv}{dx}=\frac{q(E_0-ax))}{m}$$
$$vdv=\frac{q(E_0-ax)}{m}dx$$
$$\int_{0}^{0} vdv=\int_{0}^{x_0} \frac{q(E_0-ax)}{m}dx$$
$$ 0=\frac{q(E_0x-\frac{ax^2}{2})}{m}$$
Now, $$ v=0, x=x_0$$
Hence,

$$E_0x_0=a\frac{x_0^2}{2}$$
$$x_0=\frac{2E_0}{a}$$
Distance covered by the particle before coming to rest =
$$\frac{2E_0}{a}$$
Acceleration before coming to rest will be
$$ a=\frac{-qE_0}{m}$$
The direction of the particle will be towards the negative x-axis.

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