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Learn MoreLet's discuss a beautiful problem useful for Physics Olympiad based on Motion in an Electric Field.

**The Problem: Motion in an Electric Field**

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 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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