Question

A particle of mass and charge is released from rest in uniform electric field. If there is no other force on the particle, the dependence of its speed $v$ on the distance $x$ travelled by it is correctly given by (graphs are schematic and not drawn to scale)

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The correct option is B

Step1: Given data and assumptions.

Mass of the particle $=m$

Initial velocity $u=0$

Distance covered by the particle$=x$

Step 2: Find the acceleration of the particle.

We know that from newton's second law.

$F=ma$

Where $F$ is the force, $m$ is the mass of the particle, and $a$ is the acceleration of the particle

$\because$$a=\frac{F}{m}$

Also,

Force $\left(F\right)=$Charge $\left(q\right)$$\times$Electric field intensity $\left(E\right)$

Then,

$a=\frac{qE}{m}$ $\left(i\right)$

Step3: Finding the speed $v$ on the distance $x$.

Formula used:

$v^2=u^2+2as$

Where $v$ is the final velocity, $a$ is the acceleration, $s$ is the distance and $u$ is the initial velocity.

Now,

$v^2=u^2+2\left(\frac{qE}{m}\right)x$ $\left[\because fromequation\left(i\right)\right]$

$\Rightarrow$$v^2=0+2\left(\frac{qE}{m}\right)x$ $\left[\because u=0\right]$

$\Rightarrow$$v^2=\frac{2qE}{m}x$

Therefore $v^2=\frac{2qE}{m}x$ represent the equation of the parabola.

Again,

$\Rightarrow$$v^2=\frac{2qE}{m}x$

$\Rightarrow$$v=\sqrt{\frac{2qE}{m}x}$

$\Rightarrow$$v=k\sqrt{x}$ $\left[\because k=\sqrt{\frac{2qE}{m}}\right]$

$\because v\propto \sqrt{x}$

Thus from the above velocity-distance equation after assuming the value of distance $x$ we get.

Hence, option B is correct.