Question

Electric field in a region is given as $\overrightarrow{E}=(10-5x)\hat{i}$. A charged particle of mass $5\text{kg}$ and charge $Q=1\text{C}$ is situated at origin and free to move in the given electric field. Then choose the correct options.

• A. Motion of the charged particle is oscillatory
• B. Maximum displacement of charge particle from origin is $4\text{SI units}$
• C. Maximum velocity gain by charge particle is $2\text{SI units}$ units.
• D. The position of charge particle, when velocity gained by particle is maximum, is $2\text{SI units}$.

Answer 1

Answer: (A), (B), (C), (D)

The position of charge particle, when velocity gained by particle is maximum, is $2\text{SI units}$.

Given, electric field $\overrightarrow{E}=(10-5x)\hat{i}$

Mass of charged particle $m=5\text{kg}$ and
Charge $Q=1\text{C}$.

(a) As we know that force on charged particle in electric feld is given by,

$\overrightarrow{F}=Q\overrightarrow{E}$

Substituting the value, we get

$\overrightarrow{F}=1(10-5x)\hat{i}$

$\overrightarrow{F}=(10-5x)\hat{i}$

It is clear that that $F\propto -x$

So, motion of charge particle is oscillatory.

(b) Now, at mean position $F=0\text{N}$

$\therefore 10-5x=0$

$\Rightarrow x=2$


As it oscillates is about $x=2$ so the maximum displacement from the origin is $4$.

Now, as $F=ma\Rightarrow a=\frac{F}{m}$

$\therefore a=\frac{10-5x}{5}$

$a=2-x\text{(or)}a=-(x-2)$

$\because$Acceleration in SHM is $a=-{\omega }^{2}x$

$\therefore {\omega }^2=1$

$\Rightarrow \omega =1$

(c) Maximum velocity $v_{max}=a\omega$
$v_{max}=2$

(d) Velocity will be maximum at mean position i.e., $x=2$

Hence, options $A,B,C$ and $D$ are correct.