Question

A charged particle of mass $m=1\text{kg}$ and charge $q=2\mu C$ is thrown from the ground at an angle $\theta ={45}^{\circ }$ with the horizontal, with initial speed $20\text{m/s}$. In the space a horizontal electric field of magnitude $E=2\times {10}^7\text{V/m}$ exists. Find the range covered by the projectile on the horizontal ground. $[g=10{\text{m/s}}^2]$

• A. $100\text{m}$
• B. $150\text{m}$
• C. $170\text{m}$
• D. $200\text{m}$

Answer 1

The correct option is D $200\text{m}$


The path of the particle will be a parabola as it is a projectile. But, in this case due to the presence of the electric field in $x$ direction some force will act on the particle and along $x-$ axis also the motion of the particle will be accelerated.

Using kinematic relation , $y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

Net displacement along $y-$ direction is zero (i.e) $y=0$.

$\Rightarrow 0=u_{y}T-\frac{1}{2}g{T}^2$

$T=\frac{2\times 20\sin {45}^{\circ }}{10}=2\sqrt{2}\text{s}$

Along $x-$ direction, $a_x=\frac{F}{m}=\frac{qE}{m}$

$a_x=\frac{(2\times {10}^{-6})(2\times {10}^7)}{1}=40{\text{m/s}}^2$

Now, we can use the second equation of motion to find the motion of projectile along the $x$ direction.

$\because x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

Horizontal range of the particle will thus be

$R=u_{x}T+\frac{1}{2}{a}_x{T}^2$

$\therefore R=\left(20\cos {45}^{\circ }\right)\left(2\sqrt{2}\right)+\frac{1}{2}\left(40\right)(2\sqrt{2}{)}^2$

$R=40+160=200\text{m}$

Hence, option (d) is the correct answer.