Question

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

• A. $400\text{m}$
• B. $300\text{m}$
• C. $200\text{m}$
• D. $100\text{m}$

Answer 1

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

Given that,
Mass of charged particle, $m=1\text{kg}$
Charge of particle, $q=2\mu \text{C}=2\times {10}^{-6}\text{C}$
Initial speed of the particle, $v=20\text{m/s}$
Horizontal electric field, $E=2\times {10}^7\text{V/m}$

The path of the particle will be a parabola, but the particle will accelerate along $x-$ axis also.
Let the horizontal component of the velocity is, $u_x=20\cos {45}^{\circ }$
Verticle component of the velocity, $v_y=20\sin {45}^{\circ }$

Time of flight of the projectile is
$T=\frac{2u_y}{a_y}=\frac{2\times 20\sin {45}^{\circ }}{10}=2\sqrt{2}\text{s}$

Along $x-$ axis,
acceleration, $a=\frac{F}{m}=\frac{qE}{m}$
$\Rightarrow a=\frac{2\times {10}^{-6}\times 2\times {10}^7}{1}=40{\text{m/s}}^2$
So, range of the projectile,
$s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s=u\cos \theta t+\frac{1}{2}a{t}^2$
$\Rightarrow s=20\cos {45}^{\circ }\times 2\sqrt{2}+\frac{1}{2}\times 40\times (2\sqrt{2}{)}^2$
$\Rightarrow s=200\text{m}$