Question

A charged particle of mass m=1kg and charge q=2μC is thrown from a horizontal ground at an angle θ=450 with speed 20m/s. In space a horizontal electric field E=2×106V/m exist Find the range on horizontal ground of the projectile

Answer 1

☢️PLEASE MARK AS BRAINLIEST☢️

Answer 2

Answer:

R = 47.27 m

Explanation:

Given,

• Mass of the particle = m = 1 kg
• charge on the particle = q = $2\times 10^{-6}\ C$
• Angle of projection = $\theta\ =\ 45^o$
• Initial speed of the particle = u = 20 m/s.
• Horizontal electric field = E = $2\times 10^6\ v/m$

Now the electric force is acting on the particle due to the electric field.

Let a be the acceleration of the particle.

$\therefore F\ =\ qE\\\Rightarrow ma\ =\ qE\\\Rightarrow a\ =\ \dfrac{qE}{m}\\\Rightarrow a\ =\ \dfrac{2\times 10^{-6}\times 2\times 10^6}{1}\\\Rightarrow a\ =\ 4\ m/s^2$

Now In vertical direction there is no force acting on the particle therefore the time of flight is

$t\ =\ \dfrac{2u}{g}\\\Rightarrow t\ =\ \dfrac{2\times 20}{9.812}\ =\ 4.07\ sec$

Horizontal velocity of the particle is $u_x\ =\ ucos45^o\ =\ 20cos45^o$

Let R be the range of the particle.

Now from the kinematics,

$R\ =\ u_xt\ +\ \dfrac{1}{2}at^2\\\Rightarrow R\ =\ 20cos45^o\times 4.07\ +\ 0.5\times 4\times 4.07^2\\\Rightarrow R\ =\ 47.27\ m$

Hence the range of the particle is 47.27 m