Question

A particle of mass 2 kg and charge 1 mC is projected vertically with a velocity $10m/s$. There is a uniform horizontal electric field of ${10^4}$ N/C: (This question has multiple correct options)
A. The horizontal range of the particle is 10m.
B. The time of flight of the particle is 2s.
C. The time of flight of the particle is 5s.
D. The horizontal range of the particle is 5m.

Answer 1

For solving this question, we need to use the kinematic equations of motion. The particle is projected towards the vertical direction. Therefore, only the y- axis component of force will be used as there is no movement in the x-axis direction. But the total force will be the vector superposition of these two forces. Accordingly, the acceleration of the particle can be found out. It is necessary to remember the formula of time of flight and newton’s equation of motions formulas.

Complete answer: Charge on the particle, $q = {10^{ - 3}}C$.
Mass of the particle, $m = 2Kg$.
Particle is projected vertically, therefore y-axis component of the velocity is given as,
${u_y} = 10m/s$.
X-axis component of the velocity is given as,
${u_x} = 0\,m/s$.
Uniform electric field, $E = {10^4}N/C$.
Acceleration in the vertical direction, ${a_y} = - 10\,m/{s^2}$.

Let time of flight be T and horizontal range be ${S_x}$.
Now, by using newton’s equation of motion, we get,
Vertical direction:
${S_y} = {u_y}t + \dfrac{1}{2}{a_y}{t^2}$
where Sy=0 and t=T
$0 = 10T - \dfrac{1}{2}(10){T^2}$
$\therefore T = 2s$
Horizontal direction:
Horizontal acceleration,
${a_x} = \dfrac{{qE}}{m} = \dfrac{{{{10}^{ - 3}} \times {{10}^4}}}{2} = 5\,m/{s^2}$
Horizontal range, ${S_x} = {u_x}T + \dfrac{1}{2}{a_x}{T^2}$
∴ ${S_x} = 0 + \dfrac{1}{2}(5){(2)^2} = 10m$.
Therefore, the horizontal range of the particle is 10 m. Hence, option (A) and option (B) are the correct options.

Note: One must make sure to split the acceleration down to its components. Skipping this step would lead to only the magnitudes of some quantities of the motion. But not the important components. One may notice that the Gravitational field is mathematically the same as acceleration due to gravity. But one should understand that the electric field is not acceleration.