Chapter 19, Problem 18TP

To determine

The physical difference that will give the different impact points, the sign of the charge and also to define how this compares to the gravitational projectile motion case.

Answer to Problem 18TP

The mass of the second particle is 16 times more than the first particle. As both the particle impacts on the positive plate, the charge on the particles must be negative. When two particles are launched in the gravitational field, they must impact the ground at the same point as the acceleration due to the gravity is same for any mass.

Explanation of Solution

Given:

The charge of the particle is identical, that is, $q=q$.

The impact of the second particle on the positive plate is, $X=4x$.

The vertical displacement of the particle is equal as, $Y=y$.

Formula used:

The required diagram for the first massive particle is shown in figure (1).

Figure (1)

The acceleration in the downward direction of the first particle is given by,

  $a_m=\frac{qE}{m}$

The horizontal displacement of the first particle is given by,

  $\begin{array}{l}x=vt\\ t=\frac{x}{v}\end{array}$

The displacement of the first particle in the $y$ direction by the equation of motion under uniform acceleration is given by,

  $y=ut+\frac{1}{2}{a}_m{t}^2$

The required diagram for the second massive particle is shown in figure (2).

Figure (2)

The acceleration in the downward direction of the second particle is given by,

  $a_M=\frac{qE}{M}$

The horizontal displacement of the second particle is given by,

  $\begin{array}{l}X=vt\\ t=\frac{X}{v}\end{array}$

The displacement of the second particle in the $y$ direction by equation of motion under uniform acceleration is given by,

  $Y=ut+\frac{1}{2}{a}_M{t}^2$

Calculation:

The displacement of the first particle in the $y$ direction is calculated as,

  $\begin{array}{c}y=ut+\frac{1}{2}{a}_m{t}^2\\ =\left(0\right)t+\frac{1}{2}\left(\frac{qE}{m}\right)t^2\\ =\frac{1}{2}\left(\frac{qE}{m}\right)t^2\end{array}$

The horizontal displacement of the first particle is calculated as,

  $\begin{array}{l}y=\frac{1}{2}\left(\frac{qE}{m}\right)t^2\\ y=\frac{1}{2}\left(\frac{qE}{m}\right){\left(\frac{x}{v}\right)}^2\\ x^2=\left(\frac{2mv}{qE}\right)y\end{array}$

The displacement of the second particle in the $y$ direction is calculated as,

  $\begin{array}{c}Y=ut+\frac{1}{2}{a}_M{t}^2\\ =\left(0\right)t+\frac{1}{2}\left(\frac{qE}{M}\right)t^2\\ =\frac{1}{2}\left(\frac{qE}{M}\right)t^2\end{array}$

The horizontal displacement of the second particle is calculated as,

  $\begin{array}{l}Y=\frac{1}{2}\left(\frac{qE}{M}\right)t^2\\ Y=\frac{1}{2}\left(\frac{qE}{M}\right){\left(\frac{X}{v}\right)}^2\\ X^2=\left(\frac{2Mv}{qE}\right)Y\end{array}$

The ratio of horizontal displacement of the first and the second particle is calculated as,

  $\begin{array}{l}\frac{x^2}{X^2}=\frac{\left(\frac{2mv}{qE}\right)y}{\left(\frac{2Mv}{qE}\right)Y}\\ {\left(\frac{x}{X}\right)}^2=\left(\frac{m}{M}\right)\left(\frac{y}{Y}\right)\\ {\left(\frac{x}{4x}\right)}^2=\left(\frac{m}{M}\right)\end{array}$

Solve further,

  $M=16m$

Conclusion:

Therefore, the mass of the second particle is 16 times more than the first particle. As both the particle impacts on the positive plate, the charge on the particles must be negative. When two particles are launched in the gravitational field, they must impact the ground at the same point as the acceleration due to the gravity is same for any mass.