Question

An charge with mass m and charge q is emitted from the origin,  (x, y) = (0, 0). A large, flat screen is located at x=L. There is a target on the screen at y position yh, where yh > 0 . In this problem, you will examine two different ways that the charge might hit the target. Ignore gravity in this problem.

Part A:

Assume that the charge is emitted with velocity v0 in the positive x direction. Between the origin and the screen, the charge travels through a constant electric field pointing in the positive y direction. What should the magnitude E of the electric field be if the charge is to hit the target on the screen?

Express your answer in terms of m,q,yh,v0, and L.

Part B:

Now assume that the charge is emitted with velocity v0 in the positive y direction. Between the origin and the screen, the charge travels through a constant electric field pointing in the positive x direction. What should the magnitude E of the electric field be if the charge is to hit the target on the screen?

Express your answer in terms of m,q,yh,v0, and L.

Answer 1

Concepts and reason

Use kinematics equation of motion to solve this problem.

First calculate time required to complete the linear displacement, using second equation of motion. After that calculate acceleration of particle form Newton’s second law and electric force. Plug acceleration and time in second equation of motion. Then rearrange equation for electric field. Similarly do calculation for part B.

Fundamentals

Three equations of motion for constant acceleration is as follows:

$\begin{array}{l}\\v = u + at\\\\s = ut + \frac{1}{2}a{t^2}\\\\{v^2} - {u^2} = 2as\\\end{array}$

Here, $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration, $t$ is time and $s$ is displacement travelled in time $t$.

The acceleration of body using Newton’s second law is given as follows:

$a = \frac{F}{m}$

Here, F is the force acting and m is the mas of body.

(A)

The horizontal velocity is ${v_0}$ and it will remain constant in the positive x direction. The time required to travel horizontal distance L is calculated as follows:

$\begin{array}{c}\\t = \frac{{{\rm{distance in }}x\;{\rm{direction}}}}{{{\rm{velocity in }}x{\rm{ direction}}}}\\\\ = \frac{L}{{{v_0}}}\\\end{array}$

Force experienced by particle due to electric field, in $y$ direction is $qE$ so, from Newton’s second law of motion, acceleration in y direction will be,

$a = \frac{{qE}}{m}$

Substitute, $0{\rm{ m/s}}$ for $u$, $\frac{L}{{{v_0}}}$ for $t$, $\frac{{qE}}{m}$ for $a$ and ${y_h}$ for $s$ in second equation of motion.

${y_h} = \left( {0{\rm{ m/s}}} \right)\left( {\frac{L}{{{v_0}}}} \right) + \frac{1}{2}\left( {\frac{{qE}}{m}} \right){\left( {\frac{L}{{{v_0}}}} \right)^2}$

Rearrange the above equation for electric field.

$E = \frac{{2{y_h}mv_0^2}}{{q{L^2}}}$

(B)

The vertical velocity is ${v_0}$ and it will remain constant in the positive y direction. The time required to travel vertical distance ${y_h}$ is calculated as follows,

$\begin{array}{c}\\t = \frac{{{\rm{distance in }}y\;{\rm{direction}}}}{{{\rm{velocity in }}y{\rm{ direction}}}}\\\\ = \frac{{{y_h}}}{{{v_0}}}\\\end{array}$

Force experienced by particle due to electric field, in x direction is $qE$ so, from Newton’s second law of motion, acceleration in x direction will be,

$a = \frac{{qE}}{m}$

Substitute, $0{\rm{ m/s}}$ for $u$, $\frac{{{y_h}}}{{{v_0}}}$ for $t$, $\frac{{qE}}{m}$ for $a$ and $L$ for $s$ in second equation of motion.

$L = \left( {0{\rm{ m/s}}} \right)\left( {\frac{{{y_h}}}{{{v_0}}}} \right) + \frac{1}{2}\left( {\frac{{qE}}{m}} \right){\left( {\frac{{{y_h}}}{{{v_0}}}} \right)^2}$

Now, rearrange equation for electric field.

$E = 2L\left( {\frac{m}{q}} \right){\left( {\frac{{{v_0}}}{{{y_h}}}} \right)^2}$

Ans: Part A

The amount of electric field should be $\frac{{2{y_h}mv_0^2}}{{q{L^2}}}$ so that particle could hit the target.

Part B

The amount of electric field should be $2L\left( {\frac{m}{q}} \right){\left( {\frac{{{v_0}}}{{{y_h}}}} \right)^2}$ so that particle could hit the target.