Question

A proton moves at 4.40 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.00 103 N/C. Ignore any gravitational effects.
(a) Find the time interval required for the proton to travel 6.00 cm horizontally.
Answer in ns

(b) Find its vertical displacement during the time interval in which it travels 6.00 cm horizontally. (Indicate direction with the sign of your answer.)
Answer in mm

(c) Find the horizontal and vertical components of its velocity after it has traveled 6.00 cm horizontally.

v=< i + j> km/s

Answer 1

Concepts and reason

The concepts used to solve this problem are in a uniform electric field the electric force applied is constant.

For simplicity, choose the x-direction along the velocity of the proton as it goes into the electric field, the y-direction along the electric field vector and the reference time ${t_0}$ when the proton penetrates the field. When the proton is in the uniform electric field the electric force applied on the proton is constant. Here we can assume that the other forces are smaller in magnitude compared to the electric force exerted by the electric field. So, the net force applied on the proton is roughly equal to the electric force and is constant. By using Newton’s second law, acceleration can be calculated and with the equation for position and velocity, time interval, displacement and velocity can be calculated.

Fundamentals

According to Newton's second law, the particle moves with a constant acceleration.

The expression of the force is equal to,

$F = ma$

Here, $F$ is the force, $m$ is the mass and $a$ is the acceleration.

The expression of the force in terms of the electric field is equal to,

$F = qE$

Here, $q$ is the charge, $F$ is the electrostatic force and $E$ is the electric field.

From the expression of the linear kinematic, the expression of the distance covered by the object is equal to,

$s = ut + \frac{1}{2}a{t^2}$

Here, $s$ is the distance covered by the object, $u$ is the initial speed, $a$ is the acceleration and $t$ is the time.

From the expression of the linear kinematic, the expression of the final velocity of the object is,

$v = u + at$

Here, $v$ is the final velocity.

The expression of the distance covered by the object if the acceleration of the system is constant is equal to,

$s = vt$

(a)

Calculate the horizontal distance travel by the object.

When a proton is entered into the uniform magnetic field with the constant horizontal component of the velocity then the acceleration of the object is zero.

$a = 0$

The expression of the distance travel in the horizontal direction is equal to,

$s = vt$

Rearrange the above expression in terms of the time $t$ .

$t = \frac{s}{v}$

Substitute $6.00{\rm{ cm}}$ for $s$ and $4.40 \times {10^5}{\rm{ m/s}}$ for $v$ in the above expression of the time.

$\begin{array}{c}\\t = \frac{{6.00{\rm{ cm}}}}{{4.40 \times {{10}^5}{\rm{ m/s}}}}\\\\ = \left( {\frac{{\left( {6.00{\rm{ cm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}}{{4.40 \times {{10}^5}{\rm{ m/s}}}}} \right)\\\\ = 1.36 \times {10^{ - 7}}{\rm{ s}}\\\\{\rm{ = }}\left( {1.36 \times {{10}^{ - 7}}{\rm{ s}}} \right)\left( {\frac{{1{\rm{ ns}}}}{{1 \times {{10}^9}{\rm{ s}}}}} \right)\\\end{array}$

Further, solve for time,

$\begin{array}{c}\\t = \left( {1.36 \times {{10}^{ - 7}}{\rm{ s}}} \right)\left( {\frac{{1{\rm{ ns}}}}{{1 \times {{10}^9}{\rm{ s}}}}} \right)\\\\ = 136{\rm{ ns}}\\\end{array}$

(b)

In the vertical direction, the initial speed of the object is zero.

${v_y} = 0$

From the above case compare the linear force and the electrostatic force equation for the calculation of the acceleration of the object.

$F = {F_s}$

Substitute $ma$ for $F$ and $qE$ for ${F_s}$ in the above equation.

$\begin{array}{c}\\ma = qE\\\\a = \frac{{qE}}{m}\\\end{array}$

From the linear kinematic expression of the distance travel by the object, the displacement of proton is equal to,

$y = ut + \frac{1}{2}a{t^2}$

Substitute $\frac{{qE}}{m}$ for $a$ in the above expression of the distance in the y direction.

$y = ut + \frac{1}{2}\left( {\frac{{qE}}{m}} \right){t^2}$

Substitute $0$ for $u$ , $9.00 \times {10^3}{\rm{ N/C}}$ for $E$ , $1.6 \times {10^{ - 19}}{\rm{ C}}$ for $q$ , $1.67 \times {10^{ - 27}}{\rm{ kg}}$ for $m$ and $1.36 \times {10^{ - 7}}{\rm{ s}}$ for $t$ in the above expressions of the distance equation,

$\begin{array}{c}\\y = 0 + \frac{1}{2}\left( {\frac{{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {9.00 \times {{10}^3}{\rm{ N/C}}} \right)}}{{1.67 \times {{10}^{ - 27}}{\rm{ kg}}}}} \right){\left( {1.36 \times {{10}^{ - 7}}{\rm{ s}}} \right)^2}\\\\{\rm{ = }}\left( {7.97 \times {{10}^{ - 3}}{\rm{ m}}} \right)\left( {\frac{{1{\rm{ mm}}}}{{1 \times {{10}^{ - 3}}{\rm{ m}}}}} \right)\\\\ = 7.97{\rm{ mm}}\\\\ \approx 8.00{\rm{ mm}}\\\end{array}$

(c)

Calculate the velocity component.

The expression of the final velocity in terms of the displacement is equal to,

${v^2} = {u^2} + 2as$

Substitute $0$ for $u$ in the above expression of the $v$ ,

${v^2} = 2as$

Substitute $\frac{{qE}}{m}$ for $a$ and $y$ for $s$ in the above expression and solve for v.

$v = \sqrt {\left( 2 \right)\left( {\frac{{qE}}{m}} \right)\left( y \right)}$

Substitute $9.00 \times {10^3}{\rm{ N/C}}$ for $E$ , $1.6 \times {10^{ - 19}}{\rm{ C}}$ for $q$ , $1.67 \times {10^{ - 27}}{\rm{ kg}}$ for $m$ and $7.97 \times {10^{ - 3}}{\rm{ m}}$ for $y$ in the above expressions.

$\begin{array}{c}\\v = \sqrt {\left( 2 \right)\left( {\frac{{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {9.00 \times {{10}^3}{\rm{ N/C}}} \right)}}{{\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right)}}} \right)\left( {7.97 \times {{10}^{ - 3}}{\rm{ m}}} \right)} \\\\ = 1.17 \times {10^5}{\rm{ }}{{\rm{m}}^2}{\rm{/}}{{\rm{s}}^{\rm{2}}}\\\end{array}$

From the provided question the velocity of the proton is in horizontal direction ant this is constant. Thus the horizontal component of the velocity is equal to,

${v_x} = 4.40 \times {10^5}{\rm{ m/s}}$

The expression of the velocity is equal to,

$v = {v_x}i + {v_y}j$

Substitute $4.40 \times {10^5}{\rm{ m/s}}$ for ${v_x}$ and $1.17 \times {\kern 1pt} {10^5}{\rm{ m/s}}$ for ${v_y}$ in the above expression of the velocity,

$v = \left( {4.40 \times {{10}^5}{\rm{ m/s}}} \right)\hat i + \left( {1.17 \times {\kern 1pt} {{10}^5}{\rm{ m/s}}} \right)\hat j$

Ans: Part a

The time required to travel horizontal distance is equal to $136{\rm{ ns}}$ .

Part b

The vertical displacement during this time interval is equal to $8.00{\rm{ mm}}$ .

Part c

The velocity component is equal to $\left( {4.40 \times {{10}^5}{\rm{ m/s}}} \right)\hat i + \left( {1.17 \times {\kern 1pt} {{10}^5}{\rm{ m/s}}} \right)\hat j$ .