Question

An electron with speed 2.50×10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.17×10^4 N/C.
A. How far will the electron travel before it stops?
B. How much time will elapse before it returns to its starting point?

answer either one w/ explanation for max rating. Thanks!

Answer 1

Concepts and reason

The concepts used to solve this problem are electric force, newton’s law, and equations of motion.

Find the acceleration of the electron using the newton’s second law.

Find the distance travelled by the electron before it comes to rest and time taken for it using the equations of motion.

Fundamentals

The electrons travelling in external electric field will experience electric force. Electric force is a force, which is experienced by a charged particle when it is kept in an electric field.

The expression for the electric force is,

${F_E} = qE$

Here, electric force is ${F_E}$ , charge is $q$ , and electric field is $E$ .

Newton’s second law states that, a particle experiencing a net external force will get accelerated by that force and the acceleration is proportional to the force.

Expression for newton’s second law is,

$F = ma$

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

Equations of motion are set of mathematical relations, which describe the motion of particles. These are set of equations of variables as a function of time.

The equations of motion are expressed as,

$\begin{array}{l}\\{v^2} = {u^2} + 2as\\\\v = u + at\\\\s = ut + \left( {1/2} \right)a{t^2}\\\end{array}$

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

(A)

The expression for the electric force is,

${F_E} = qE$

Expression for Newton’s second law is,

$F = ma$

Electric force is the only force acting on electron. Applying newton’s second law to electron, we get

${F_E} = ma$

Substitute $qE$ for ${F_E}$ in the above expression.

$qE = ma$

Form the above equation, the acceleration of electron is expressed as,

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

Substitute $1.602 \times {10^{ - 19}}\,{\rm{C}}$ for $q$ , for $E$ , and $9.11 \times {10^{ - 31}}\,{\rm{kg}}$ for $m$ .

$\begin{array}{c}\\a = \frac{{\left( {1.602 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\left( {1.17 \times {{10}^4}\,{\rm{N/C}}} \right)}}{{\left( {9.11 \times {{10}^{ - 31}}\,{\rm{kg}}} \right)}}\\\\ = 20.575 \times {10^{14}}\,{\rm{m/}}{{\rm{s}}^2}\\\end{array}$

As the electron stops after some time, this acceleration should be negative.

Therefore, the deceleration of electron is $a = - 20.575 \times {10^{14}}\,{\rm{m/}}{{\rm{s}}^2}$ .

The electron gets decelerated and it stops eventually. Equation of motion required to calculate the distance it travels before getting stopped is,

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

Substitute $0\,{\rm{m/s}}$ for $v$ , $2.50 \times {10^7}\,{\rm{m/s}}$ for $u$ , and $- 20.575 \times {10^{14}}\,{\rm{m/}}{{\rm{s}}^2}$ for $a$ .

$\begin{array}{c}\\s = - \frac{{{{\left( {2.50 \times {{10}^7}\,{\rm{m/s}}} \right)}^2}}}{{2\left( { - 20.575 \times {{10}^{14}}\,{\rm{m/}}{{\rm{s}}^2}} \right)}}\\\\ = 0.152\,{\rm{m}}\\\end{array}$

[Part A]

Part A

Part A

(B)

The time taken by the electron to come to rest is,

$v = u + at$

Substitute $0\,{\rm{m/s}}$ for $v$ , $2.50 \times {10^7}\,{\rm{m/s}}$ for $u$ , and $- 20.575 \times {10^{14}}\,{\rm{m/}}{{\rm{s}}^2}$ for $a$ .

$\begin{array}{c}\\t = - \frac{{\left( {2.50 \times {{10}^7}\,{\rm{m/s}}} \right)}}{{\left( { - 20.575 \times {{10}^{14}}\,{\rm{m/}}{{\rm{s}}^2}} \right)}}\\\\ = 1.215 \times {10^{ - 8}}\,{\rm{s}}\\\end{array}$

As the electron returns back to the starting point, the total time taken is,

$T = 2t$

Here, the time period of motion of electron is $T$ and the time taken by the electron before it came to rest is $t$ .

Substitute $1.215 \times {10^{ - 8}}\,{\rm{s}}$ for $t$

$\begin{array}{c}\\T = 2\left( {1.215 \times {{10}^{ - 8}}\,{\rm{s}}} \right)\\\\ = 2.43 \times {10^{ - 8}}\,{\rm{s}}\\\end{array}$

Ans: Part A

Thus, the distance travelled by electron before it comes to rest is ${\bf{0}}{\bf{.152}}\,{\bf{m}}$ .