Question

An electron and a proton are each placed at rest in a uniform electric field of magnitude 570 N/C. Calculate the speed of each particle 47.6 ns after being released.

electron	m/s
proton	m/s

Answer 1

Concepts and reason

The concepts required to solve the given question are Newton’s second law and the relation between the force and the electric field.

Initially, calculate the force acting on the charge due to the electric field. Later, using the Newton’s second law find the acceleration of the electron related to force.

Fundamentals

The expression for the relation between the electric field and the electrostatic force is as follows:

$F = qE$

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

The expression for the Newton’s second law is as follows:

${F_{{\rm{net}}}} = ma$

Here, ${F_{{\rm{net}}}}$ is the net force, m is the mass, and a is the acceleration of an electron.

The first equation of motion is as follows:

$v = u + at$

Here, v is the final speed, u is the initial speed, a is the acceleration, and t is the time taken.

Substitute $570{\rm{ N/C}}$ for E and $1.602 \times {10^{ - 19}}{\rm{ N/C}}$ for q in the equation $F = qE$.

$\begin{array}{c}\\F = \left( {1.602 \times {{10}^{ - 19}}{\rm{ N/C}}} \right)\left( {570{\rm{ N/C}}} \right)\\\\ = 913.14 \times {10^{ - 19}}{\rm{ N}}\\\end{array}$

Rearrange the equation ${F_{{\rm{net}}}} = ma$ for a as follows:

$a = \frac{{{F_{{\rm{net}}}}}}{m}$

Substitute $913.14 \times {10^{ - 19}}{\rm{ N}}$ and $9.11 \times {10^{ - 31}}{\rm{ kg}}$ for m in the equation $a = \frac{{{F_{{\rm{net}}}}}}{m}$.

$\begin{array}{c}\\a = \frac{{913.14 \times {{10}^{ - 19}}{\rm{ N}}}}{{9.11 \times {{10}^{ - 31}}{\rm{ kg}}}}\\\\ = 100.23 \times {10^{12}}{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Substitute 0.00 m/s for u,$100.23 \times {10^{12}}{\rm{ m/}}{{\rm{s}}^2}$ for a, and 47.6 ns for t in the equation $v = u + at$.

$\begin{array}{c}\\v = 0.00{\rm{ m/s }} + \left( {100.23 \times {{10}^{12}}{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {47.6{\rm{ ns}}} \right)\\\\ = \left( {100.23 \times {{10}^{12}}{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {47.6{\rm{ ns}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{ s}}}}{{1.00{\rm{ ns}}}}} \right)\\\\ = 4770.948 \times {10^3}{\rm{ m/s}}\\\\ = {\rm{4}}{\rm{.7}} \times {10^6}{\rm{ m/s}}\\\end{array}$

Rearrange the equation ${F_{{\rm{net}}}} = ma$ for a as follows:

$a = \frac{{{F_{{\rm{net}}}}}}{m}$

Substitute $913.14 \times {10^{ - 19}}{\rm{ N}}$ and $1.672 \times {10^{ - 27}}{\rm{ kg}}$ for m in the equation $a = \frac{{{F_{{\rm{net}}}}}}{m}$.

$\begin{array}{c}\\a = \frac{{913.14 \times {{10}^{ - 19}}{\rm{ N}}}}{{1.672 \times {{10}^{ - 27}}{\rm{ kg}}}}\\\\ = 546.13 \times {10^8}{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Substitute 0.00 m/s for u,$546.13 \times {10^8}{\rm{ m/}}{{\rm{s}}^2}$ for a, and 47.6 ns for t in the equation $v = u + at$.

$\begin{array}{c}\\v = 0.00{\rm{ m/s }} + \left( {546.13 \times {{10}^8}{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {47.6{\rm{ ns}}} \right)\\\\ = \left( {546.13 \times {{10}^8}{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {47.6{\rm{ ns}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{ s}}}}{{1.00{\rm{ ns}}}}} \right)\\\\ = 2599.60{\rm{ m/s}}\\\end{array}$

Ans:

The speed of an electron after being released is equal to ${\rm{4}}{\rm{.7}} \times {10^6}{\rm{ m/s}}$.

The speed of an electron after being released is equal to $2599.60{\rm{ m/s}}$.