Question

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, travelling 4.50 m in the first 3.00 μs after it is released. a. what are the magnitude and direction of the electric field? b. are we justified in ignoring the effects of gravity? Justify your answer quantitatively.

Answer 1

Concepts and reason

The concept used to solve this problem is the Kinematic equation of motion, Newton’s second law and force due to electric field.

Initially, from the kinematic equation to calculate the acceleration of the electron.

Later, substitute the force due to electric field in the Newton’s second law and solve for the electric field. Substitute the values in the electric field equation to find the magnitude of the electric field.

Finally, compare the acceleration with acceleration due to gravity.

Fundamentals

The Kinematics equation of motion is,

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

Here, $a$is the acceleration, $t$ is the time, $s$ is the travelled distance and $u$ is the initial velocity.

The expression of Newton Second law of motion is,

$F = ma$

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

The electric force as,

$F = qE$

Here, $E$is the electric field and $q$is the charge of the electron

The direction of electric field is opposite to the direction of acceleration of the negative charge.

The acceleration due to gravity is,

$g = 9.8{\rm{ m/}}{{\rm{s}}^2}$.

(a)

Substitute $0$ for $u$ in the kinematics equation $s = ut + \frac{1}{2}a{t^2}$to solve for the acceleration of the electron.

$\begin{array}{c}\\s = 0 + \frac{1}{2}a{t^2}\\\\a = \frac{{2s}}{{{t^2}}}\\\end{array}$

Substitute 4.5 m for $s$, 3 ${\rm{\mu s}}$ for $t$ in the above equation.

$\begin{array}{c}\\a = \frac{{2\left( {4.5{\rm{ m}}} \right)}}{{{{\left( {3{\rm{ \mu s}}} \right)}^2}}}\\\\ = \frac{{2\left( {4.5{\rm{ m}}} \right)}}{{{{\left( {3{\rm{ \mu s}}\left( {\frac{{{{10}^{ - 6}}{\rm{ s}}}}{{1{\rm{ \mu s}}}}} \right)} \right)}^2}}}\\\\\;\; = 1.0 \times {10^{12}}\;{\rm{m/}}{{\rm{s}}^2}\\\end{array}$

Substitute $qE$ for $F$ in the Newton’s second law $F = ma$ and solve for electric field.

$\begin{array}{c}\\qE = ma\\\\\;E = \frac{{ma}}{q}\\\end{array}$

Substitute $1.0 \times {10^{12}}\;{\rm{m/}}{{\rm{s}}^2}$for $a$, $9.1 \times {10^{ - 31}}{\rm{kg}}$ for $m$, and $1.6 \times {10^{ - 19}}{\rm{C}}$ for $q$.

$\begin{array}{l}\\E = \frac{{\left( {9.1 \times {{10}^{ - 31}}{\rm{ kg}}} \right)\left( {1.0 \times {{10}^{12}}{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{1.6 \times {{10}^{ - 19}}{\rm{ C}}}}\\\\\;\;\; = 5.687\;{\rm{N/C}}\\\end{array}$

The electron is accelerating upward, so the direction of electric field is downward.

(b)

The acceleration of the electron is $a = {10^{12}}{\rm{ m/}}{{\rm{s}}^2}$ which is very large as compared to the acceleration due to gravity $g = 9.8{\rm{ m/}}{{\rm{s}}^2}$. So, the effect of gravity is ignored correctly as it will barely effect the value of acceleration.

Ans: Part a

The magnitude of the electric field is $E = 5.687\;{\rm{N/C}}$ and direction is downward.