Question

Two large parallel copper plates are 4.86 cm apart and have a uniform electric field of magnitude E = 5.60 N/C between them (see Figure). An electron is released from the negative plate at the same time that a proton is released from the positive plate. Neglect the force of the particles on each other and find their distance from the positive plate when they pass each other.

Answer 1

Concepts and reason

Use the concept of electric force, kinematic equations of motion, and Newton’s second law of motion to solve this problem.

Use the concept of electric force and Newton’s second law of motion to obtain the acceleration of the particle. Later use the expression for acceleration and the kinematic equation of motion to calculate the distance of the proton and electron from the positive plate and the negative plate respectively. Finally, use the expressions for the displacement of the proton and the electron and the distance between the two plates to obtain the distance from the positive plate at which the electron and proton meet.

Fundamentals

Newton’s second law of motion states that the net force on an object is directly proportional to the acceleration of the object and is given by,

$F = ma$

Here, m is the mass of the object and a is the acceleration.

The electric force is the product of charge and electric field. The electric force is given by,

$F = qE$

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

The kinematic equation that relates displacement and acceleration is,

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

Here, s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

The electric force is given by,

$F = qE$

Newton’s second law of motion is given by,

$F = ma$

Substitute $ma$ for F in $F = qE$.

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

The magnitude of charge of both the proton and electron is equal. So, substitute e for q where $e$ is the magnitude of the charge.

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

The acceleration of the proton is,

${a_p} = \frac{{eE}}{{{m_p}}}$

The kinematic equation that relates displacement and acceleration is,

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

Substitute 0 for u, ${x_p}$for s, and ${a_p}$ for a.

$\begin{array}{c}\\{x_p} = \left( 0 \right)t + \frac{1}{2}{a_p}{t^2}\\\\{x_p} = \frac{1}{2}{a_p}{t^2}\\\end{array}$

Substitute $\frac{{eE}}{{{m_p}}}$ for ${a_p}$.

${x_p} = \frac{1}{2}\left( {\frac{{eE}}{{{m_p}}}} \right){t^2}$ …… (1)

The kinematic equation that relates displacement and acceleration is,

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

Here, s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

The acceleration of the electron is,

${a_e} = \frac{{eE}}{{{m_e}}}$

The initial velocity of the electron is zero.

Substitute 0 for u, ${x_e}$ for s, and ${a_e}$ for a.

$\begin{array}{c}\\{x_e} = \left( 0 \right)t + \frac{1}{2}{a_e}{t^2}\\\\ = \frac{1}{2}{a_e}{t^2}\\\end{array}$

Substitute $\frac{{eE}}{{{m_e}}}$ for ${a_e}$.

${x_e} = \frac{1}{2}\left( {\frac{{eE}}{{{m_e}}}} \right){t^2}$

When the electron and proton meet each other,

$\begin{array}{c}\\{x_p} + {x_e} = D\\\\{x_e} = D - {x_p}\\\end{array}$

Substitute $D - {x_p}$ for ${x_e}$ in ${x_e} = \frac{1}{2}\left( {\frac{{eE}}{{{m_e}}}} \right){t^2}$.

$D - {x_p} = \frac{1}{2}\left( {\frac{{eE}}{{{m_e}}}} \right){t^2}$ …… (2)

Divide equation (2) by equation (1).

$\begin{array}{c}\\\frac{{D - {x_p}}}{{{x_p}}} = \frac{{\left( {\frac{1}{2}\left( {\frac{{eE}}{{{m_e}}}} \right){t^2}} \right)}}{{\left( {\frac{1}{2}\left( {\frac{{eE}}{{{m_p}}}} \right){t^2}} \right)}}\\\\\frac{D}{{{x_p}}} - 1 = \frac{{\left( {\frac{1}{{{m_e}}}} \right)}}{{\left( {\frac{1}{{{m_p}}}} \right)}}\\\\\frac{D}{{{x_p}}} = \left( {\frac{{{m_p}}}{{{m_e}}}} \right) + 1\\\\{x_p} = \frac{D}{{\frac{{{m_p}}}{{{m_e}}} + 1}}\\\end{array}$

The distance of the particles from the positive plate when they meet is,

${x_p} = \frac{D}{{\frac{{{m_p}}}{{{m_e}}} + 1}}$

Substitute 4.86 cm for D, $9.11 \times {10^{ - 31}}{\rm{ kg}}$ for ${m_e}$, and $1.672 \times {10^{ - 27}}{\rm{ kg}}$ for ${m_p}$.

$\begin{array}{c}\\x = \frac{{\left( {4.86{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)} \right)}}{{\left( {\frac{{1.672 \times {{10}^{ - 27}}{\rm{ kg}}}}{{9.11 \times {{10}^{ - 31}}{\rm{ kg}}}} + 1} \right)}}\\\\ = 2.646 \times {10^{ - 5}}{\rm{ m}}\\\end{array}$

Round off to three significant figures.

The distance from the positive plate at which the proton and the electron meet is $2.65 \times {10^{ - 5}}{\rm{ m}}$.

Ans:

The distance from the positive plate at which the proton and the electron meet is ${\bf{2}}{\bf{.65 \times 1}}{{\bf{0}}^{{\bf{ - 5}}}}{\bf{ m}}$.