Question

An electron and a proton are fixed at a separation distance of 949 nm. Find the magnitude and direction of the electric field at their midpoint.

Which is the magnitude

Answer 1

Concepts and reason

The concept required to solve the given problem is electric field due to a point charge and super position principle of electric field.

Calculate the net electric field at the point midway between the electron and the proton by calculating the vector sum of the two fields at that point.

Fundamentals

Electric field: It is the region around a charge where its influence can be felt.

The expression for the electric field at a distance r from a charge is,

$E = \frac{{{k_e}q}}{{{r^2}}}$

Here, ${k_e}$ is the Coulomb’s constant, $q$ is the charge and $r$ is the distance of the point from the charge where the electric field needs to be calculated.

Super position principle of electric field:

The net electric field at a point due to two or more charges is equal to the vector sum of the electric fields at that point due to each charge.

The electric field midway between the charges due to the proton is,

${E_A} = \frac{{{k_e}{q_A}}}{{{r_A}^2}}$

Here, ${q_A}$ is the charge on proton, ${r_A}$ is the distance of the proton from the point where the electric field wish to calculate.

The electric field midway between the charges due to the electron is,

${E_B} = \frac{{{k_e}{q_B}}}{{{r_A}^2}}$

The direction of the electric field due to electron and the proton at a point midway between the particles has same direction. Hence, the two fields will be added up at that point.

The magnitude of the net electric field at a point midway between the electron and proton is,

$E = {E_A} + {E_B}$

Substitute $\frac{{{k_e}{q_A}}}{{{r_A}^2}}$ for ${E_A}$ and $\frac{{{k_e}{q_B}}}{{{r_B}^2}}$ for ${E_B}$ in the above equation.

$E = \frac{{{k_e}{q_A}}}{{{r_A}^2}} + \frac{{{k_e}{q_B}}}{{{r_A}^2}}$

Since the two charges are equal, substitute $q$ for${q_A}{\rm{ and }}{q_B}$and $r$ for${r_A}{\rm{ and }}{r_B}$.

$\begin{array}{c}\\E = \frac{{{k_e}q}}{{{r^2}}} + \frac{{{k_e}q}}{{{r^2}}}\\\\ = \frac{{2{k_e}q}}{{{r^2}}}\\\end{array}$

Since the point where the electric field needs to be calculated is midway between the charges, hence the distance $r$ is,

$\begin{array}{c}\\r = \frac{{{\rm{949 nm}}}}{2}\\\\ = 474.5{\rm{ nm}}\left( {\frac{{{{10}^{ - 9}}{\rm{ m}}}}{{1{\rm{ nm}}}}} \right)\\\\ = 474.5 \times {10^{ - 9}}{\rm{ m}}\\\end{array}$

Substitute $1.6 \times {10^{ - 19}}{\rm{ C}}$ for $q$, $474.5 \times {10^{ - 9}}{\rm{ m}}$ for $r$ and $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/ }}{{\rm{C}}^2}$for ${k_e}$ in the above equation.

$\begin{array}{c}\\E = \frac{{2\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/ }}{{\rm{C}}^2}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}}{{{{\left( {474.5 \times {{10}^{ - 9}}{\rm{ m}}} \right)}^2}}}\\\\ = 1.28 \times {10^4}{\rm{ N / C}}\\\end{array}$

Ans:

The magnitude of electric field at a point midway between electron and proton is $1.28 \times {10^4}{\rm{ N / C}}$.