Question

A.) If two electrons are each 2.50×10⁻¹⁰ m from a proton, as shown in the figure, find the magnitude and of the net electrical force they will exert on the proton.

B.) Find the direction of the net electrical force electrons will exert on the proton.

Answer 1

Concepts and reason

The concept used to solve the problem is Coulomb’s law.

The magnitude of the net electrical force that will be exert on the proton can be calculated by using the Coulomb’s law. The magnitude of the net electric force can be calculated by using the relationship among the charges, force constant and the inverse of the square of the distance between the charges.

The x and y components of the force exerted on the proton can be calculated by using the vector resolution method.

The net force exerted on the proton can be calculated by using the Pythagoras theorem.

The direction of the net force exerted on the proton can be calculated by using the arctan of the ration of y- component of force to the x-component of force.

Fundamentals

Coulomb’s Law: The magnitude of net electrical force between charges is directly proportional to the product of charges and inversely proportional to the square of the distance between charges.

The magnitude of the net electrical force between charges is calculated as follows,

$F = \left( {\frac{{k{q_1}{q_2}}}{{{r^2}}}} \right)$

Here, k is the force constant, q₁ is the charge of proton, q₂ is the charge of electron on the x-axis, and r is the distance between the charges.

The direction of the net force vector can be calculated as follows:

$\tan \theta = \frac{{{F_y}}}{{{F_x}}}$

Here, F_y is the component of force in y-direction and F_x is the component of force in x-direction.

(A)

Consider the formula for calculating the magnitude of force acting on the proton.

The formula for calculating the magnitude of force acting on the proton from electron lies in the x direction as follows:

${F_{21}} = \frac{{k\left( {{q_1}{q_2}} \right)}}{{{r^2}}}$

Here, k is the force constant, q₁ is the charge of proton, q₂ is the charge of electron on the x-axis, F₂₁ is the force acting on proton due to electron lying on x-axis, and r is the distance between the charges.

Substitute $\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)$for k, $\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)$ for q₁, and $\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)$for q₂ in equation as follows:

$\begin{array}{l}\\{F_{21}} = \frac{{k\left( {{q_1}{q_2}} \right)}}{{{r^2}}}\\\\ = \left( {\frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}}{{{{\left( {2.50 \times {{10}^{ - 10}}{\rm{ m}}} \right)}^2}}}} \right)\\\\ = 3.69 \times {10^{ - 9}}{\rm{ N}}\\\end{array}$

Find the magnitude of force acting on the proton from electron which is placed at $65^\circ$.

The formula for calculating the magnitude of force acting on the proton from electron lies at $65^\circ$:

${F_{31}} = \frac{{k\left( {{q_1}{q_3}} \right)}}{{{r^2}}}$

Here, k is the force constant, q₁ is the charge of proton, q₃ is the charge of electron lies at $65^\circ$, F₃₁ is the force acting on proton due to electron, and r is the distance between the charges.

Substitute $\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)$for k, $\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)$ for q₁, and $\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)$for q₂ in equation as follows:

$\begin{array}{l}\\{F_{31}} = \frac{{k\left( {{q_1}{q_2}} \right)}}{{{r^2}}}\\\\ = \left( {\frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}}{{{{\left( {2.50 \times {{10}^{ - 10}}{\rm{ m}}} \right)}^2}}}} \right)\\\\ = 3.69 \times {10^{ - 9}}{\rm{ N}}\\\end{array}$

Find the x component of forces exerted on the proton by the vector resolution method.

The x component of force exerted on the proton by the electron placed at $65^\circ$ is,

${F_{31}}_{,{\rm{x}}} = {F_{31}}\cos \left( {{{65}^ \circ }} \right)$

Here, ${F_{31,x}}$is the x-component of force exerted on the proton by the electron placed at $65^\circ$ and F₃₁ is the force exerted on the proton by the electron.

Substitute $\left( {3.6864 \times {{10}^{ - 9}}{\rm{ N}}} \right)$ for ${F_{31}}$ and $\left( {0.422618^\circ } \right)$ for $\left( {{\rm{cos}}65^\circ } \right)$in equation as follows:

$\begin{array}{c}\\{F_{31}}_{,{\rm{x}}} = {F_{31}}\cos \left( {{{65}^ \circ }} \right)\\\\ = \left( {3.6864 \times {{10}^{ - 9}}\,{\rm{N}}} \right)\left( {0.422618^\circ } \right)\\\\ = 1.56 \times {10^{ - 9}}{\rm{ N}}\\\end{array}$

Find the y component of force acting on the proton by using vector resolution method.

The y-component of force exerted on the proton by the electron placed at $65^\circ$ is,

${F_{31}}_{,{\rm{y}}} = {F_{31}}\sin \left( {{{65}^ \circ }} \right)$

Here, ${F_{31,y}}$is the y-component of force exerted on the proton by the electron placed at $65^\circ$ and F₃₁ is the force exerted on the proton by the electron.

Substitute $\left( {3.6864 \times {{10}^{ - 9}}{\rm{ N}}} \right)$ for ${F_{31}}$ and $\left( {0.9063^\circ } \right)$ for $\left( {{\rm{sin }}65^\circ } \right)$in equation as follows:

$\begin{array}{c}\\{F_{31}}_{,{\rm{y}}} = {F_{31}}\sin \left( {{{65}^ \circ }} \right)\\\\ = \left( {3.6864 \times {{10}^{ - 9}}\,{\rm{N}}} \right)\left( {0.422618^\circ } \right)\\\\ = 3.34 \times {10^{ - 9}}{\rm{ N}}\\\end{array}$

Find the x component of force acting on the proton by using vector resolution method.

The x-component of force exerted on the proton by the electron placed along x-axis is as follows:

${F_{21,{\rm{x}}}} = {F_{21}}\cos \left( {0^\circ } \right)$

Here, ${F_{21}}$ is the force exerted on the proton due to electron placed along x axis.

Substitute $\left( {3.6864 \times {{10}^{ - 9}}{\rm{ N}}} \right)$in above equation,

$\begin{array}{c}\\{F_{21,{\rm{x}}}} = {F_{21}}\cos \left( {0^\circ } \right)\\\\ = \left( {3.6864 \times {{10}^{ - 9}}{\rm{ N}}} \right)\\\end{array}$

The y component of force exerted on the proton by the electron placed along x-axis is as follows:

$\begin{array}{c}\\{F_{21,{\rm{y}}}} = {F_{21}}\sin \left( {0^\circ } \right)\\\\ = 0{\rm{ N}}\\\end{array}$

Calculate the x and y components of the net force exerted on the proton.

The magnitude of the net force exerted on the proton in the x direction is calculated as follows:

${F_{1,{\rm{x}}}} = {F_{31,{\rm{x}}}} + {F_{21,{\rm{x}}}}$

Here, ${F_{1,x}}$is the magnitude of total force exerted on the proton in the x direction, ${F_{31,{\rm{x}}}}$is the magnitude of total force exerted on the proton by the electron placed at $65^\circ$, and ${F_{21,{\rm{x}}}}$ is the magnitude of force exerted on the proton by the electron placed along x axis.

Substitute $\left( {1.558 \times {{10}^{ - 9}}{\rm{ N}}} \right)$ for ${F_{31,{\rm{x}}}}$ and $\left( {3.6864 \times {{10}^{ - 9}}{\rm{ N}}} \right)$ for ${F_{21,{\rm{x}}}}$ as follows:

$\begin{array}{c}\\{F_{1,{\rm{x}}}} = {F_{31,{\rm{x}}}} + {F_{21,{\rm{x}}}}\\\\ = \left( {1.558 \times {{10}^{ - 9}}{\rm{ N}}} \right) + \left( {3.6864 \times {{10}^{ - 9}}{\rm{ N}}} \right)\\\\ = 5.2444 \times {10^{ - 9}}{\rm{ N}}\\\end{array}$

The magnitude of the y components of the net force exerted on the proton is calculated as follows:

${F_{1,y}} = {F_{31,y}} + {F_{21,y}}$

Here, ${F_{1,y}}$is the magnitude of total force exerted on the proton in the y direction, ${F_{31,y}}$is the magnitude of total force exerted on the proton by the electron placed at $65^\circ$, and ${F_{21,y}}$ is the magnitude of force exerted on the proton by the electron placed along y axis.

Substitute $\left( {3.34 \times {{10}^{ - 9}}{\rm{ N}}} \right)$ for ${F_{31,y}}$ and $\left( {0{\rm{ N}}} \right)$ for $\left( {{F_{21,{\rm{y}}}}} \right)$ as follows:

$\begin{array}{c}\\{F_{1,{\rm{y}}}} = {F_{31,{\rm{y}}}} + {F_{21,{\rm{y}}}}\\\\ = \left( {3.34 \times {{10}^{ - 9}}{\rm{ N}}} \right) + \left( {0{\rm{ N}}} \right)\\\\ = \left( {3.34 \times {{10}^{ - 9}}{\rm{ N}}} \right)\\\end{array}$

Calculate the magnitude of net force exerted on the proton.

The magnitude of net force exerted on the proton by using Pythagoras theorem is,

${F_1} = \sqrt {{{\left( {{F_{1,x}}} \right)}^2} + {{\left( {{F_{1,y}}} \right)}^2}}$

Here, ${F_{1,x}}$is the magnitude of total force exerted on the proton in the x direction and ${F_{1,y}}$is the magnitude of total force exerted on the proton in the y direction.

Substitute $\left( {5.2444 \times {{10}^{ - 9}}{\rm{ N}}} \right)$ for $\left( {{F_{1,x}}} \right)$ and $\left( {3.34 \times {{10}^{ - 9}}{\rm{ N}}} \right)$ for $\left( {{F_{1,y}}} \right)$ as follows,

$\begin{array}{c}\\{F_1} = \sqrt {{{\left( {{F_{1,x}}} \right)}^2} + {{\left( {{F_{1,y}}} \right)}^2}} \\\\ = \sqrt {{{\left( {5.2444 \times {{10}^{ - 9}}{\rm{ N}}} \right)}^2} + {{\left( {3.34 \times {{10}^{ - 9}}{\rm{ N}}} \right)}^2}} \\\\ = 6.22 \times {10^{ - 9}}{\rm{ N}}\\\end{array}$

(b)

Calculate the direction of net force exerted on the proton.

The direction of net force exerted on the proton is calculated by dividing the y- component of net force to the x- component of net force.

The direction of net force is,

$\tan \theta = \left( {\frac{{{F_{1,y}}}}{{{F_{1,x}}}}} \right)$

Here, ${F_{1,y}}$ is the magnitude of net force exerted on the proton in y direction and ${F_{1,x}}$ is the magnitude of net force exerted on the proton in the x-direction.

Substitute $\left( {3.34 \times {{10}^{ - 9}}{\rm{ N}}} \right)$ for ${F_{1,y}}$ and $\left( {5.2444 \times {{10}^{ - 9}}{\rm{ N}}} \right)$ for ${F_{1,x}}$ as follows,

$\begin{array}{c}\\\tan \theta = \left( {\frac{{{F_{1,y}}}}{{{F_{1,x}}}}} \right)\\\\ = \left( {\frac{{3.34 \times {{10}^{ - 9}}{\rm{ N}}}}{{5.2444 \times {{10}^{ - 9}}{\rm{ N}}}}} \right)\\\\ = 0.6369\\\end{array}$

$\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {0.6369^\circ } \right)\\\\ = 32.5^\circ \\\end{array}$

Ans: Part a

The magnitude of the net electrical force that will be exerted on the proton is $\left( {6.22 \times {{10}^{ - 9}}{\rm{ N}}} \right)$.

Part b

The direction of the net force exerted on the proton is $32.5^\circ$.