Question

An electron is located at the origin of the coordinates, and a second electron is brought to a position 2 Å from the origin in the 1y direction.

(a) Calculate the force on the electron at the origin, giving the components Fx and Fy.

(b) Calculate the potential energy of the two electrons.

Short answer

(a)

The force between the electrons at the origin of the components${\text{F}}_{\text{x}}{\text{andF}}_{\text{y}}$is$5.77\times {10}^{-9}\text{N}$

Step-by-step solution

a) Force.5.5.

The force is the external cause that tends tochange the state of the bodyonce applied. Force is thevector quantity. It hasdirectionas well asmagnitude.

The formula to calculate the force between the two charged particles separated by a distance is given below.

$\text{F(r)}=\frac{{\text{q}}_1{\text{q}}_2}{4\pi {\epsilon }_0{\text{r}}^2}$

Here,

F(r) is the force.

${\text{q}}_{\text{1}}{\text{,q}}_{\text{2}}$are the charges of the two particles of the components ${\text{F}}_{\text{x}}{\text{andF}}_{\text{y}}$, respectively.

${\epsilon }_o$is the epsilon value$8.854\times {10}^{-12}{\text{C}}^2{\text{J}}^{-1}{\text{m}}^{-1}$

r is the distance between the electrons.

Calculate the electron's force at the origin.

a

Put the values in the above formula.

$\begin{array}{rcl}\text{F(r)}& =& \frac{(-1.602\times {10}^{-19}{\text{C)}}^2}{\left(4\right)(3.146)(8.854\times {10}^{-12}{\text{C}}^2{\text{J}}^{-1}\text{m)(}2.0\times {10}^{-10}{\text{m)}}^2}\\ & =& \frac{2.57\times {10}^{-38}{\text{C}}^2}{4.45\times {10}^{-30}{\text{C}}^2{\text{J}}^{-1}\text{m}}\\ & =& 5.77\times {10}^{-9}{\text{Jm}}^{-1}\text{(Since1N.m=1J)}\\ & & {\text{=5.77×10}}^{-9}\text{-N}\end{array}$

Therefore, the force between the electrons is$\mathbf{5}\mathbf{.}\mathbf{77}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{\text{N}}$