# Two electrons are fixed 2cm apart. Another electron is shot from infinity and comes to rest...

## Question:

Two electrons are fixed 2cm apart. Another electron is shot from infinity and comes to rest midway between the two fixed electrons.

A) Find the energy of the system initially ( unknown quantity leave as a variable)

B) Find the final energy of the system? ( unknown quantity leave as a variable)

C. find the initial speed of the electron.

## Kinetic Energy:

When an object is traveling from one position to another position due to this movement, it posses some energy into it. This energy of the object due to his motion is known as the kinetic energy of the object. The kinetic energy is directly related to the speed of the object.

Given data:

• Distance between two electrons, {eq}d = 2\;{\rm{cm}} = 0.02\;{\rm{m}} {/eq}

{eq}r = \dfrac{{0.02}}{2} = 0.01\;{\rm{m}} {/eq}

Part A)

The initial energy of the system is:

{eq}{E_i} = \dfrac{1}{2}m{v^2} {/eq}

Here, {eq}m = 9.1 \times {10^{ - 31}}\;{\rm{kg}} {/eq} is the mass of the electron.

{eq}\begin{align*} {E_i} &= \dfrac{{9.1 \times {{10}^{ - 31}}}}{2}{v^2}\\ {E_i} &= 4.55 \times {10^{ - 31}}{v^2}\;{\rm{J}} \end{align*} {/eq}

Therefore, the initial energy of the system is {eq}4.55 \times {10^{ - 31}}{v^2}\;{\rm{J}} {/eq} .

Part B)

The final energy of the system is:

{eq}{E_f} = KE + PE {/eq}

The final kinetic energy of the system will be zero because the final velocity of the electron is zero. Thus the final energy of the system is:

{eq}\begin{align*} {E_f} &= PE\\ {E_f} &= \dfrac{{k{q^2}}}{r} \end{align*} {/eq}

Here, {eq}q = - 1.6 \times {10^{ - 19}}\;{\rm{C}} {/eq} is the charge on electron and {eq}k = 9 \times {10^9}\;{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}} {/eq} is the proportionality constant.

{eq}\begin{align*} {E_f} &= \dfrac{{9 \times {{10}^9} \times {{\left( { - 1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{0.01}}\\ {E_f} &= 2.3 \times {10^{ - 26}}\;{\rm{J}} \end{align*} {/eq}

Therefore, the final energy of the system is {eq}2.3 \times {10^{ - 26}}\;{\rm{J}} {/eq} .

Part C)

By applying conservation of energy at final condition:

{eq}K{E_i} = \Delta PE {/eq}

Here, change in potential energy is:

{eq}\begin{align*} \Delta PE &= \dfrac{{k2{q^2}}}{r}\\ \Delta PE &= \dfrac{{9 \times {{10}^9} \times 2 \times {{\left( { - 1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{0.01}}\\ \Delta PE &= 4.6 \times {10^{ - 26}}\;{\rm{J}} \end{align*} {/eq}

Thus,

{eq}\begin{align*} \dfrac{1}{2}m{v_i}^2 &= 4.6 \times {10^{ - 26}}\\ {v_i}^2 &= \dfrac{{2 \times 4.6 \times {{10}^{ - 26}}}}{m}\\ {v_i} &= \sqrt {\dfrac{{2 \times 4.6 \times {{10}^{ - 26}}}}{{9.1 \times {{10}^{ - 31}}}}} \\ {v_i} &= 317.79\;{\rm{m/s}} \simeq {\rm{3}}{\rm{.2}} \times {\rm{1}}{{\rm{0}}^2}\;{\rm{m/s}} \end{align*} {/eq}

Therefore, the initial speed of the electron is {eq}{\rm{3}}{\rm{.2}} \times {\rm{1}}{{\rm{0}}^2}\;{\rm{m/s}} {/eq}. 