Question

Two electrons are fixed $2.0cm$ apart. Another electron is shot from infinity and stops midway between the two. What is its initial speed?

• A. $3.18\times {10}^{2}mm/s$
• B. $3.18\times {10}^{2}cm/s$
• C. $3.18\times {10}^{2}m/s$
• D. $3.18\times {10}^{2}km/s$

Answer 1

The correct option is C $3.18\times {10}^{2}m/s$

Given two electrons are $2$ cm apart and are fixed.

Another electron is shot from infinity and stops midway between the two.

We have to find the initial speed of this electron.

Since the third electron is infinitely far away from the other electrons, it posses only kinetic energy. Since this electron comes to rest so, $V=0$ m/s (final).

So in this position it will have electrical potential energy,

$V_e=\frac{k{q}_1{q}_2}{d/2}$; $V_e=\frac{k{q}_1{q}_2}{d/2}$

Since, $q_1=q_2=q=$ charge on electron

$d=$ distance $=2$ cm between two fixed electrons.

So, overall finall potential energy $=\frac{k{q}^2}{d/2}+\frac{k{q}^2}{d/2}=\frac{4k{q}^2}{d}$

From conservation of energy $K.E=P.E$

So, $\frac{1}{2}m{v}^2=\frac{4k{q}^2}{d}$

Initial velocity $=V_0=\sqrt{\frac{8k{q}^2}{md}}$; $m=$ mass of electron

So, $=V_0=\sqrt{\frac{8\times (1.6\times {10}^{-9}{)}^2\times 9\times {10}^9}{9.1\times {10}^{-31}\times 2\times {10}^{-2}}}$ m/s

$V_0=3.18\times {10}^2$ m/s