Question

9.
Two electrons are fixed 2 cm apart. Another electron is shot from infinity and stops midway between
the two. What is the initial speed?​

Answer 1

✪ᴀɴsᴡᴇʀ✪

• $\boxed{\red{v_0 = \sqrt{\frac{8ke^2}{m_er}}}}$

• $\boxed{\blue{v_0 = 318.24\:m/s}}$

☆ɢɪᴠᴇɴ☆

• Two electrons are fixed at $r = 2$ cm apart.

• Another electron is shot from infinity.

• The electron stops midway between the two fixed electrons

✫ᴛᴏ ғɪɴᴅ✫

• Initial Speed of the electron shot.

⍟ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ⍟

Let

• $v_0$ be the initial speed of electron.

• $m_e$ be mass of electron.

• $e$ be charge of an electron.

Assume that the electron is moving at non-relativistic speed.

Expression of Initial Velocity

By conservation of energy,

$\implies (K. E.)_i+(P. E.)_i$

$\:\:\:\:\:\:= (K. E.)_f + (P. E.)_f$

$\implies \frac{1}{2}m_e{v_0}^2 + 0$

$\:\:\:\:\:\: = 0+ \frac{k(e)(e)}{(\frac{r}{2})} +\frac{k(e)(e)}{(\frac{r}{2})}$

$\implies \frac{1}{2}m_e{v_0}^2 = \frac{2ke^2}{r} +\frac{2ke^2}{r}$

$\implies \frac{1}{2}m_e{v_0}^2 = \frac{4ke^2}{r}$

$\implies m_e{v_0}^2 = \frac{8ke^2}{r}$

$\implies {v_0}^2 = \frac{8ke^2}{m_er}$

$\implies v_0 = \sqrt{\frac{8ke^2}{m_er}}$

Numerical Value

Here,

• $k\:\: = \:\:9\times 10^9\:Nm^2/C^2$
• $e\:\:\:= - 1.6\times 10^{-19}\:C$
• $m_e = \:\:9.1\times 10^{-31}\:Kg$
• $r \:\:\:=\:\: 2 \times 10^{-2}\:m$

Putting these values in the expression,

$\implies v_0 = \sqrt{8\frac{(9\times 10^9)\times (- 1.6\times 10^{-19})^2}{(9.1\times 10^{-31})\times (2 \times 10^{-2}) }}$

$\implies v_0 = \sqrt{\frac{4\times 9\times 2.56}{9.1}\times 10^{9-38+31+2}}$

$\implies v_0 = \sqrt{10.1275\times 10^{4}}$

$\implies v_0 = 3.1824\times 10^{2}$

$\implies v_0 = 318.24\: m/s$