Question

What potential difference must be applied to produce an electric field that can accelerate an electron to one length the velocity of light-

Answer 1

Given:

An electron is accelerated to velocity of light in one metre length.

To find:

Potential difference applied across the terminals.

Calculation:

Let Potential Difference be V.

So, Electrostatic Field Intensity be E

$\therefore \: E = \dfrac{V}{l}$

Now force experienced will be ;

$\therefore \: force = Ee = \dfrac{Ve}{l}$

Hence , acceleration of electron will be :

$\therefore \: a = \dfrac{force}{mass} = \dfrac{Ve}{ml}$

Now , applying the third equation of kinematics:

${v}^{2} = {u}^{2} + 2al$

$= > {c}^{2} = {(0)}^{2} + \{ 2 \times (\dfrac{Ve}{ml} ) \times l \}$

$= > {c}^{2} = 2 \times (\dfrac{Ve}{ml} ) \times l$

$= > {c}^{2} = \dfrac{2Ve}{m}$

$= > V = \dfrac{m {c}^{2} }{2e}$

So, final answer is:

$\boxed{ \sf{ \blue{ \large{V = \dfrac{m {c}^{2} }{2e} }}}}$