Answer:
Velocity of the electron mid way = v = 6.47×10⁶ m/s
Explanation:
Electric potential varies with distance as [tex]V(x) = C x^\frac{4}{3}[/tex]
At a distance 5 mm , the constant C is evaluated.
C = [tex]V/x^\frac{4}{3}[/tex]
= [tex]300/(0.005)^\frac{4}{3}[/tex]
= 3.50×10⁵
Now at the mid point, x = 2.5 mm = 0.0025 m
Potential = V' = [tex]C x^\frac{4}{3}[/tex]
= [tex]3.50\times 10^5\times (0.0025)^\frac{4}{3}[/tex]
= 119 V
Kinetic energy per unit charge is equal to the electric potential.
or 0.5 m v^2 = V' q
Here m is the mass of the electron and q is the charge of the electron.
m= 9.1×10⁻³¹ kg
q = 1.6×10⁻¹⁹ coulombs
⇒ [tex]v^2 = \frac{V q}{0.5 m}[/tex]
⇒ [tex]v^2 = \frac{(119)(1.6\times 10⁻¹⁹)}{0.5 (9.1\times 10⁻³¹)}[/tex]
⇒ v = 6.47×10⁶ m/s
⇒ Velocity of the electron mid way = v = 6.47×10⁶ m/s
