Answer:
Therefore, the velocity of the electron is 1.22 × 10⁷m/s
Explanation:
The velocity of the electron can be found using the conservation of energy. The conservation of energy for the electron passing from one plate to another plate is,
[tex]\frac{1}{2} mv^2=qV[/tex]
Expression for velocity
= [tex]v = \sqrt{\frac{2q \sigma d}{\varepsilon_0 m} }[/tex]
[tex]v = \sqrt{\frac{2(1.6\times10^{-19}\times2.5\times10^{-7}(0.015)}{(8.854\times10^{-12}\times(9.1\times10^{-31}))} }[/tex]
[tex]v =\sqrt{\frac{1.2\times10^{-27}}{8.05714\times10^{-42}} } \\v = \sqrt{1.48936\times10^{14}}[/tex]
v = 1.22 × 10⁷m/s
Therefore, the velocity of the electron is 1.22 × 10⁷m/s
Answer:
Speed = 1.22 x 10^(7) m/s
Explanation:
Electric Field has a formula of;
E = σ/ε_o
Where,
σ is surface
εσ is vacuum permittivity
Also, acceleration is given by;
a = qE/m
Thus, replace E with σ/ε_o;
a = (qσ/ε_o)/m
Where;
q is charge of electron = 1.6 x 10^(-19) C
m is mass of electron = 9.11 x 10^(-31) kg
While ε_o has a value of 8.85 x 10^(-12) N.m²/C²
Thus, plugging in the relevant values to obtain ;
a = [((1.6 x 10^(-19))x 2.5 x 10^(-7)]/(8.85 x 10^(-12) x 9.11 x 10^(-31)) = 4.96 x 10^(15) m/s²
From Newton's 3rd law of motion,
V² = U² + 2as
Thus,plugging in the relevant values,
V² = 0² + 2(4.96 x 10^(15) x 1.5 x 10^(-2))
V² = 14.88 x 10^(13)
V = √14.88 x 10^(13) = 1.22 x 10^(7) m/s
