GIven data:
Distance between the plates = 1.5 mm
Potential difference V = 600V
Charge on electron q = -1.6× [tex]10^{-19}[/tex] C
mass on electron = m = 9.1×[tex]10^{-31}[/tex] Kg
Solution:
First we will find the change in potential energy of the charge while moving through the potential difference of 600V.
ΔU = qΔV
= (-1.6×[tex]10^{-19}[/tex])(600)
= -9.6×[tex]10^{-19}[/tex]J
By the law of conservation of mechanical energy, as there is no external force acting, so the sum of the kinetic and potential energies will be a constant.
K + U = E
ΔK + ΔU = 0
ΔK = -ΔU
1/2mv² = -ΔU
v² = -2ΔU/m
= [tex]\frac{-2(-9.6*10^{-19}) }{9.1*10^{-31} }[/tex]
v = [tex]\sqrt{2.11*10^{14} }[/tex]
v = 1.45×[tex]10^{7}[/tex] m/s
1.45 x 10⁷m/s
In any system, the conservation of mechanical energy is always conserved. In other words, the sum of potential ([tex]E_{P}[/tex]) and kinetic energy ([tex]E_{K}[/tex]) is constant. i.e
[tex]E_{P}[/tex] + [tex]E_{K}[/tex] = k
=> Δ[tex]E_{P}[/tex] + Δ[tex]E_{K}[/tex] = 0
=> Δ[tex]E_{P}[/tex] = -Δ[tex]E_{K}[/tex] ---------------------(i)
The system considered here is an electric field.
(i) We know that the change in potential energy, Δ[tex]E_{P}[/tex], of a charge Q as it moves between two points in an electric field is given as;
Δ[tex]E_{P}[/tex] = Q x Δ V ---------------(ii)
Where;
ΔV = change in electric potential as the charge moves negative to positive plate = 600V
Q = charge of an electron = -1.6 x 10⁻¹⁹C
Substitute these values into equation (ii)
Δ[tex]E_{P}[/tex] = -1.6 x 10⁻¹⁹C x 600V = -9.6 x 10⁻¹⁷J
(ii) Also, since the electron is starting from rest, the change in kinetic energy is given by;
[tex]E_{K}[/tex] = [tex]\frac{1}{2}[/tex] x m x v² -------------------(ii)
Where;
m = mass of the electron = 9.1 x 10⁻³¹ kg
v = final velocity of the electron
Substitute these values into equation (ii) as follows;
[tex]E_{K}[/tex] = [tex]\frac{1}{2}[/tex] x 9.1 x 10⁻³¹ x v² J
(iii) Now substitute the values of [tex]E_{K}[/tex] = [tex]\frac{1}{2}[/tex] x 9.1 x 10⁻³¹ x v² J and Δ[tex]E_{P}[/tex] = -9.6 x 10⁻¹⁷J into equation (i) as follows;
-9.6 x 10⁻¹⁷ = - [tex]\frac{1}{2}[/tex] x 9.1 x 10⁻³¹ x v²
2 x 9.6 x 10⁻¹⁷ = 9.1 x 10⁻³¹ x v²
v² = [tex]\frac{2 * 9.6 * 10^{-17} }{9.1 * 10^{-31} }[/tex]
v² = 2.11 x 10¹⁴
v = [tex]\sqrt{2.11 * 10^{14} }[/tex]
v = 1.45 x 10⁷ m/s
The electron is going as fast as 1.45 x 10⁷m/s
