Answer: 7.37×10^-15 J
Explanation: according to the work- energy theorem.
The work done in moving the electron from the negative plate to the positive equals it kinetic energy.
Mathematically, we have that
Kinetic energy = work done = qV
The distance traveled by the electron is simply the the distance between the plates of the capacitor (d) = 2cm = 0.02m.
Strength of electric field (E) = 2.3×10^6 v/m
We need to get the potential difference between the plates first, this is gotten by using the formulae below
V = Ed
Where V = potential difference =?
E = strength of electric field = 2.3×10^6 v/m
d = distance between plates = 0.02m
By substituting the parameters, we have that
V = 2.3×10^6 × 0.02
V = 0.046× 10^6
V = 4.6×10^4 v
But work done = qV
Where q = magnitude of electronic charge = 1.602×10^-19c
Work done = 1.602×10^-19 × 4.6×10^4
Work done = 7.37×10^-15 J
Answer:
7.36×10⁻¹⁵ J
Explanation:
From the question,
Then kinetic energy of the electron = electric potential × charge of the electron
Ek = q'V......................... Equation 1
Where Ek = kinetic Energy of the electron, q' = charge of an electron, V = Electric potential.
But,
V = E×d.................... Equation 2
Where E = Electric Field, d = distance of separation of the plates of the capacitor
Given: E = 2.3×10⁶ V/m, d = 2.0 cm = (2/100) m = 0.02 m.
Substitute into equation 2
V = 2.3×10⁶(0.02)
V = 4.6×10⁴ V.
Constant: q' = 1.6×10⁻¹⁹ C
Substitute into equation 1
Ek = 4.6×10⁴(1.6×10⁻¹⁹)
Ek = 7.36×10⁻¹⁵ J
