Answer:
a) W = 1.63 10⁻²⁸ J, b) W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,
d) W = - 4.93 10⁻²⁸ J
Explanation:
a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m
If we use the law of conservation of energy, work is the change in energy of the system
W = ΔU = U_∞ -U
the potential energy for point charges is
U =k
in this case we only have two particles
U = k
the distance is
r₁₂ =
r₁₂ =Ra 0 + (2-0)
r₁₂ = √2= 1.4142 m
we substitute
W = k \sum \frac{q_i q_j}{r_{ij} }
let's calculate
W = 9 109 1.6 10-19 1.6 10-19 / 1.4142
W = 1.63 10⁻²⁸ J
b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0
in this case we have two fixed electrons
U = k
in this case all charges are electrons
q₁ = q₂ = q₃ = q
W = U = k q²
the distances are
r₁₃ = RA (3.00 -0) 2 + 0
r₁₃ = 3
r₂₃ = Ra (3 0) 2 + (2 0) 2
r₂₃ = √13
r₂₃ = 3.606 m
let's look for the job
W = U
let's calculate
W =
W = 1.407 10⁻²⁷ J
c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,
y₄ = 4.00 m
W = U = k
all charges are equal q₁ = q₂ = q₃ = q₄ = q
W = k q²
let's look for the distances
r₁₄ =
r₁₄ = 5 m
r₂₄ =
r₂₄ = √13 = 3.606 m
r₃₄ =
r₃₄ = 4 m
we calculate
W = 9 10⁹ (1.6 10⁻¹⁹)²
W = 1.68 10⁻²⁸ J
d) we take the proton to the location x5 = 1m y5 = 1m
W = U = k
in this case the charges have the same values but charge 5 is positive and the others negative, so the products of the charges give a negative value
W = - k q²
we look for distances
r₁₅ = Ra (1-0) 2 + (1-0) 2
r₁₅ = √ 2 = 1.4142 m
r₂₅ =
r₂₅ = √2 = 1.4142 m
r₃₅ =
r₃₅ = √5 = 2.236 m
r₄₅ =
r₄₅ = √13 = 3.606 m
we calculate
W = - 9 10⁹ (1.6 10⁻¹⁹)²
W = - 4.93 10⁻²⁸ J