Answer:
1) U₁ = -2.8648 10⁻¹⁵ J 2) U₂ = 1.15 10⁻¹⁶ J
Explanation:
The electrical potential for point charges is given by the formula
V = k q / r
Where k is the Coulomb constant that value 8.99 109 Nm2 / C2, q the load and r the distance to the point of interest. Since the potential is a scalar when there are several charges, we can add the potentials
V = V₁ + V₂
The electric potential energy is the electric potential for the test load.
U = k q₁ q₂ / r₁₂
Where q₁ and q₂ are the charge and the test charge and r₁₂ is the distance between these two charges, you determined that the electric potential is also a scalar
Let's apply this last equation to our case.
The data they give are the charges
q₁ = 3.00 nc = 3.00 10⁻⁹ C
q₂ = 2.00 nC = 2.00 10⁻⁹ C
d = 50.0 cm = 50.0 10⁻² m
Case 1
The test charge is an electron
q3 = e = - 1.6 10-19 C
We look for the potential electric energy at the midpoint
x = d / 2 = 25.0 10⁻² m
U = U₁₃ + U₂₃
U = k q₁q₃ / r₁₃ + k q₂q₃ / r₂₃
In this case
r₁₃ = r₂₃ = r = 25.0 10⁻² m
U₁ = k / r q₃ (q₁ + q₂)
Let's calculate
U₁ = 8.99 10⁹ (-1.6 10⁻¹⁹) / 25.0 10⁻² (3.00 10⁻⁹ + 2.00 10⁻⁹)
U₁ = -2.8648 10⁻¹⁵ J
Case 2
Distance
r₁₃ = 10.0 cm = 10.0 10⁻² m
The other distance r2.3 is measured from charge 2
r₂₃ = d -r₁₃
r₂₃ = 50 - 10 = 40 cm = 40.0 10⁻² m
Let's write the formula
U₂ = k q₃ (q₁₃ / r₁₃ + q₂₃ / r₂₃)
U₂ = 8.99 10⁹ (-1.6 10⁻¹⁹) (3.00 10⁻⁹ / 10.0 10⁻² + 2.00 10⁻⁹ / 40.0 10⁻²)
U₂ = 14,384 10⁻¹⁰ (0.3 10⁻⁷ + 0.5 10⁻⁷)
U₂ = 11.5072 10⁻¹⁷ J
U₂ = 1.15 10⁻¹⁶ J
