Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the charges and moves along the line connecting them. What is the electric potential energy of the electron when it is1. at the midpoint?2. 10.0 cm from the 3.00 nC charge?I tried to do the first part of this problem by observing another example, but I still got it all wrong. Could you please show me the step-by-steps of how to do this problem and how you came up with all the equations?

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