Answer:
a) V = 45 10³ V, b) U = 4.59 J
Explanation:
a) The electric potential for a series of point charges is
V = k ∑ [tex]\frac{q_i}{r_i}[/tex]
in this case point P is at a distance of 1 m from each charge, so the point is located perpendicular to the charges at its midpoint
V = k ( [tex]\frac{q_1}{r} + \frac{q_2}{r}[/tex])
V = 9 10⁹ (10 - 5/ 1) 10⁻⁶
V = 45 10³ V
b) the potential energy is
U = k ( [tex]\frac{q_1q}{r} + \frac{q_2q}{r} + \frac{q_1q_2}{r_2}[/tex] )
where r = 1m and r₂ is the distance between the two charges r₂ = 0.10 m
U = 9 10⁹ (10 2 / 1 - 5 2/1 - 10 5 /0.10) 10⁻¹²
U = 9 10⁻³ 510
U = 4.59 J
