Answer:
V = 2.52 10³ V
Explanation:
The electric potential for a point charge is
V = k ∑ [tex]\frac{q_i}{r_i}[/tex]
In this case the formula is
V = k ([tex]\frac{q_1}{r_1} + \frac{q_2}{r_2}[/tex])
distances are the absolute value
r₁ =√ (7.1 -0)² = 7.1 m
r₁ =√ (7.1 - 14.3)² = 7.2 m
we substitute
V = 9 10⁹ (q₁ / 7.1 + q₂ / 7.2)
we have two possibilities
* different charges
V = 9 10⁹ (q₁ / 7.1 + q₂ / 7.2)
* equal charges and same sign
q₁ = q₂ = q
V = 9 10⁹ q (1 / 7.1 + 1 / 7.2) = p 9 10⁹ 0.2797
V = 2.52 10⁹ q
if we assume a value of the charge, for example q = 1 10⁻⁶ c
V = 2.52 10⁹ 1 10⁻⁶
V = 2.52 10³ V
