Answer:
A) V = -136.36 V , B) V = 4.85 10³ V , C) V = 1.62 10⁴ V
Explanation:
To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is
V = k ∑
where and
are the loads and the point distances.
A) We apply this equation to our case
V = k (q₁ / r₁ + q₂ / r₂)
They ask us for the potential at the midpoint of separation
r = 3.30 / 2 = 1.65 m
this distance is much greater than the radius of the spheres
let's calculate
V = 9 10⁹ (1.1 10⁻⁸ / 1.65 + (-3.6 10⁻⁸) / 1.65)
V = 9 10¹ / 1.65 (1.10 - 3.60)
V = -136.36 V
B) The potential at the surface sphere A
r₂ is the distance of sphere B above the surface of sphere A
r₂ = 3.30 -0.02 = 3.28 m
r₁ = 0.02 m
we calculate
V = 9 10⁹ (1.1 10⁻⁸ / 0.02 - 3.6 10⁻⁸ / 3.28)
V = 9 10¹ (55 - 1,098)
V = 4.85 10³ V
C) The potential on the surface of sphere B
r₂ = 0.02 m
r₁ = 3.3 -0.02 = 3.28 m
V = 9 10⁹ (1.10 10⁻⁸ / 3.28 - 3.6 10⁻⁸ / 0.02)
V = 9 10¹ (0.335 - 180)
V = 1.62 10⁴ V
