Respuesta :
Answer:
a) ΔV = 0 , b) ΔV = 1.08 10² V , c) ΔV = -5,394 10¹ V, d) ΔV = 5.394 / r 4 <r <7 , e) ΔV₂ = 1,156 10² V The outer sphere has greater potential
Explanation:
The electrical potential can be calculated
ΔV = - ∫ E. ds
-Within a conductive body the electric potential and the electric field are zero
-Conductive surface strength in electric field is
ΔV = -∫ E .ds
ΔV = -∫ k Q / r² dr
ΔV = k Q / r
Let's apply this to our case
A) distance r = 0
Since all equipotential surfaces are external, the electrical potential is
ΔV = 0
B) the distance of r = 5 cm
In this case the tip is inside the outer sphere, which is why it gives a contribution to the potential of zero
The point is outside the small sphere, so the power is
ΔV = k Q₁ / r
ΔV = 8.99 10⁹ 6 10⁻⁹ / 5 10⁻²
ΔV = 1.08 10² V
C) distance r = 8.00 cm
In this case, the point is outside the two spheres, so both contribute to the potential
ΔV = k q₁ / r + k q₂ / r
ΔV = 8.99 10⁸ (6 -9) 10⁻⁹ / 5 10⁻²
ΔV = -5,394 10¹ V
D) the potential between the two spheres is created by the inner sphere only
ΔV = K q₁ / r 4 <r 7
ΔV = 8.99 10⁸ 6 10⁻⁹ / r
ΔV = 5.394 / r 4 <r <7
E) The potential of the inner sphere is
ΔV₁ = 8.99 10⁸ 6 10⁻⁹ / 5 10⁻²
ΔV₁ = 1.0788 10² V
Potential of the outer sphere
ΔV₂ = 8.99 10⁸ 9 10⁻⁹ / 7 10⁻²
ΔV₂ = 1,156 10² V
The outer sphere has greater potential
