Question

Two equipotential surfaces surround a +1.70 x 10-8-C point charge. How far is the 120-V surface from the 54.0-V surface?

Answer 1

1.55 m

Step-by-step explanation:

The potential produced by a point charge, is inversely proportional to the distance from the charge to the point where the potential is being calculated, as follows:

`V =(k*q)/(r)`

As it only depends from the distance r, we can conclude that if the potential is the same for any point to a distance r from the point charge, the equipotencial surface must be a sphere of radius r.

Replacing q = +1.7*10⁻⁸ C, and k = 9*10⁹ N*m²/C², and V, by 120 V and 54 V, we can find the distance from the charge, to the points where we are calculating the potential V, as follows:

`r1 =(k*q)/(V1) = (9e9 N*m2/C2*1.7e-8C)/(120 V) = 1.28 m`

`r2 =(k*q)/(V2) = (9e9 N*m2/C2*1.7e-8C)/(54V) = 2.83 m`

The distance between both points, is just the difference between the radius of both spheres, as follows:

r₂ - r₁ = 1.55 m