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

Answer:

1.55 m

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:

$V =\frac{k*q}{r}$

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:

$r1 =\frac{k*q}{V1} = \frac{9e9 N*m2/C2*1.7e-8C}{120 V} = 1.28 m$

$r2 =\frac{k*q}{V2} = \frac{9e9 N*m2/C2*1.7e-8C}{54V} = 2.83 m$

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

r₂ - r₁ = 1.55 m