Question

(II) Two small charged spheres are 6.52 cm apart. They are moved, and the force each exerts on the other is found to have tripled. How far apart are they now?

Short answer

The distance between the charged spheres becomes \(3.76\,{\rm{cm}}\).

Step-by-step solution

Understanding the relation between electric force and separation distance

The electric force exerted between two charges is directly proportional to the magnitude of charges and inversely proportional to the square of the separation distance.

The expression for the electric force is given as:

\(F = k\frac{{{Q_1}{Q_2}}}{{{r^2}}}\) … (i)

Here, k is the Coulomb’s constant,\({Q_1},\;{Q_2}\)are the charges and r is the separation between the charges.

Given Data

The initial distance between the charged spheres is, \(r = 6.52\;{\rm{cm}}\)

Calculation of distance moved by the charged spheres

From equation (i), the electric force is,

\(F \propto \frac{1}{{{r^2}}}\)

When the force is multiplied by the factor 3, the value of rwill reduce by factor \(\sqrt 3 \).

So, the new distance between the charged spheres is,

\(\begin{aligned}{l}r' = \frac{r}{{\sqrt 3 }}\\r' = \frac{{6.52\,{\rm{cm}}}}{{\sqrt 3 }}\\r' = 3.76\,{\rm{cm}}\end{aligned}\)

Thus, the distance between the charged spheres becomes \(3.76\,{\rm{cm}}\).