Question

Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the charge on B one-third the charge on A. How far apart would the two spheres then have had to be for A to have had the same deflection that it had before

Answer 1

The new separation is
`\bf{(d/√(3))}`.

Step-by-step explanation:

The expression of the force between two spheres is given by

`F = k(q_(A)q_(B))/(d^(2))~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)`

where,
`q_(A)` is the charge on sphere A,
`q_(B)` is the charge on sphere B,
`k` is constant and
`d` i the separation between two spheres.

The new value of charge on sphere B is
`q_(B)^(n) = (q_(A))/(3)`. Consider the new separation between the spheres be
`d'`. Under the new configuration the force between the spheres is given by

`F = k (q_(A)(q_(A)/3))/(d'^(2))~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)`

Equating equation (1) and equation (2), we have

`~~~~&& (1)/(d'^(2)) = (1)/(3d^(2))\\&or,& d' = (d)/(√(3))`

So, the new separation is
`(d/√(3))`.