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

Respuesta :

Answer:

The new separation is [tex]\bf{(d/\sqrt{3})}[/tex].

Explanation:

The expression of the force between two spheres is given by

[tex]F = k\dfrac{q_{A}q_{B}}{d^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]

where, [tex]q_{A}[/tex] is the charge on sphere A, [tex]q_{B}[/tex] is the charge on sphere B, [tex]k[/tex] is constant and [tex]d[/tex] i the separation between two spheres.

The new value of charge on sphere B is [tex]q_{B}^{n} = \dfrac{q_{A}}{3}[/tex]. Consider the new separation between the spheres be [tex]d'[/tex]. Under the new configuration the force between the spheres is given by

[tex]F = k \dfrac{q_{A}(q_{A}/3)}{d'^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]

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

[tex]~~~~&& \dfrac{1}{d'^{2}} = \dfrac{1}{3d^{2}}\\&or,& d' = \dfrac{d}{\sqrt{3}}[/tex]

So, the new separation is [tex](d/\sqrt{3})[/tex].