Question

Two identical conducting spheres $ A $ and $ B $ carry equal charge. They are initially separated by a distance much larger than their diameters and the force between them is $ F. A $ third identical conducting sphere $ C $ is uncharged. Sphere $ C $ is first touched to $ A $ , then to $ C $ and removed. As a result, the force between $ A $ and $ B $ now is

• A. $ F/16 $
• B. $ F/4 $
• C. $ 3 F/8 $
• D. $ F/2 $

Answer 1

Answer: (C)

$r$ = distance between two identical spheres,
and $Q$ = charge on spheres $A$ and $B$
$F=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{r^{2}} \ldots\left(i\right)$
When a third identical sphere 'C' is first touches with $A$, then the value of charges on $A$ and $C$ will be $\frac{Q}{2}$
Again, when $C$ touches with $B$, then value of charge on $B$ and $C=\frac{\frac{Q}{2}+Q}{2}=\frac{3Q}{4}$
Now, the electrostatic force between spheres $A$ and $B$
$F'=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{\frac{Q}{2}\cdot\frac{3Q}{4}}{r^{2}}=\frac{3}{8}\cdot\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{Q^{2}}{r^{2}}$
$=\frac{3}{8}F$ [from Eq. $\left(i\right)$]