Question

Two similar very small conducting spheres having charges $40C$ and $-20C$ are at some distance apart. Now they are touched and then kept at a same initial distance. The ratio of the initial to the final force between them is:

• A. $8:1$
• B. $4:1$
• C. $1:8$
• D. $1:1$

Answer 1

The correct option is A $8:1$

Initially force between conducting spheres $F_1=\frac{K{q}_1{q}_2}{r^2}=\frac{K\left(40\right)(-20)}{r^2}$(i)
After contact total charge$=40-20=20\mu C$
Now charge will be equally distributed
so $q_1=q_2=\frac{20}{2}=10\mu C$
Finally force between spheres$F_2=\frac{K\left(10\right)\left(10\right)}{r^2}$(ii)
Ratio of forces=$\frac{F_1}{F_2}=\frac{40\times 20}{10\times 10}=8$