Question

Two similar very small conducting spheres having charges $40\mu C$ and $-20\mu C$ are some distance apart. Now they are touched and kept at the same 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 $8:1$

Initial force $=\frac{K(40\times {10}^{-6})(-20\times {10}^{-6})}{r^2}=\frac{8K\times {10}^{-10}}{r^2}N$

when they are touched the charge redistrubuted among them & final charge on both of them is average of them.

$\Rightarrow$ Charge on one sphere $=\left(\frac{40-20}{2}\right)=10\mu c$

$2^{nd}$ sphere $=10\mu c$

Final force $=\frac{\left(K\right)(10\times {10}^{-6})(10\times {10}^{-6})}{r^2}=\frac{\left(K\right)\left({10}^{-10}\right)}{r^2}$

$\Rightarrow$ Initial force : Final force $=8:1$