Question

Two equally charged identical metal spheres A and B repel other with a force of $2\times {10}^{-5}$.Another identical uncharged sphere C is touched to B and then placed at the mid point between A and B.The net electric force on C is.

• A. $2\times {10}^{-5}Nalong\overrightarrow{AB}$
• B. $2\times {10}^{-5}Nalong\overrightarrow{BA}$
• C. $4\times {10}^{-5}Nalong\overrightarrow{AB}$
• D. $4\times {10}^{-5}Nalong\overrightarrow{BA}$

Answer 1

The correct option is A $2\times {10}^{-5}Nalong\overrightarrow{AB}$

$F_{AB}=\frac{K{Q}^2}{r^2}=2\times {10}^{-5}$

Now,

'C' is touched to 'B' and both are identical so the charge on 'B' will be equally distributed on both 'B' and 'C' i.e,

So,

$F_{AC}=\frac{K\left(Q\right)\left(Q/2\right)}{(r/2{)}^2}=\frac{2K{Q}^2}{r^2}{F}_{BC}=\frac{K\left(Q/2\right)\left(Q/2\right)}{(r/2{)}^2}=\frac{K{Q}^2}{r^2}$

So, Net force on 'C' is given by

$\Rightarrow F_{AC}-F_{BC}\Rightarrow \frac{2K{Q}^2}{r^2}-\frac{K{Q}^2}{r^2}\Rightarrow \frac{K{Q}^2}{r^2}=F_{AB}\Rightarrow 2\times {10}^{-5}Nalong\overrightarrow{AB}$