Question

Two identical tiny metal balls carry charges of $+3nC$ and $-12nC$. They are $3cm$ apart. The balls are now touched together and then separated to $3cm$. The force on them is :

• A. $20.25\times {10}^{-4}N$
• B. $2.025\times {10}^{-3}N$
• C. $2.025\times {10}^{-4}N$
• D. $2025N$

Answer 1

The correct option is C $2.025\times {10}^{-4}N$

When the balls are touched charge flows from high potential to low potential till potential on both becomes equal $(V=\frac{Kq}{R})$
As balls are identical, finally they both have equal charges i.e, $\frac{totalcharge}{2}$ $=\frac{+3-12}{2}=4.5nc$

force on them (repulsion)$F=\frac{K\left(4.5nC\right)\left(4.5nC\right)}{(3\times {10}^{-2}{)}^2}$

$=\frac{9\times {10}^9\times 4.5\times {10}^{-9}\times 4.5\times {10}^{-9}}{9\times {10}^{-4}}$

$=4.5\times 4.5\times {10}^{-5}$
$=20.25\times {10}^{-5}N$
$=2.025\times {10}^{-4}N$