Question

Two point charges + 3 $ \mu C $ and + 8 $ \mu C $ repel each other with a force of 40 N. If a charge of - 5 $ \mu C $ is added to each of them, then the force between them will become:

• A. +10 N
• B. +20 N
• C. -20 N
• D. -10 N

Answer 1

Answer: (D)

Two charges $ +\,3\mu C $ and $ 8\mu C $ are given which repel each other when a charge $ -5\mu C $ is added then in second case changes becomes $ -2\mu C $ and $ 3\mu C $ . From the relation $ F\propto {{q}_{1}}{{q}_{2}} $ Hence $ \frac{{{F}_{1}}}{{{F}_{1}}}=\frac{{{q}_{1}}\times {{q}_{2}}}{{{q}_{1}}{{q}_{2}}} $ or $ \frac{40}{{{F}_{1}}}=\frac{3\times 8}{-2\times 3}=-\,4 $ so, $ {{F}_{1}}=-\frac{40}{4}=10\,\,\text{attractive} $