Electrostatics Question 191
Question: The force between two charges $ 0.06m $ apart is $ 5N $ . If each charge is moved towards the other by $ 0.01m $ , then the force between them will become [SCRA 1994]
Options:
A) $ 7.20N $
B) $ 11.25N $
C) $ 22.50N $
D) $ 45.00N $
Show Answer
Answer:
Correct Answer: B
Solution:
$ F\propto \frac{1}{r^{2}}\Rightarrow \frac{F _{1}}{F _{2}}={{( \frac{r _{2}}{r _{1}} )}^{2}} $
therefore $ \frac{5}{F _{2}}={{( \frac{0.04}{0.06} )}^{2}}=F _{2}=11.25N $
