Electrostatics Question 193
Question: Two point charges $ +3\mu C $ and $ +8\mu C $ repel each other with a force of $ 40N $ . If a charge of $ -5\mu C $ is added to each of them, then the force between them will become [SCRA 1998; JIPMER 2000]
Options:
A) $ -10N $
B) $ +10N $
C) $ +20N $
D) $ -20N $
Show Answer
Answer:
Correct Answer: A
Solution:
In second case, charges will be $ -2\mu C $ and $ +3\mu C $
Since $ F\propto Q _{1}Q _{2} $ i.e. $ \frac{F}{{{F}’}}=\frac{Q _{1}Q _{2}}{Q{’ _{1}}Q{’ _{2}}} $ $ \frac{40}{{{F}’}}=\frac{3\times 8}{-2\times 3}=-4 $
therefore $ {F}’=10N $ (Attractive)
