Electrostatics Question 168
Question: A total charge Q is broken in two parts $ Q _{1} $ and $ Q _{2} $ and they are placed at a distance $ R $ from each other. The maximum force of repulsion between them will occur, when [MP PET 1990]
Options:
A) $ Q _{2}=\frac{Q}{R},\ Q _{1}=Q-\frac{Q}{R} $
B) $ Q _{2}=\frac{Q}{4},\ Q _{1}=Q-\frac{2Q}{3} $
C) $ Q _{2}=\frac{Q}{4},\ Q _{1}=\frac{3Q}{4} $
D) $ Q _{1}=\frac{Q}{2},\ Q _{2}=\frac{Q}{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ Q _{1}+Q _{2}=Q $ ….. (i) and $ F=k\frac{Q _{1}Q _{2}}{r^{2}} $ …..(ii)
From (i) and (ii) $ F=\frac{kQ _{1}(Q-Q _{1})}{r^{2}} $
For F to be maximum $ \frac{dF}{dQ _{1}}=0 $
therefore $ Q _{1}=Q _{2}=\frac{Q}{2} $
