Electrostatics Question 220
Question: Two point charges $ \text{3}\times \text{1}{{0}^{\text{6}}}C $ and $ \text{8}\times \text{1}{{0}^{\text{6}}}C $ repel each other by a force of $ \text{6}\times \text{1}{{0}^{\text{3}}}N $ . If each of them is given an additional charge $ \text{ 6}\times \text{1}{{0}^{\text{6}}}C $ , the force between them will be [DPMT 2003]
Options:
A) $ 2.4\times 10^{3}N $ (attractive)
B) $ 2.4\times 10^{9}N $ (attractive)
C) $ \text{1}.\text{5}\times \text{1}{{0}^{\text{3}}}N $ (repulsive)
D) $ \text{1}.\text{5}\times \text{1}{{0}^{\text{3}}}N $ (attractive)
Show Answer
Answer:
Correct Answer: D
Solution:
$ F\propto Q _{1}Q _{2} $
therefore.
$ \frac{F _{1}}{F _{2}}=\frac{Q _{1}Q _{2}}{Q _{1}‘Q _{2}’} $
$ =\frac{3\times {{10}^{-6}}\times 8\times {{10}^{-6}}}{(3\times {{10}^{-6}}-6\times {{10}^{-6}})(8\times {{10}^{-6}}-6\times {{10}^{-6}})}=\frac{3\times 8}{-3\times 2}=-\frac{4}{1} $ therefore $ F _{2}=-\frac{F _{1}}{4}=-\frac{6\times {{10}^{-3}}}{4}=-1.5\times {{10}^{-3}}N $ (Attractive)
BETA
