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 \text{1}0^{3}N $ (attractive)

B) $ 2.4\times \text{1}0^{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)

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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)



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