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Question

The sum of two point charges is $$7\mu C$$. They repel each other with a force of $$1\ N$$ when kept $$30\ cm$$ apart in free space. Calculate the value of each charge.

Solution
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Given,
$$q_{1} + q_{2} = 7\times 10^{-6} C .... (1)$$

$$\dfrac {1}{4\pi \epsilon_{0}} \dfrac {q_{1}q_{2}}{(0.30)^{2}} = 1\Rightarrow q_{1}q_{2} = (4\pi \epsilon_{0})(0.30)^{2}$$

or $$q_{1}q_{2} = \dfrac {1}{9\times 10^{9}}\times 9\times 10^{2} = 10^{-11} ... (2)$$
$$(q_{1} - q_{2})^{2} = (q_{1} + q_{2})^{2} - 4q_{1}q_{2}$$

$$= (7\times 10^{-6})^{2} - 4\times 10^{-11}$$

$$= 49\times 10^{-12} - 40\times 10^{-12} = 9\times 10^{-12}$$

$$q_{1} - q_{2} = 3\times 10^{-6}C ... (3)$$

Solving (1) and (3), we get
$$q_{1} = 5\times 10^{-6} C, q_{2} = 2\times 10^{-6} C$$

$$\Rightarrow q_{1} = 5\mu C, q_{2} = 2\mu C$$.

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