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Two $Q_{1} and Q_{2}$ are $3 m$ apart and their combined charge is $20 μ C .$
If one attracts the other with force of $0.525 N$ . Find the magnitude of the charges.

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$- 35 μ C, - 15 μ C$

$35 μ C, - 15 μ C$

$35 μ C, 15 μ C$

$- 35 μ C, 15 μ C$

answer is A, C.

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Detailed Solution

Force is attractive, so one charge is negative so force equation is

$- 0.525 = 9 \times 10^{9} \frac{Q_{1} Q_{2}}{9}$

$\Rightarrow Q_{1} Q_{2} = - 525 μ C^{2}$

$\Rightarrow Q_{1} (20 - Q_{1}) = - 525$

$\Rightarrow Q_{1}^{2} - 20 Q_{1} - 525 = 0$

From here $Q$ values are $35 μ C and - 15 μ C or - 35 μ C and 15 μ C$ .

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