The sum of two point charges is \( 7\,\mu\mathrm{C} \). They repel each other with a force of 1 N when kept 30 cm apart in free space. What is the value of each charge?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2\,\mu\mathrm{C} \) and \( 5\,\mu\mathrm{C} \)
By Coulomb’s law, \( F=kq_1q_2/r^{2} \), so the force fixes the product of the charges, while the sum is given. Two numbers with a known sum and product follow from a quadratic. Take \( k=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \) and \( r=30\,\mathrm{cm}=0.3\,\mathrm{m} \).
The product of the charges is:
\( \begin{aligned} q_1q_2 &= \frac{Fr^{2}}{k} \\ &= \frac{1\,\mathrm{N}\times(0.3\,\mathrm{m})^{2}}{9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}} \\ &= 1\times10^{-11}\,\mathrm{C^{2}} \\ &= 10\,(\mu\mathrm{C})^{2} \end{aligned} \)
With \( q_1+q_2=7\,\mu\mathrm{C} \), the charges (in \( \mu\mathrm{C} \)) are the roots of:
\( x^{2}-7x+10=0 \)
\( (x-2)(x-5)=0 \)
\( \Rightarrow\; x=2 \text{ or } 5 \)
Therefore, the charges are \( 2\,\mu\mathrm{C} \) and \( 5\,\mu\mathrm{C} \). Both are positive, which agrees with the repulsion.