b) (i) State coloumb 's law (1 mk) (ii) What must be the distance between point charge point charge q_(1)=26mu C point charge q_(2)=-47mu C for the electrostatic force between them to have a magnitude of 5.70 N?(3 mks)

(i) Coulomb's law states that the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as:<br /><br />\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]<br /><br />where:<br />- \( F \) is the electrostatic force between the charges,<br />- \( k \) is the Coulomb's constant (\( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)),<br />- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,<br />- \( r \) is the distance between the charges.<br /><br />(ii) To find the distance between the point charges \( q_1 = 26 \mu C \) and \( q_2 = -47 \mu C \) for the electrostatic force between them to have a magnitude of 5.70 N, we can rearrange Coulomb's law to solve for \( r \):<br /><br />\[ r = \sqrt{\frac{k \cdot |q_1 \cdot q_2|}{F}} \]<br /><br />Substituting the given values:<br /><br />\[ r = \sqrt{\frac{(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \cdot |(26 \times 10^{-6} \, \text{C}) \cdot |(-47 \times 10^{-6} \, \text{C})|}{5.70 \, \text{N}}} \]<br /><br />\[ r = \sqrt{\frac{(8.99 \times 10^9) \cdot (26 \times 10^{-6}) \cdot (47 \times 10^{-6})}{5.70}} \]<br /><br />\[ r = \sqrt{\frac{(8.99 \times 10^9) \cdot (1212 \times 10^{-12})}{5.70}} \]<br /><br />\[ r = \sqrt{\frac{10.857 \times 10^{-2}}{5.70}} \]<br /><br />\[ r = \sqrt{1.902 \times 10^{-2}} \]<br /><br />\[ r \approx 0.138 \, \text{m} \]<br /><br />Therefore, the distance between the point charges must be approximately 0.138 meters for the electrostatic force between them to have a magnitude of 5.70 N.