There are two point charges. The charges are Q_(1)=5.00mu C and Q_(2)=-5.00mu C Charge q_(1) is 6.00 cm from charge 92 . times 10^4Nm^2/C^2) Using Coulome's law, the distance between the center of charges, and scientific notation, calculate the mapitude of the net electric force on charge of. k=8.99 5.6times 10^2N 8.99times 10^2N c 4.45times 10^2N 1.49times 10^2N

To solve this problem, we can use Coulomb's law, which states that the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.<br /><br />Given information:<br />- Charge 1: $Q_1 = 5.00 \mu C$<br />- Charge 2: $Q_2 = -5.00 \mu C$<br />- Distance between charges: $r = 6.00 \text{ cm} = 0.06 \text{ m}$<br />- Coulomb's constant: $k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2$<br /><br />Step 1: Calculate the magnitude of the electric force between the charges.<br />$F = \frac{k \cdot |Q_1 \cdot Q_2|}{r^2}$<br />$F = \frac{(8.99 \times 10^9 \text{ Nm}^2/\text{C}^2) \cdot (5.00 \times 10^{-6} \text{ C}) \cdot (5.00 \times 10^{-6} \text{ C})}{(0.06 \text{ m})^2}$<br />$F = \frac{(8.99 \times 10^9) \cdot (25.00 \times 10^{-12})}{(0.0036)}$<br />$F = 6.25 \times 10^2 \text{ N}$<br /><br />Therefore, the magnitude of the net electric force on charge 1 is $6.25 \times 10^2 \text{ N}$.<br /><br />The correct answer is: $6.25 \times 10^2 \text{ N}$.