Final answer:
The magnitude of the net force on q3 from q1 and q2, given that q1=q2=q3 and they are equidistant from each other, is found to be zero. This is due to the symmetric arrangement of the charges resulting in equal but oppositely directed forces from q1 and q2 on q3, which cancel each other out.
Step-by-step explanation:
To calculate the magnitude of the net force on q3 from q1 and q2 where q1 = q2 = q3 = 5.67×10⁻⁸ C, and the distance d between each charge is 75.0 cm, we can use Coulomb's Law. The force between two point charges is given by F = k×(|q1×q2|)/(d²), where k is Coulomb's constant (8.987×10⁹ Nm²/C²), q1 and q2 are the magnitudes of the charges, and d is the separation between the charges. Since the question asks for the net force on q3 and the charges q1 and q2 are equal and opposite from q3 at the same distance, the forces on q3 from q1 and q2 will be of equal magnitude but in opposite directions. Hence they will add up vectorially, taking into account their directions. Assuming q1 is to the left of q3 and q2 to the right (or vice versa), the net force will lie along the line connecting q1, q3, and q2. The magnitudes of the forces from each charge on q3 can be calculated separately and then combined vectorially to determine the net force. Since the distance should be converted from cm to meters, d = 0.75 meters for the calculation. Plugging into Coulomb's Law: F1 = F2 = (8.987×10⁹ Nm²/C²) × ((5.67×10⁻⁸ C)²) / (0.75 m)². After calculating F1 and F2, we take the vector sum of these two forces to find the net force on q3. Since the forces have the same magnitude but are in opposite directions, the net force will be zero.