Final answer:
The net electric field at x=-4.0 cm due to the presence of a 6.0 µC charge at the origin and a -9.6 µC charge at x=10.0 cm can be found by calculating the electric field due to each charge separately and then adding them vectorially considering their directions.
Step-by-step explanation:
The student is asking about the net electric field at a point on the x-axis due to two point charges also located on the x-axis. To find the net electric field at x=-4.0cm due to a charge of 6.0 µC at the origin (x=0cm) and a charge of -9.6 µC at x=10.0cm, we must calculate the electric field due to each charge separately and then vectorially add them to get the net electric field.
The electric field due to a point charge is given by E = k * q / r^2, where E is the electric field, k is Coulomb's constant (8.99 x 10^9 N·m^2/C^2), q is the charge, and r is the distance from the charge to the point in question.
For the charge at the origin (6.0 µC), the electric field at x=-4.0cm would be directed to the right (positive x-direction), since the charge is positive. For the charge at x=10.0cm (-9.6 µC), the electric field at x=-4.0cm would be directed to the left (negative x-direction), because the charge is negative. Calculating these fields separately and adding them while considering their directions will yield the net electric field at x=-4.0cm.
However, to provide an exact numerical answer, calculations must be made using the given values and formula.