Final answer:
The electric field at a point on the x-axis at x = 0.200 m can be found by calculating the electric fields due to each point charge and adding them together using vector addition.
Step-by-step explanation:
The electric field created by a point charge can be calculated using the equation E = kQ/r², where E is the electric field strength, k is Coulomb's constant (9.0 x 10^9 N·m²/C²), Q is the charge, and r is the distance from the charge. In this case, we have two charges, a -4.00 nC charge at the origin and a -6.50 nC charge at x = 0.800 m on the x-axis.
To find the electric field at a point on the x-axis at x = 0.200 m, we need to calculate the electric field for each charge separately and then add them together using vector addition. Let's calculate it step by step.
For the -4.00 nC charge at the origin (x=0), the distance to the point at x=0.200 m is 0.200 m. Plugging the values into the equation, we get E1 = kQ1/r² = (9.0 x 10^9 N·m²/C²)(-4.00 x 10^-9 C)/(0.200 m)².
Now, for the -6.50 nC charge at x = 0.800 m, the distance to the point at x=0.200 m is 0.600 m. Using the same equation, we get E2 = kQ2/r² = (9.0 x 10^9 N·m²/C²)(-6.50 x 10^-9 C)/(0.600 m)².
Finally, we can find the total electric field at the point by adding the magnitudes and considering the direction. Since both charges are negative, the electric fields will point towards the positive x-axis.