Final answer:
To find the electric field at a point on the x-axis at x = 0.200 m, calculate the electric field contributions from each point charge and sum them up. The resulting electric field magnitude is approximately 2.43 × 10^4 N/C towards the positive x-axis.
Step-by-step explanation:
To find the electric field at a point on the x-axis at x = 0.200 m, we can use the formula for the electric field due to a point charge:
E = kQ / r^2
where E is the electric field magnitude, k is Coulomb's constant, Q is the charge, and r is the distance between the charge and the point.
In this case, there are two point charges: -4.00 nC at the origin and -7.00 nC at x = 0.800 m. To find the electric field at x = 0.200 m, we need to calculate the electric field due to each charge and then add them up.
The electric field due to the first charge can be calculated as:
E1 = kQ1 / r1^2
where Q1 = -4.00 nC and r1 = 0.200 m (distance from the first charge to the point).
Similarly, the electric field due to the second charge can be calculated as:
E2 = kQ2 / r2^2
where Q2 = -7.00 nC and r2 = 0.800 - 0.200 = 0.600 m (distance from the second charge to the point).
Finally, the total electric field at the point is:
E = E1 + E2
Using the values for k and performing the calculations, we find that the electric field magnitude is approximately 2.43 × 10^4 N/C pointing towards the positive x-axis.