In this scenario, we are analyzing the electric field created by two charges: one positive and one negative. To determine the electric field at a specific point of interest, we first need to understand the direction of the electric fields produced by each charge. The electric field generated by a positive charge points away from the charge, while the electric field from a negative charge points towards it.

To calculate the electric field, we use the formula:

$$E = k \frac{Q}{r^2}$$

where:

• E is the electric field strength,
• k is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\),
• Q is the charge, and
• r is the distance from the charge to the point of interest.

In this case, we have a right triangle formed by the distances from the charges to the point of interest, with one leg measuring 5 cm and the other 8 cm. Using the Pythagorean theorem, we find the hypotenuse:

$$r = \sqrt{(5^2 + 8^2)} = 9.4 \, \text{cm}$$

Given the symmetry of the problem, we can simplify our calculations. The electric fields from both charges will have equal magnitudes but opposite vertical components, which will cancel each other out. Therefore, we only need to consider the horizontal components of the electric fields.

To find the horizontal component, we use the cosine of the angle θ formed by the triangle:

$$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{9.4} \approx 0.53$$

Now, we can express the net electric field at the point of interest as:

$$E_{\text{net}} = 2 \cdot E \cdot \cos(\theta)$$

Substituting the values into the equation:

$$E_{\text{net}} = 2 \cdot \left(8.99 \times 10^9 \, \frac{\text{N m}^2}{\text{C}^2} \cdot \frac{1 \, \text{C}}{(0.094 \, \text{m})^2}\right) \cdot 0.53$$

After performing the calculations, we find that the net electric field is:

$$E_{\text{net}} \approx 1.8 \times 10^{12} \, \text{N/C}$$

This net electric field points purely in the horizontal (X) direction. Understanding the symmetry in this problem allows for a more efficient calculation of the electric field, highlighting the importance of recognizing patterns in physics problems.