To solve this problem, we need to calculate the electric field at the third corner of the equilateral triangle due to both the proton and the electron.

Step-by-Step Breakdown:

1. Formula for Electric Field:
   The electric field ($E$) due to a point charge is given by Coulomb's law: $E = \frac{k_e |q|}{r^2}$ where:

   • $k_e$ is Coulomb's constant ($8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2$),
   • $q$ is the charge (in this case, for a proton $q = +e = 1.6 \times 10^{-19} \, \text{C}$, and for an electron $q = -e = -1.6 \times 10^{-19} \, \text{C}$),
   • $r$ is the distance from the charge to the point where the field is being calculated, which is the side length of the triangle in this case ($r = 5.6 \times 10^{-6} \, \text{m}$).

2. Electric Fields from Proton and Electron:
   Since the charges are opposite, the electric field vectors from the proton and the electron will point in opposite directions. However, because they form an equilateral triangle, the direction of the fields will not perfectly cancel each other out.

3. Magnitude of the Electric Field:
   We calculate the magnitudes of the electric fields from both the proton and the electron. Since they are equidistant from the third corner, the magnitudes of the electric fields will be the same: $E_{\text{proton}} = E_{\text{electron}} = \frac{k_e e}{r^2}$ Plugging in the values: $E = \frac{8.99 \times 10^9 \times 1.6 \times 10^{-19}}{(5.6 \times 10^{-6})^2}$

4. Resulting Electric Field:
   The resultant electric field will be the vector sum of the two fields. Because of the symmetry of the equilateral triangle, the fields will add vectorially in a specific direction.

Let me calculate the magnitude for you.The magnitude of the electric field produced by either the proton or the electron at the third corner is approximately $45.87 \, \text{N/C}$.

Now, since the proton and the electron produce fields in different directions, the resultant electric field will not be a simple sum but a vector sum. In an equilateral triangle, the angle between the two electric field vectors will be 60°.

To compute the net electric field at the third corner, we use vector addition. Let's calculate that next.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can calculate the resultant electric field by using the formula for vector addition:

$E_{\text{net}} = \sqrt{E_{\text{proton}}^2 + E_{\text{electron}}^2 + 2 \cdot E_{\text{proton}} \cdot E_{\text{electron}} \cdot \cos(\theta)}$ where $\theta = 60^\circ$ (the angle between the two vectors). Since the magnitudes of both fields are the same, you can substitute $E_{\text{proton}} = E_{\text{electron}} = 45.87 \, \text{N/C}$ into the equation.

Let me know if you'd like more details or have further questions!

Here are some related questions to think about:

1. What would happen to the electric field if the triangle's side length were doubled?
2. How would the direction of the resultant electric field change if one of the charges was doubled?
3. Can we calculate the electric potential at the third corner due to both charges?
4. How does the principle of superposition help in solving problems with multiple charges?
5. What would the electric field look like if the third corner had a charge as well?

Tip: The electric field is a vector quantity, so direction matters!