Problem 71-13    Source:  Problem elaborated by the author of the site.

Being Q₁ = - 50 µC and Q₂ = + 50 µC, find the electric field at points A and B. Use the distances shown in the Figura 71-13.1.

Solution of the Problem 71-13   

Initially we will calculate the electric field at point A due to charge Q₁, which we will call E_A1. The distance from the point A to the charge Q₁, applying the Pythagorean theorem, is d(AQ1)² = 30² + 80². Then d(AQ1) = 85.44 cm = 0.8544 m. So

Carrying out the calculation, we find

Now let's calculate the electric field at point A due to charge Q₂, which we will call E_A2. The distance from point A to load Q₂ is direct from Figure 71-13.1, or d (AQ2) = 30 cm = 0.3 m. Using the same equation as before, substituting the numerical values and performing the calculation, we find

Note that the electric field at point A due to charge Q₂ is about 10 times greater than due to load Q₁.

The Figure 71-13.2 anticipates the scheme of forces acting at points A and B. Initially, let's fix our attention on the point A. We must determine the angle that the fields E_A1 and E_A2 do with the horizontal. The angle of E_A2 with the horizontal is 90°. The angle that E_A1 makes with the horizontal is negative, as it is below the horizontal. The angle is given by:

Carrying out the calculation, we find:

With this value we can calculate the component on the axis x of E_A1, or

And we can calculate the component on the axis y of E_A1, or

Now we can find the field value on the axis y, or

Carrying out the calculation, we find:

We now have two fields that make an angle of 90° to each other. Then, applying the theorem of Pythagoras, we will find the value of E_A at the point A.

Carrying out the calculation, we find:

This same value would be found using the law of cosines, remembering that the angle between the fields is equal to 90° + 20.56° = 110.56°. And to find the angle φ that E_A makes with the axis x, just calculate

Carrying out the calculation, we find:

And for the calculation of the electric field at the point B, we realize that there is a symmetry in relation to the charges, since the distances between the point B and the charges are equal. Then, in module, the value of E_B1 and E_B2 will be the same, changing only the direction as indicated by the Figure 71-13.2. And based on Figure 71-13.1 we can calculate the distance d² (BQ1) = 0.3² + 0.4², which results in the value d (BQ1) = 0.5 m. Then, we have:

Carrying out the calculation, we find:

To find the angle that the fields make with the horizontal, we just need to calculate the arctangent of the opposite side by the adjacent side, or:

As shown in Figure 71-13.2 we see that the vertical components will cancel out, because one will be negative and the other positive. Since, in modulus, they are equal, so the result is zero. Then, only the horizontal components are left which, in module, are equal. Therefore, we can write:

From this information we conclude that:

Carrying out the calculation, we find:

Note that the field E_B is parallel to the axis x, that is, it makes an angle of 0° with the horizontal.