**Problem statement:**

Three point charges q_{1}, q_{2} and q_{3} lie at the vertices of an equilateral triangle of side length *a* as shown in the figure below. Calculate the electric field due to q_{1}, q_{2} and q_{3} at the centroid (A) of the triangle.

__Givens__: q_{1} = q_{2} = 4 μC; q_{3} = -2 μC; a = 0.5 m; k = 9 10^{9} Nm^{2}/C^{2}

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**Solution:**

First we are going to draw the electric field vector at point A due to each of the point charges. The **centroid** of a triangle is the point of intersection of its three medians (represented as dotted lines in the figure).

To find the direction of the electric field vector at any point due to a point charge we perform a “thought experiment” which consists in placing a **positive test charge** at this point. The direction of the electric field is the direction of the force the positive test charge would experience.

A positive test charge located at point A would be repelled as q_{1} is positive, and therefore **E**_{1} goes outwards q_{1}. Recall that positive charges are sources of electric field lines. We deduce the direction of vector **E**_{2} by performing the same “thought experiment” for q_{2}.

As the point charge q_{3} is negative, a positive test charge located at point A would be attracted by it. Therefore the electric field **E**_{3} is oriented towards the charge q_{3}. Recall that negative charges are sinks of electric field lines. The three electric fields are represented in the next figure.

The net field is the sum of the three fields represented in the next figure. The sum of the vectors **E**_{1} and **E**_{2} lies on the vertical axis, as seen in the next figure, because the horizontal components of both vectors (in red) cancel out:

Furthermore, the vertical projections of **E**_{1} and **E**_{2} (in light green) have both the same value, which is:

Therefore, the resultant of vectors **E**_{1} y **E**_{2} expressed in vector form is:

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The field **E**_{3} is directed towards the positive direction of the vertical axis:

The magnitudes of **E**_{1} and **E**_{3} given by (Coulomb’s law) are respectively:

Where r_{1} and r_{3} are the distances between each of the charges and point A. As you can see on the left side of the next figure, they both have the same value:

The distance r_{3} is equal to the triangle height *h* minus the apothem *d*. Using some trigonometry we can calculate both magnitudes:

As the triangle is equilateral the angle α is equal to 30^{0}, therefore:

The net field at point A is:

And after substituting the numerical values, we obtain:

You will find more electric field solved problems at the bottom of this page.

You can learn more about the multiples and submultiples of the SI units in the units of measurement page.

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