Electric field at the centroid of an equilateral triangle

Problem statement:

Three point charges q1, q2 and q3 lie at the vertices of an equilateral triangle of side length a as shown in the figure below. Calculate the electric field due to q1, q2 and q3 at the centroid (A) of the triangle.

Givens: q1 = q2 = 4 μC; q3 = -2 μC; a = 0.5 m; k = 9 109 Nm2/C2

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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 q1 is positive, and therefore E1 goes outwards q1. Recall that positive charges are sources of electric field lines. We deduce the direction of vector E2 by performing the same “thought experiment” for q2.

As the point charge q3 is negative, a positive test charge located at point A would be attracted by it. Therefore the electric field E3 is oriented towards the charge q3. 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 E1 and E2 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 E1 and E2 (in light green) have both the same value, which is:

Therefore, the resultant of vectors E1E2 expressed in vector form is:

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The field E3 is directed towards the positive direction of the vertical axis:

The magnitudes of E1 and E3 given by (Coulomb’s law) are respectively:

Where r1 and r3 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 r3 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 300, 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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