Final answer:
To find the magnitude of the electric field at the center of an equilateral triangle with charges q, 2q, and 3q at its vertices, calculate the electric field due to each charge and then sum the vertical components using vector addition.
Step-by-step explanation:
To calculate the magnitude of the electric field at the geometric center of an equilateral triangle formed by three point charges, q, 2q, and 3q, you must consider each charge's contribution to the field individually and then use vector addition to find the net field. Given that q = 16.8 nC and a = 11.2 cm, each charge creates an electric field at the center, and because of symmetry, the horizontal components will cancel out. The vertical components, due to the arrangement of the charges, will add up. Using Coulomb's Law and trigonometry, you can solve for the resultant electric field.
The electric field due to a point charge is given by E = k * |q| / r^2, where k is Coulomb's constant (8.988 x 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge to the point of interest. For an equilateral triangle, the distance from each vertex to the center is (a/√3). Each charge will contribute to the total electric field, and because the geometry is symmetrical, the horizontal components will cancel out. Only the vertical components of the electric fields due to the charges will add together.
The net electric field can then be found using vector addition of the vertical components of the individual fields produced by the charges q, 2q, and 3q.