Three negative charges with charge –q occupy the vertices of an equilateral triangle with sides of length L as shown below
a) What is the magnitude of the electric field at the center of the triangle? The center is the point that is equally distant from all three vertices. (zero)
b) One of the –q charges is replaced by a +q charge as shown below:
What is the magnitude of the electric field in units of k e q / L2 at the center of the
triangle? (The center is the point that is equidistant from all three vertices) (6 ke q ¿ L2
What is the direction of the electric field?
a) The magnitude of the electric field at the center of the triangle is zero. This is because the electric fields created by the three negative charges cancel each other out due to the symmetry of the equilateral triangle.
b) The magnitude of the electric field at the center of the triangle, when one of the negative charges is replaced by a positive charge, becomes 6 ke q / L^2. This is because the positive charge adds an additional electric field that is in the same direction as the existing electric fields created by the negative charges.
The direction of the electric field at the center of the triangle is radially outward from the center. This is because the positive charge creates an electric field that points away from it, and the negative charges create electric fields that also point away from them. The resulting electric field at the center is the vector sum of these individual electric fields, which points outward.