the Imagine a right angle triangle with two equal sides of 10 cm. Two equal charges of +10μC are placed in the vertices corresponding to the 45° angles. A negative, -20μC charge is placed at the 90° vertex. Find the electric field in the middle hypotenuse. You need to provide the magnitude of the field and the direction. For the direction use the angle between the electric field and the hypotenuse (the one less than 90°). ▾ Part A - What is the magnitude of the field at the point in the midle of the hypothenuse? ? Value Units Submit Request Answer ▸ Part B. What is the angle between the total field and the hypothenuse? (If the angle is not 90 degress, give the angle which is less than 90 degrees)

Two equal and opposite charges are placed on the x-y plane as shown. The positive charge is at location (L, 0) and the negative charge at (0, 2L). +2L 6 +L Consider the electric field at the origin. On the graph draw: a) the electric field due to the positive charge, label that E, b) the electric field due to the negative charge, label that E- c) the net electric field at the origin, label that Emet Draw appropriately the length of each vector to take into account their relative magnitudes.

A small rubber sphere has a mass of 3.64 g and a charge of -18.6 µC. It levitates, motionless, when placed in a uniform electric field perpendicular to the ground. What is the magnitude of the electric field (in N/C)? N/C What is the direction of the electric field? O upward O downward O east west

Two equal positive charges (10 μC) are arranged at the base of an equilateral triangle of sides 20 cm. What is the electric field at the top vertex of the triangle? What is the direction of the electric field?