Question

A triangle with equal sides of length 14 cm has -2.5-nC charged objects at each corner.

Part A
Determine the direction of the electrical force exerted on the object at the top corner due to the two objects at the horizontal base of the triangle.

A) vertically downward
B) horizontally rightward
C) horizontally leftward
D) vertically upward

Part B
Determine the magnitude of the electrical force exerted on the object at the top corner due to the two objects at the base of the triangle.

Answer 1

The electrical force exerted on the object at the top corner of the triangle is horizontally leftward, and its magnitude is approximately 3.487 N.

Step-by-step explanation:

Part A: The direction of the electrical force exerted on the object at the top corner due to the two objects at the horizontal base of the triangle is horizontally leftward. Since the charges at the corners of the triangle are negative and the charges at the base are also negative, the electrical force will repel the top object in the opposite direction.

Part B: To determine the magnitude of the electrical force, we can use Coulomb's Law. The formula for Coulomb's Law is F = k * (q1 * q2) / r^2, where F is the force, k is the electric constant, q1 and q2 are the charges, and r is the distance between them.

In this case, both charges at the base of the triangle are -2.5 nC, and the distance between them and the top corner is 14 cm. Plugging in the values, we get F = (9 * 10^9 Nm^2/C^2) * ((-2.5 * 10^-9 C) * (-2.5 * 10^-9 C)) / (0.14 m)^2 ≈ 3.487 N.