AP Physics 2 Exam Question:

Two point charges, Q1 and Q2, are placed 3 meters apart in a vacuum. Q1 has a charge of +4.0 microcoulombs and Q2 has a charge of -6.0 microcoulombs.

a) Calculate the magnitude of the electric force between Q1 and Q2. b) Determine the direction of the electric force between Q1 and Q2. c) Calculate the electric field strength at a point located 2 meters from Q1. d) Determine the direction of the electric field at the point in part (c).

Answer:

a) To find the magnitude of the electric force between Q1 and Q2, we can use Coulomb's Law:

Coulomb's Law: F = k * |Q1 * Q2| / r^2

where: F is the magnitude of the electric force, k is Coulomb's constant (9 × 10^9 N m²/C²), Q1 and Q2 are the charges of the two point charges, and r is the distance between the charges.

Substituting the given values:

Q1 = +4.0 microcoulombs = 4.0 × 10^(-6) C Q2 = -6.0 microcoulombs = -6.0 × 10^(-6) C r = 3 m

Plugging these values into Coulomb's Law:

F = (9 × 10^9 N m²/C²) * |(4.0 × 10^(-6) C) * (-6.0 × 10^(-6) C)| / (3 m)²

Simplifying:

F = (9 × 10^9 N m²/C²) * (24 × 10^(-12) C²) / (9 m²)

Converting the units:

F = (9 × 10^9 N m²/C²) * 2.67 × 10^(-12) C² / (9 m²) F = (1 N m²) * 2.67 × 10^(-12) / (1)

Therefore, the magnitude of the electric force between Q1 and Q2 is 2.67 × 10^(-12) N.

b) The direction of the electric force between Q1 and Q2 will be attractive because the charges have opposite signs. Thus, the force will act along the line joining the two charges from Q2 towards Q1.

c) To calculate the electric field strength at a point located 2 meters from Q1, we can use the equation for electric field:

Electric Field (E) = k * |Q| / r^2

where: E is the electric field strength, k is Coulomb's constant (9 × 10^9 N m²/C²), Q is the charge experiencing the electric field, and r is the distance between the charge and the point at which the field is measured.

In this case, the charge experiencing the electric field is Q1 = +4.0 × 10^(-6) C and the distance is r = 2 m.

Plugging in the values:

E = (9 × 10^9 N m²/C²) * (4.0 × 10^(-6) C) / (2 m)²

Simplifying:

E = (9 × 10^9 N m²/C²) * (4.0 × 10^(-6) C) / (4 m²)

Converting the units:

E = (9 × 10^9 N m²/C²) * (1.0 × 10^(-6)) / (1 m²) E = (9 × 10^3 N/C)

Therefore, the electric field strength at a point located 2 meters from Q1 is 9 × 10^3 N/C.

d) The direction of the electric field at the point in part (c) is radially outward from Q1. This is because Q1 is positive and the electric field lines always point away from positive charges.