Certainly! Below is an AP Physics 1 exam question on electric force and field, along with the answer and step-by-step explanation.

Question: Two point charges, q1 = +3.0 μC and q2 = -8.0 μC, are placed 10.0 cm apart in a vacuum. Calculate the magnitude and direction of the electric force exerted on q1 by q2.

Answer: To find the magnitude and direction of the electric force exerted on q1 by q2, we can use Coulomb's law, which states that the magnitude of the electric force between two point charges is given by:

Where:

• F = magnitude of the electric force
• k = Coulomb's constant (8.99 × 10^9 N·m^2/C^2)
• q1, q2 = magnitudes of the two point charges
• r = separation distance between the two charges

First, let's convert the charges to Coulombs: [ q_1 = 3.0 , \mu C = 3.0 \times 10^{-6} , C ] [ q_2 = -8.0 , \mu C = -8.0 \times 10^{-6} , C ]

The separation distance r is given as 10.0 cm, or 0.10 m.

Now, plug the values into Coulomb's law: [ F = (8.99 \times 10^9) \frac{|3.0 \times 10^{-6} \times -8.0 \times 10^{-6}|}{(0.10)^2} ]

Thus, the magnitude of the electric force exerted on q1 by q2 is 215.76 N.

Now, to determine the direction of the electric force, we consider the fact that q1 and q2 have opposite charges, so the force is attractive. Therefore, the direction of the force is along the line joining the two charges, from q1 to q2.

So, the magnitude of the electric force exerted on q1 by q2 is 215.76 N, directed from q1 to q2.