AP Physics 2: Algebra-Based Practice Quiz: Electric Fields

Correct Answer: C

Based on the equation $\vec{E}=\frac{\vec{F}_{E}}{q}$, the electric field is the ratio of the electric force exerted on a test charge to the charge of the test charge.

Correct Answer: C

The provided content states that electric fields are produced by a "charged object or configuration of point charges."

Correct Answer: A

The content explicitly states that "the electric field within the conductor is zero" when it is in electrostatic equilibrium.

Correct Answer: C

The content specifies that in electrostatic equilibrium, "the excess charge of a solid conductor is distributed on the surface of the conductor."

Correct Answer: B

Using the equation $\vec{E}=\frac{\vec{F}_{E}}{q}$, we have $E = \frac{4 \times 10^{-6} \text{ N}}{2 \times 10^{-9} \text{ C}} = 2000$ N/C. Since the test charge is positive, the direction of the electric field is the same as the direction of the force, which is to the right.

Correct Answer: D

The field from Q1 at the origin points to the right. The field from Q2 at the origin points to the left. The net field is the vector sum. If Q1 > Q2, the net field is to the right. If Q2 > Q1, the net field is to the left. If Q1 = Q2, the net field is zero. Therefore, the direction depends on their relative magnitudes.

Correct Answer: B

The electric field at a given point is a property of the source charges that create the field, not the test charge placed there to measure it. The field E is independent of the test charge q. While the force would change, the field created by the source charges remains the same.

Correct Answer: D

The electric field from +Q at point P will point away from +Q (down and to the right). The electric field from -Q at point P will point towards -Q (down and to the left). The horizontal (x-components) of these two field vectors will cancel out due to symmetry. The vertical (y-components) will both point downwards, so their vector sum will be in the negative y-direction.

Correct Answer: B

The provided content states that for a solid conductor in equilibrium, excess charge resides on the surface. For an insulator, charges are not free to move, so if charge is placed within its volume, it will generally remain there. The content allows for describing the field generated by insulators, implying their charge can be distributed differently than in conductors.

Correct Answer: C

From the definition $\vec{E}=\frac{\vec{F}_{E}}{q}$, we can rearrange to find the force: $\vec{F}_{E}=q\vec{E}$. If the field E is tripled to 3E, the new force will be $q(3E) = 3(qE) = 3F$.

Correct Answer: A

The rules for a conductor in electrostatic equilibrium apply to all parts of the conductor, including the material of a hollow sphere. The electric field within the conducting material must be zero. The free charges within the conductor rearrange on the surfaces to cancel the external field within the conductor's bulk.

Correct Answer: B

By convention, the direction of the electric field is the direction of the force on a positive test charge. A positive test charge would be repelled by the source positive charge, so the force and the field lines point radially outward. This aligns with the content about describing the field from a point charge.

Correct Answer: C

The definition of a conductor is that it has mobile charges. The definition of electrostatic equilibrium is that there is no net motion of charge. If there were an electric field inside the conductor, the mobile charges would feel a force ($\vec{F}=q\vec{E}$) and accelerate. Since they are not moving (in equilibrium), the net force, and thus the net electric field, must be zero.

Correct Answer: C

The content explicitly states that "The net electric field at a given location is the vector sum of individual electric fields created by nearby charged objects." This is the principle of superposition. Because electric fields have both magnitude and direction, they must be added as vectors.