AP Physics 2 Quiz: Electric Fields

What this quiz covers

This quiz focuses on Electric Fields, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. It doubles because a larger test charge produces a larger field.
2. It reverses direction because the test charge is larger.
3. It is unchanged because it depends only on the source charge and location. (correct answer)
4. It becomes zero because the test charge cancels the field at $P$.

Explanation: This question tests understanding of electric fields. The electric field is a property of space determined solely by source charges and their positions, existing whether or not any test charge is present. At any point P, the field E = kQ/r² depends only on the source charge Q and distance r, not on any test charge placed there. Replacing the test charge with one of different magnitude does not alter the field at that location. Choice A incorrectly assumes the test charge contributes to the field it experiences, confusing the roles of source and test charges. To avoid this error, always distinguish between source charges (which create fields) and test charges (which experience forces in those fields).

Question 2

1. It points from $+Q$ toward $-Q$ along the line joining them. (correct answer)
2. It points from $-Q$ toward $+Q$ along the line joining them.
3. It is zero because equal and opposite charges always cancel fields.
4. It depends on whether the test charge is positive or negative.

Explanation: This question tests understanding of electric fields. The electric field is the vector sum of contributions from all source charges, with positive charges creating fields pointing away and negative charges creating fields pointing toward them. At the midpoint between +Q and -Q, the field from +Q points rightward (away from +Q) while the field from -Q also points rightward (toward -Q). Since both charges have equal magnitude and are equidistant, their fields have equal magnitude and same direction, resulting in a net field pointing from +Q toward -Q. Choice C incorrectly assumes opposite charges create canceling fields, confusing this with the zero field between identical charges. When analyzing fields from multiple charges, carefully determine each field's direction before adding vectors.

Question 3

1. It is zero because the charges are equal in magnitude.
2. It points in the $-x$ direction, because the positive source repels the test charge.
3. It depends on whether the test charge is positive or negative.
4. It points in the $+x$ direction, from the positive source toward the negative source. (correct answer)

Explanation: Electric fields are vector quantities that must be added using vector addition when multiple source charges are present. The positive charge at x = -0.10 m creates a field at the origin pointing in the +x direction (away from the positive charge), while the negative charge at x = +0.10 m also creates a field at the origin pointing in the +x direction (toward the negative charge). Since both individual fields point in the same direction (+x), they add constructively, giving a net field in the +x direction. Choice C incorrectly assumes equal magnitude charges always produce zero net field, ignoring that the charges have opposite signs. The strategy is to determine each field's direction separately, then add them as vectors.

Question 4

1. It is zero because there is no charge located at point $P$.
2. It depends on the sign of the test charge $+q$ placed at $P$.
3. It points in the $+x$ direction, away from the source charge. (correct answer)
4. It points in the $-x$ direction, toward the source charge.

Explanation: This question tests understanding of electric fields. Electric fields are properties of space created by source charges, existing whether or not a test charge is present to detect them. A positive source charge +Q creates an electric field that points radially outward from the charge at all points in space. Since point P is on the +x axis (to the right of the origin), the field at P points in the +x direction, away from the positive source. Choice C incorrectly suggests the field direction depends on the test charge sign, confusing the field (which exists independently) with the force on a test charge (which does depend on the test charge's sign). To avoid this error, always determine the electric field first as a property of space due to source charges, then consider any forces on test charges placed in that field.

Question 5

1. one-ninth as large. (correct answer)
2. three times as large.
3. one-third as large.
4. zero because the source charge is negative.

Explanation: This question tests understanding of electric fields. Electric field magnitude depends on the source charge and inversely on the square of distance: E = k|Q|/r². Point B is three times farther from the source than point A (rB = 3rA), so the field at B is proportional to 1/(3rA)² = 1/9rA². This makes the field magnitude at B one-ninth as large as at A, regardless of the source charge's sign (negative here). Choice D incorrectly suggests the field is zero because the source is negative, confusing field magnitude (always positive) with field direction. To compare field magnitudes correctly, focus on the 1/r² relationship and remember that magnitude is independent of the source charge's sign.

Question 6

1. It points in the $+y$ direction, away from the positive source charge. (correct answer)
2. It points in the $-y$ direction, toward the positive source charge.
3. It is zero because the test charge is positive.
4. It points in the $+y$ direction only if the test charge is negative.

Explanation: Electric fields are properties of space created by source charges that point away from positive charges and toward negative charges. The positive source charge at y = +0.50 m creates an electric field that points radially outward at all surrounding points. Since point P is at y = +1.00 m (above the source on the y-axis), the field at P points away from the source in the +y direction. Choice D incorrectly suggests the field direction depends on the test charge sign, confusing the electric field (a property of space) with the force on a charge. The strategy is to visualize field lines emanating outward from positive charges to determine field direction at any point.

Question 7

1. It is one-fourth as strong and points in the same direction. (correct answer)
2. It is one-half as strong and points in the same direction.
3. It is the same strength because the test charge is unchanged.
4. It is zero because there is no charge located at $R$.

Explanation: This question tests understanding of electric fields. The electric field is a property of space created by source charges, existing at every point regardless of whether a test charge is present. For a point charge +Q, the electric field magnitude follows E = kQ/r², decreasing with the square of distance from the source. When the distance doubles from P to R, the field becomes E = kQ/(2r)² = kQ/4r² = E_P/4, making it one-fourth as strong. Choice C incorrectly assumes the field depends on the test charge, confusing the field itself with the force it would exert. To solve electric field problems, always calculate the field based solely on source charges and position, before considering any forces on test charges.

Question 8

1. It points toward one of the charges because fields add.
2. It is zero because the fields from the two charges cancel. (correct answer)
3. It is nonzero only if the test charge is negative.
4. It is zero because there is no source charge at the midpoint.

Explanation: This question tests understanding of electric fields. The electric field at any point is the vector sum of fields from all source charges, following the principle of superposition. Each +Q charge creates a field pointing radially outward from itself, with magnitude E = kQ/d² at the midpoint. Since the charges are identical and equidistant from the midpoint, their field magnitudes are equal but directions are opposite (one points right, one points left). These equal and opposite vectors sum to zero at the midpoint. Choice D incorrectly assumes fields only exist where charges are located, misunderstanding that fields permeate all space. To find the net field, always add field vectors from each source charge, considering both magnitude and direction.

Question 9

1. $E_P > E_S$ because the field is stronger along the $+x$-axis.
2. $E_P < E_S$ because the field depends on the direction from the charge.
3. $E_P = E_S$ because both points are the same distance from the source charge. (correct answer)
4. $E_P = 0$ and $E_S = 0$ because neither point contains a charge.

Explanation: This question tests understanding of electric fields. The electric field from a point source charge depends only on the distance from that charge, following E = kQ/r² for the magnitude. Since points P and S are both on a circle of radius r centered on the source charge +Q, they are equidistant from the source. Therefore, the field magnitudes at both points are identical: E_P = E_S = kQ/r². Choice A incorrectly assumes field strength varies with direction, confusing the vector nature of fields (direction changes) with scalar magnitude (which depends only on distance). To determine field magnitude from a point charge, use only the distance from the source, recognizing that all points equidistant from a point charge experience the same field strength.

Question 10

1. It is zero because the two source charges have equal magnitude.
2. It points right, from $+Q$ toward $-Q$. (correct answer)
3. It depends on the value of the test charge $+q$.
4. It points left, toward the $+Q$ source charge.

Explanation: This question tests understanding of electric fields. Electric fields from multiple sources add vectorially at each point in space. At the midpoint M between +Q (left) and -Q (right), the positive charge creates a field pointing right (away from +Q), while the negative charge also creates a field pointing right (toward -Q). Both individual fields point in the same direction—to the right—so they add constructively, resulting in a net field pointing right from +Q toward -Q. Choice C incorrectly suggests the field depends on the test charge value, confusing the field (determined by source charges only) with forces on test charges. To find net fields correctly, always determine the direction of each individual field first, then add them vectorially.

Question 11

1. It points in the $+x$ direction, away from the source charge.
2. It points in the $-x$ direction, toward the source charge. (correct answer)
3. It depends on whether the test charge is positive or negative.
4. It is zero because the source charge is negative.

Explanation: This question tests understanding of electric fields. Electric fields exist in space due to source charges, regardless of whether a test charge is present. A negative source charge -Q creates an electric field that points radially inward toward the charge at all points in space. Since point P is on the +x axis (to the right of the origin where -Q sits), the field at P points in the -x direction, toward the negative source. Choice C incorrectly suggests the field depends on the test charge's sign, conflating the field itself with the force that would act on a test charge. Remember: define the electric field based solely on the source charges and their positions, treating it as an intrinsic property of space before considering any test charges.

Question 12

1. four times as large. (correct answer)
2. one-fourth as large.
3. the same because the source charge is $+2Q$ at both locations.
4. twice as large.

Explanation: This question tests understanding of electric fields. The electric field magnitude from a point charge follows E = kQ/r², showing an inverse square relationship with distance. Point A is at distance r while point B is at distance 2r from the source charge +2Q. The field at A is EA = k(2Q)/r², while at B it's EB = k(2Q)/(2r)² = k(2Q)/4r² = EA/4. Therefore, the field at A is four times as large as at B. Choice D incorrectly claims the fields are equal because the source charge is the same, ignoring how field strength decreases with distance. When comparing fields at different distances, always apply the 1/r² relationship to determine how the field changes with position.

Question 13

1. It is zero because the fields from the two source charges cancel. (correct answer)
2. It is nonzero only if a test charge is placed at $M$.
3. It points toward the right source charge because both sources are positive.
4. It points toward the left source charge because $+q$ is positive.

Explanation: This question tests understanding of electric fields. Electric fields obey the principle of superposition: the net field at any point is the vector sum of fields from all source charges. At the midpoint M between two identical positive charges +Q, each creates a field pointing away from itself. The field from the left charge points right at M, while the field from the right charge points left at M. Since the charges are equal and equidistant from M, these fields have equal magnitudes but opposite directions, resulting in zero net field. Choice D incorrectly suggests the field is nonzero only if a test charge is placed there, misunderstanding that fields exist independently of test charges. Always calculate the electric field as a property of space before considering any test charges.

Question 14

1. the same because both points are the same distance from the source charge. (correct answer)
2. zero at both points because there is no charge located at $A$ or $B$.
3. greater because the field depends on direction as well as distance.
4. smaller because the test charge is moved to a different location.

Explanation: This question tests understanding of electric fields. The electric field magnitude from a point charge depends only on the distance from the source: E = kQ/r². Points A and B are both at the same distance r from the source charge +Q at the origin, just in different directions. Since electric field magnitude depends only on distance (not direction), the field magnitudes at A and B are equal. Choice A incorrectly suggests the field magnitude depends on direction, possibly confusing the vector nature of fields (direction matters for the full vector) with scalar magnitude (which depends only on distance). When comparing field magnitudes at equal distances from a point charge, remember that magnitude is independent of direction.

Question 15

1. four times as large.
2. twice as large.
3. one-fourth as large. (correct answer)
4. the same because the same test charge is used.

Explanation: This question tests understanding of electric fields. The electric field created by a point charge decreases with the square of the distance from the source: E = kQ/r². Point B is twice as far from the source charge as point A (rB = 2rA), so the field at B is proportional to 1/(2rA)² = 1/4rA². This makes the field at B one-fourth as large as at A. Choice D incorrectly suggests the field is the same because the same test charge is used, confusing the field (which depends only on source charges and position) with measurements made by a test charge. To solve field problems correctly, always write the field equation E = kQ/r² first, focusing on how the field depends on distance from the source.

Question 16

1. It is zero because the field is uniform.
2. It points left, because the test charge is negative.
3. It points right only when the test charge begins moving.
4. It points right, set by the source charges on the plates. (correct answer)

Explanation: Electric fields are properties of space created by source charges on the parallel plates, existing independently of any test charges. The uniform field between parallel plates points from positive to negative charges, which in this case is to the right as stated in the problem. This field direction is entirely determined by the configuration of charges on the plates and exists at every point in the region, including where the test charge is placed. Choice A incorrectly assumes the field direction depends on the sign of the test charge, confusing the field with the force on a charge. The strategy is to identify the electric field direction first, then use F = qE to find the force direction on any test charge.

Question 17

1. It is zero everywhere inside the cavity.
2. It is tangential to the shell's inner surface.
3. It points inward because induced charges reverse the field.
4. It is radially outward from the center. (correct answer)

Explanation: Electric fields are properties of space created by source charges, and inside a conductor in electrostatic equilibrium, the field must be zero. However, point P is inside the cavity (not inside the conducting material), where the field from the central charge +Q exists unaffected by the conductor. The field at P points radially outward from the positive source charge at the center, following E = kQ/r². Choice B incorrectly applies the zero-field rule for inside conductors to the cavity region, which is not conducting material. The strategy is to distinguish between regions inside conducting material (E = 0) and cavities within conductors (E determined by charges in the cavity).

Question 18

1. It points in the $-x$ direction, because both sources are positive.
2. It is zero because the sources are equidistant from $P$.
3. It points in the $+x$ direction, because the larger source on the left dominates. (correct answer)
4. It depends on the magnitude of the test charge placed at $P$.

Explanation: Electric fields are vector quantities that must be added when multiple sources are present. The +2Q charge at x = -0.30 m creates a field at the origin pointing in the +x direction (away from the positive charge), while the +Q charge at x = +0.30 m creates a field at the origin pointing in the -x direction (also away from its positive charge). Since both sources are equidistant from P, we compare magnitudes: E₁ = k(2Q)/r² pointing right and E₂ = kQ/r² pointing left. The net field is E₁ - E₂ = kQ/r² pointing in the +x direction because the larger charge on the left dominates. Choice C incorrectly assumes equal distances mean zero net field, ignoring the different charge magnitudes. The strategy is to calculate both the magnitude and direction of each field contribution before adding vectors.

Question 19

1. one-fourth as large.
2. twice as large.
3. four times as large. (correct answer)
4. the same because the test charges are tiny.

Explanation: Electric fields. The electric field created by a point charge depends on distance according to the inverse square law: E = kQ/r². This relationship exists at every point in space, independent of any test charge. Since point A is at half the distance of point B (0.20 m vs 0.40 m), and field strength varies as 1/r², the field at A is (0.40/0.20)² = 4 times stronger than at B. Choice D incorrectly suggests the field magnitude doesn't change with distance, confusing the independence of field from test charge with independence from position. Always apply the inverse square law when comparing field strengths at different distances.

Question 20

1. It points in the $+x$ direction.
2. It points in the $-x$ direction.
3. It is zero because the contributions cancel by symmetry. (correct answer)
4. It depends on whether the test charge is positive or negative.

Explanation: Electric fields. The electric field at any point is the vector sum of fields from all source charges, existing as a property of space independent of test charges. Each +Q charge creates a field pointing radially outward from itself. At the origin, the left charge creates a field pointing right (+x direction) while the right charge creates a field pointing left (-x direction). Since the charges are equal and equidistant from P, these fields have equal magnitudes but opposite directions, resulting in complete cancellation and zero net field. Choice D incorrectly assumes the field depends on the test charge's sign, missing that fields exist independently. Always analyze symmetry to identify when field contributions cancel.