Electric Fields
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AP Physics 2 › Electric Fields
A fixed source charge $+Q$ creates an electric field. A test charge is placed at point $P$ a distance $r$ from $+Q$, then replaced by a test charge of twice the magnitude at the same point. Which statement best describes the electric field at point $P$ after the replacement?
It becomes zero because the test charge cancels the field at $P$.
It doubles because a larger test charge produces a larger field.
It reverses direction because the test charge is larger.
It is unchanged because it depends only on the source charge and location.
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).
Two fixed source charges, $+Q$ and $-Q$, are separated by a distance $2d$. A test charge is placed at the midpoint between them. Which statement best describes the direction of the electric field at the midpoint?
It depends on whether the test charge is positive or negative.
It points from $+Q$ toward $-Q$ along the line joining them.
It points from $-Q$ toward $+Q$ along the line joining them.
It is zero because equal and opposite charges always cancel fields.
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.
Two fixed source charges lie on the $x$ axis: $+Q$ at $x=-0.10,\text{m}$ and $-Q$ at $x=+0.10,\text{m}$. Point $P$ is at the origin. A tiny positive test charge is placed at $P$ to probe the field. Which statement best describes the direction of the net electric field at $P$ due to the source charges?
It depends on whether the test charge is positive or negative.
It points in the $-x$ direction, because the positive source repels the test charge.
It points in the $+x$ direction, from the positive source toward the negative source.
It is zero because the charges are equal in magnitude.
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.
A fixed source charge $+Q$ is at the origin. Point $P$ is on the $+x$ axis. A small test charge $+q$ is placed at $P$ to probe the field without changing it. Which statement best describes the direction of the electric field at $P$ due to the source charge?
It points in the $+x$ direction, away from the source charge.
It depends on the sign of the test charge $+q$ placed at $P$.
It is zero because there is no charge located at point $P$.
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.
A fixed source charge $-Q$ is at the origin. Points $A$ and $B$ lie on the $+x$ axis with $r_B=3r_A$. A small positive test charge is placed at each point, one at a time, to probe the field. Compared to the electric field magnitude at $A$, the electric field magnitude at $B$ is
zero because the source charge is negative.
one-third as large.
three times as large.
one-ninth as large.
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.
A fixed source charge $+Q$ is on the $y$ axis at $y=+0.50,\text{m}$. Point $P$ is on the $y$ axis at $y=+1.00,\text{m}$. A small positive test charge is momentarily placed at $P$ to probe the field. Which statement best describes the direction of the electric field at $P$ due to the source charge?
It points in the $-y$ direction, toward the positive source charge.
It points in the $+y$ direction, away from the positive source charge.
It is zero because the test charge is positive.
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.
A small source charge of $+Q$ is fixed at the origin. A positive test charge $+q$ is placed at point $P$ on the $+x$-axis, and then moved to point $R$ twice as far from the origin along the same axis. Which statement best describes the electric field at $R$ compared to at $P$?
It is one-fourth as strong and points in the same direction.
It is one-half as strong and points in the same direction.
It is the same strength because the test charge is unchanged.
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.
Two identical fixed source charges, each $+Q$, are separated by a distance $2d$. A test charge is placed at the midpoint between them. Which statement best describes the electric field at the midpoint?
It is nonzero only if the test charge is negative.
It points toward one of the charges because fields add.
It is zero because there is no source charge at the midpoint.
It is zero because the fields from the two charges cancel.
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.
A fixed source charge $+Q$ is at the origin. Points $P$ and $S$ are on the same circle of radius $r$ centered on the origin, at different angles. A small positive test charge $+q$ can be placed at either point. Which statement best describes the electric field magnitudes at $P$ and $S$?
$E_P = 0$ and $E_S = 0$ because neither point contains a charge.
$E_P > E_S$ because the field is stronger along the $+x$-axis.
$E_P < E_S$ because the field depends on the direction from the charge.
$E_P = E_S$ because both points are the same distance from the source 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.
Two fixed source charges, $+Q$ and $-Q$, are separated by a distance $d$ on a line, with $+Q$ on the left. Point $M$ is the midpoint. A small positive test charge $+q$ is placed at $M$ to sense the field. Which statement best describes the direction of the net electric field at $M$ due to the source charges?
It points right, from $+Q$ toward $-Q$.
It is zero because the two source charges have equal magnitude.
It points left, toward the $+Q$ source charge.
It depends on the value of the test charge $+q$.
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.