Physics 2 Quiz: Electric Field From Point Charges

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Electric Field From Point ChargesQuestion 1 of 7

Charge $+Q$ is fixed at the origin and charge $-2Q$ is fixed at $(d, 0)$ . Which of the following correctly describes the leading-order behavior of the electric field along the positive x-axis for $x \gg d$ ?

E \approx -\dfrac{kQ}{x^2}

, directed in the

-x

direction, because the system has a nonzero net charge of

-Q

, so the dominant far-field contribution is the monopole term from that net charge, which falls off as

x^{-2}

and points toward the net negative charge.

E \approx -\dfrac{3kQd}{x^3}

, directed in the

-x

direction, because the contributions from

+Q

and

-2Q

cancel to leading order, leaving the dipole term as the dominant far-field behavior; the dipole field falls as

x^{-3}

with a coefficient determined by the dipole moment

p = 2Qd

and the separation

d

E \approx -\dfrac{kQ}{x^2}\left(1 - \dfrac{2d}{x}\right)

, directed in the

-x

direction, because expanding both charge contributions in powers of

d/x

shows that the leading monopole term

-kQ/x^2

is corrected at order

x^{-3}

by a dipole contribution; however, this dipole correction is a subleading term, not the leading behavior.

E \approx +\dfrac{kQd^2}{x^4}

, directed in the

+x

direction, because the monopole and dipole contributions from this two-charge system cancel each other exactly for

x \gg d

, leaving the quadrupole term as the leading far-field behavior, which is positive and falls as

x^{-4}

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Physics 2 Quiz

Physics 2 Quiz: Electric Field From Point Charges

Practice Electric Field From Point Charges in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Electric Field From Point Charges, giving you a quick way to practice the rules, question types, and explanations that matter most for 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.

All questions

Question 1

$E \approx -\dfrac{kQ}{x^2}$ , directed in the $-x$ direction, because the system has a nonzero net charge of $-Q$ , so the dominant far-field contribution is the monopole term from that net charge, which falls off as $x^{-2}$ and points toward the net negative charge. (correct answer)
$E \approx -\dfrac{3kQd}{x^3}$ , directed in the $-x$ direction, because the contributions from $+Q$ and $-2Q$ cancel to leading order, leaving the dipole term as the dominant far-field behavior; the dipole field falls as $x^{-3}$ with a coefficient determined by the dipole moment $p = 2Qd$ and the separation $d$ .
$E \approx -\dfrac{kQ}{x^2}\left(1 - \dfrac{2d}{x}\right)$ , directed in the $-x$ direction, because expanding both charge contributions in powers of $d/x$ shows that the leading monopole term $-kQ/x^2$ is corrected at order $x^{-3}$ by a dipole contribution; however, this dipole correction is a subleading term, not the leading behavior.
$E \approx +\dfrac{kQd^2}{x^4}$ , directed in the $+x$ direction, because the monopole and dipole contributions from this two-charge system cancel each other exactly for $x \gg d$ , leaving the quadrupole term as the leading far-field behavior, which is positive and falls as $x^{-4}$ .

Explanation: When analyzing far-field electric behavior from a charge distribution, your first step should always be to check the net charge. The multipole expansion tells us that the slowest-decaying (dominant) term at large distances corresponds to the lowest non-vanishing moment: monopole

(x^{-2})

, then dipole

(x^{-3})

, then quadrupole

(x^{-4})

. Here, the net charge is

+Q + (-2Q) = -Q

. Since the monopole moment is nonzero, the far-field is dominated by that net charge acting like a point charge

-Q

at the origin. Along the positive x-axis for

x \gg d

, this gives

E \approx -\dfrac{kQ}{x^2}

, directed in the

-x

direction (toward the net negative charge). That's exactly what A states — making it correct. B is wrong because it claims the leading contributions cancel, leaving a dipole term. Monopole contributions only cancel when the net charge is zero; here it isn't. The dipole term is real but subleading, not dominant. C is actually the most physically honest expansion — it correctly identifies the monopole as leading and the dipole as a correction — but it misrepresents itself as describing only a "correction," and its full expression blends leading and subleading terms together. More importantly, the question asks for leading-order behavior, which is simply

-kQ/x^2

, so C overcomplicates the answer. D is doubly wrong: neither the monopole nor dipole cancel here, so invoking a quadrupole as the leading term is unjustified. Your strategy: always compute the net charge first. If it's nonzero, the monopole term dominates at large distances — no further expansion needed.

Question 2

Two point charges are fixed in space. Charge $q_1 = +4\,\mu\text{C}$ is located at the origin, and charge $q_2 = -1\,\mu\text{C}$ is located at $x = 3\,\text{m}$ . A test charge is placed somewhere along the x-axis.

At which location along the x-axis is the net electric field equal to zero?

$x = 6\,\text{m}$ , because the test point must lie beyond $q_2$ where both fields point in the same direction and can only cancel if the larger charge is farther away, giving a distance ratio consistent with the charge magnitudes. (correct answer)
$x = 1\,\text{m}$ , because the test point lies between the two charges where the fields from the positive and negative charge both point in the negative x-direction, allowing cancellation at the ratio determined by the charge magnitudes.
$x = -3\,\text{m}$ , because the test point must lie on the opposite side of the larger charge from the smaller charge, where both fields are directed in the same direction and the inverse-square falloff produces equal magnitudes at that separation.
$x = 9\,\text{m}$ , because the test point must lie beyond $q_2$ on the far side, and balancing $kq_1/r_1^2 = kq_2/r_2^2$ with the appropriate distances from each charge yields this position along the positive x-axis.

Explanation: When asked where the electric field is zero between two point charges, your first move should be identifying the region where cancellation is possible. The electric field from a positive charge points away from it, and from a negative charge points toward it. For the fields to cancel, they must point in opposite directions at the test point. Between the charges (0 < x < 3 m), both fields point toward

q_2

(the negative charge) — they reinforce rather than cancel, so no zero exists there. On the far side of

q_1

(x < 0), both fields point away from the region between the charges, again reinforcing. The only viable region is beyond

q_2

(x > 3 m), where

q_1

pushes rightward and

q_2

pulls leftward — opposite directions. Setting magnitudes equal:

\frac{k|q_1|}{r_1^2} = \frac{k|q_2|}{r_2^2}

, where

r_1 = x

and

r_2 = x - 3

. Substituting:

\frac{4}{x^2} = \frac{1}{(x-3)^2}

, so

2(x-3) = x

, giving

x = 6\,\text{m}

. That confirms A. Choice B is wrong because between the charges, both fields point in the same direction — no cancellation is possible there. Choice C places the zero on the far side of the larger charge, but there both fields point the same way (away from

q_1

), so they reinforce. Choice D misapplies the distance equation, likely confusing

r_2 = x - 3

with a larger separation and arriving at an incorrect position. Strategy tip: Always determine the correct region before solving algebraically — it eliminates distractors immediately and prevents sign errors in your distance expressions.

Question 3

A point charge $+Q$ is fixed at the origin. A second point charge $+4Q$ is fixed at $(L, 0)$ . A third charge $q$ (sign and magnitude unknown) is placed at $\left(\frac{L}{3}, 0\right)$ . For the electric field to be zero at the point $\left(\frac{L}{3}, 0\right)$ due to ALL THREE charges (including $q$ itself), what can be concluded about $q$ ?

$q$ must be negative with magnitude $\frac{4Q}{9}$ , determined by requiring the field contributions from $+Q$ and $+4Q$ to cancel at $x = L/3$ , since a point charge cannot contribute to the field at its own location and the sign of $q$ is irrelevant to the zero-field condition.
$q$ must equal zero, because the only way for a charge to be in a region where the total electric field is zero is if it contributes no field itself, and a nonzero charge at $x=L/3$ would always produce a field that prevents cancellation at that point.
The field at $x = L/3$ cannot be zero regardless of $q$ , because $+Q$ and $+4Q$ both produce fields pointing in the $+x$ direction at $x = L/3$ and no charge placed there can contribute a field at its own location to cancel them.
$q$ can be any value (including zero), because a point charge does not contribute to the electric field at its own position, and the field from $+Q$ and $+4Q$ at $x = L/3$ are equal and opposite by the inverse-square law given the $1:2$ distance ratio matching the $1:4$ charge ratio. (correct answer)

Explanation: When analyzing electric fields from multiple charges at a specific point, always ask two questions: (1) what fields do the other charges produce there, and (2) does the charge at that point contribute to the field there? That second question is the key to unlocking this problem. A point charge does not contribute to the electric field at its own location — this is a fundamental principle of classical electrostatics. The field of a point charge is evaluated at field points other than the source itself. So whatever value

q

takes, it contributes nothing to the field at

x = L/3

. This means the zero-field condition depends entirely on the fields from

+Q

and

+4Q

canceling each other at

x = L/3

. The charge

+Q

is at distance

L/3

, producing a field

\frac{kQ}{(L/3)^2} = \frac{9kQ}{L^2}

in the

+x

direction. The charge

+4Q

is at distance

2L/3

, producing a field

\frac{4kQ}{(2L/3)^2} = \frac{9kQ}{L^2}

in the

-x

direction. These are equal and opposite — they cancel perfectly! So the field is already zero there regardless of

q

, making D correct. A incorrectly claims

q

must be a specific negative value, confusing this with a force-balance problem where

q

's sign does matter. B incorrectly claims

q

must be zero — this would only matter if

q

actually contributed to the field at its own location, which it doesn't. C wrongly asserts the fields from

+Q

and

+4Q

both point in

+x

, ignoring that

+4Q

is to the right of the test point, pushing leftward. Study tip: Always sketch the geometry first — identifying which direction each charge pushes the field at your test point prevents sign errors that lead to traps like C.

Question 4

Two charges, $+q$ at $(0, 0, 0)$ and $-q$ at $(0, 0, d)$ , form a dipole along the z-axis. By the standard convention, the dipole moment vector points from the negative charge to the positive charge, giving $\mathbf{p} = -qd\hat{z}$ (in the $-z$ direction). A third charge $+Q$ is placed at $(0, 0, R)$ with $R \gg d$ . Using the leading-order dipole approximation, what is the force on $+Q$ ?

$F = \dfrac{2kqQd}{R^3}$ directed in $-z$ , because the axial dipole field at distance $R$ from the center points in the same direction as $\mathbf{p}$ (which is $-z$ ), and the force on the positive charge $+Q$ is in the direction of this field. (correct answer)
$F = \dfrac{2kqQd}{R^3}$ directed in $+z$ , because the $-q$ charge at $(0,0,d)$ is closer to $+Q$ than the $+q$ charge is, so $+Q$ is attracted upward toward $-q$ , which is in the $+z$ direction; the dipole approximation reproduces this net attractive force pointing away from the dipole.
$F = \dfrac{6kqQd}{R^4}$ directed in $+z$ , because the force on a charge in a dipole field is found by differentiating the dipole field with respect to position; differentiating $E \propto R^{-3}$ introduces an extra factor of $R^{-1}$ and a coefficient of 6, giving a force that falls as $R^{-4}$ .
$F = \dfrac{2kqQd}{R^3}$ directed in $+z$ , because on the positive z-axis the axial dipole field always points in the $+z$ direction for any dipole oriented along the z-axis, regardless of which end is positive or negative, and a positive charge $+Q$ is pushed in the field direction.

Explanation: Dipole problems require careful attention to two things: the direction of the dipole moment vector and the form of the axial electric field. When a test charge sits far away on the dipole's axis, you need both to find the force correctly. The dipole moment here is

\mathbf{p} = -qd\hat{z}

, pointing in the

-z

direction (from

+q

at the origin toward

-q

(0,0,d)

— wait, re-read: the convention says from negative to positive, so from

(0,0,d)

(0,0,0)

, confirming

\mathbf{p} = -qd\hat{z}

). The axial dipole field at a point on the

+z

axis is

\mathbf{E} = \frac{2k\mathbf{p}}{R^3}

, which here gives

\mathbf{E} = \frac{2k(-qd)}{R^3}\hat{z}

, pointing in

-z

. The force on

+Q

\mathbf{F} = Q\mathbf{E}

, so the magnitude is

\frac{2kqQd}{R^3}

directed in

-z

. This confirms A is correct: the field on the axis aligns with

\mathbf{p}

, and

+Q

is pushed in that same

-z

direction. Choice B gets the physical intuition right (

-q

is closer, so

+Q

is attracted toward it in

+z

) but assigns the wrong direction — it confuses the geometry because

-q

is at higher

z

, so attraction would be

+z

, yet the dipole field calculation shows

-z

. The resolution is that

\mathbf{p}

points

-z

, and the axial field formula correctly captures that

+q

being farther away wins. Choice C wrongly differentiates the field to get force — that step applies only when the external field is nonuniform across the dipole itself, not for a point charge in a dipole field. Choice D incorrectly claims the axial field direction is always

+z

regardless of dipole orientation — the field direction absolutely depends on which way

\mathbf{p}

points. Study tip: Always write out

\mathbf{p}

with its sign and direction before plugging into any dipole field formula — the sign of

\mathbf{p}

is where most mistakes happen.

Question 5

A physicist models a water molecule by placing a charge of $-2e$ at the origin and two charges of $+e$ each at positions $\mathbf{r}_1$ and $\mathbf{r}_2$ symmetric about the y-axis, each at distance $d$ from the origin and at angle $\theta$ above the negative x-axis (i.e., the molecule opens toward $+y$ ). The net dipole moment points in the $+y$ direction.

At a field point on the positive y-axis far from the molecule (distance $R \gg d$ ), which expression best describes the leading-order behavior of the electric field magnitude?

$E \propto R^{-2}$ , because the configuration has a nonzero net charge of zero with a nonzero dipole moment, and the dominant far-field contribution from any neutral charge distribution with a net dipole moment falls off as the inverse square of the distance.
$E \propto R^{-3}$ , because the net charge of the system is zero, making the monopole term vanish, and the leading contribution comes from the dipole moment, which produces a field that decreases as the inverse cube of the distance from the charge distribution. (correct answer)
$E \propto R^{-4}$ , because the charge arrangement is symmetric about the y-axis, which causes the dipole term to vanish by symmetry, leaving the quadrupole as the dominant multipole contribution with its characteristic $R^{-4}$ far-field dependence.
$E \propto R^{-1}$ , because the three-charge system acts collectively like a line charge along the y-axis for distant field points, and the field from a line charge falls off inversely with perpendicular distance regardless of the detailed charge distribution.

Explanation: When analyzing the far-field behavior of any localized charge distribution, your first move should always be to identify the leading multipole term — monopole, dipole, or quadrupole — because each successive term falls off one power faster with distance. The monopole term dominates when the net charge is nonzero, producing

E \propto R^{-2}

. Here, the total charge is

-2e + e + e = 0

, so the monopole term vanishes entirely. Next comes the dipole term. The two

+e

charges are displaced symmetrically about the y-axis but both sit above the negative x-axis, so their y-components add constructively. The central

-2e

charge sits at the origin. The net dipole moment

\mathbf{p}

points in the

+y

direction and is nonzero. A nonzero dipole produces a field that falls off as

E \propto R^{-3}

, making B the correct answer. On the positive y-axis (the dipole axis), the field is

E = \frac{1}{4\pi\epsilon_0}\frac{2p}{R^3}

, confirming the

R^{-3}

dependence. A is wrong because

E \propto R^{-2}

describes a monopole field — it requires nonzero net charge, which this system does not have. The phrase "net charge of zero with a nonzero dipole moment" is contradictory to the conclusion drawn. C incorrectly claims the dipole vanishes by symmetry. Symmetry about the y-axis cancels x-components, but the y-component of the dipole moment survives — it does not vanish. D is a fabricated analogy. Nothing about three point charges justifies treating them as a line charge; this reasoning has no physical basis. Study tip: Always check net charge first (monopole), then dipole moment. Only if both vanish do you need the quadrupole. This hierarchy —

R^{-2}, R^{-3}, R^{-4}

— is a reliable pattern for multipole far-field problems.

Question 6

Two point charges $+Q$ and $-Q$ are separated by distance $2d$ , forming an electric dipole centered at the origin with the positive charge at $+d\hat{x}$ and the negative charge at $-d\hat{x}$ . Point P is located at $(0, R, 0)$ on the y-axis with $R \gg d$ . Point S is located at $(R, 0, 0)$ on the x-axis with the same $R$ . Which statement correctly compares the electric field magnitudes at P and S?

$E_P = E_S$ , because both points are at the same distance $R$ from the dipole center, and the dipole field magnitude depends only on the radial distance from the center, not on the direction relative to the dipole axis.
$E_S = 2E_P$ , because point S lies along the dipole axis ( $\hat{x}$ -direction) where the axial field is $E_S = 2kp/R^3$ , while point P lies on the perpendicular bisector (equatorial plane) where the field is $E_P = kp/R^3$ , so the on-axis field is exactly twice the equatorial field at the same distance. (correct answer)
$E_P = 2E_S$ , because point P on the y-axis is perpendicular to the charge separation, which places it along the direction of the dipole moment vector; points along the dipole moment direction experience the stronger axial field, while points along the x-axis (perpendicular to the moment) experience the weaker equatorial field.
$E_S = \sqrt{3}\,E_P$ , because the general dipole field magnitude at angle $\theta$ from the dipole axis is $E = \frac{kp}{R^3}\sqrt{1+3\cos^2\theta}$ , and evaluating this ratio at $\theta = 0°$ (for S) versus $\theta = 90°$ (for P) yields $\sqrt{4}/\sqrt{1} = 2$ ... actually $\sqrt{3}$ if the angular factors are instead evaluated for the perpendicular components only, without accounting for the radial term.

Explanation: When analyzing electric dipole fields, the key is recognizing that the field magnitude depends on both distance from the center and the angle relative to the dipole axis. The dipole moment here points along

\hat{x}

(from

-Q

+Q

), so point S at

(R,0,0)

lies along the dipole axis (

\theta = 0°

), while point P at

(0,R,0)

lies on the perpendicular bisector (

\theta = 90°

). The two standard dipole field results you must know are: the axial field (on-axis)

E_{axis} = \frac{2kp}{R^3}

, and the equatorial field (perpendicular bisector)

E_{eq} = \frac{kp}{R^3}

. Since S is on-axis and P is equatorial, we get

E_S = 2E_P

, confirming answer B is correct. A is wrong because it assumes the dipole field is spherically symmetric — it isn't. The angular dependence is a defining feature of dipole fields, and ignoring it is a classic misconception. C confuses which axis is which: the dipole moment points along

\hat{x}

, so points along

\hat{x}

(like S) experience the stronger axial field, not points along

\hat{y}

(like P). The phrasing "along the dipole moment direction gives stronger field" is backwards in its geometric assignment. D references the correct general formula

E = \frac{kp}{R^3}\sqrt{1+3\cos^2\theta}

, which actually gives

\sqrt{4}/\sqrt{1} = 2

— the answer contradicts itself and arrives at

\sqrt{3}

through faulty partial reasoning. Your study tip: memorize the ratio

E_{axis} = 2E_{eq}

and always identify the dipole axis direction before labeling any point as "axial" or "equatorial."

Question 7

Five identical charges $+q$ are placed at the vertices of a regular pentagon centered at the origin. A sixth charge $-q$ replaces one of the five $+q$ charges, so the configuration now has four $+q$ charges and one $-q$ charge at one vertex.

What is the electric field at the center of the pentagon due to this five-charge configuration (four $+q$ and one $-q$ )?

Zero, because the four remaining positive charges and the one negative charge still maintain enough symmetry around the center of the pentagon to produce mutually canceling field contributions, just as five identical charges would.
Equal to the field that a single charge $+2q$ would produce at the center, directed away from the vertex holding $-q$ , because replacing $+q$ with $-q$ at that vertex changes that vertex's contribution by $-2kq/R^2$ (a reversal plus original removal), and this equals the field of $+2q$ pointing away from that vertex.
Equal to the field that a single charge $+2q$ would produce at the center, directed toward the vertex holding $-q$ , because the net effect of replacing $+q$ with $-q$ is to remove the outward contribution and add an inward contribution, giving a net inward field of magnitude $2kq/R^2$ at the center. (correct answer)
Equal to the field that a single charge $-q$ would produce at the center, directed toward the vertex holding $-q$ , because the four positive charges cancel each other by symmetry (they do not form a perfectly symmetric group of four) and only the single $-q$ charge contributes a net field at the center pointing toward it.

Explanation: When you see a configuration with apparent symmetry broken by one altered charge, the most powerful tool is superposition. Rather than computing each field vector individually, think about what changed relative to a simpler, known configuration. Start with all five charges being

+q

. By the perfect symmetry of a regular pentagon, the electric field at the center is exactly zero — every field vector is canceled by contributions from the other charges. Now, replacing one

+q

with

-q

is mathematically equivalent to removing the original

+q

(which contributed a field pointing away from that vertex) and adding a

-q

(which contributes a field pointing toward that vertex). The net change at the center is therefore two contributions of magnitude

kq/R^2

both directed inward — toward the vertex holding

-q

— giving a total field of

2kq/R^2

directed toward that vertex. This is exactly the field a single charge

+2q

would produce at distance

R

, so C is correct. Choice A is wrong because the symmetry is genuinely broken — four

+q

charges at pentagon vertices do not cancel each other; that cancellation required all five to be identical. Choice B correctly identifies the magnitude

2kq/R^2

but gets the direction backwards. Removing the outward

+q

contribution and adding an inward

-q

contribution both point the net field toward the negative vertex, not away from it. Choice D incorrectly assumes the four

+q

charges cancel among themselves — they don't form a symmetric group, so their net contribution is nonzero and cannot be ignored. Your takeaway: when symmetry is broken by one altered charge, use the "add and subtract" superposition trick — compare the new configuration to a perfectly symmetric one you already understand.