Physics 2 Quiz: Force On Charge In E Field

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Force On Charge In E FieldQuestion 1 of 10

A proton and an alpha particle are placed at rest at the same location in a uniform electric field of magnitude $E_0$ directed in the positive x-direction. The alpha particle has charge $+2e$ and mass $4m_p$ , where $m_p$ is the proton mass.

What is the ratio of the magnitude of the force on the alpha particle to the magnitude of the force on the proton in this field?

\frac{1}{2}

, because the greater mass of the alpha particle reduces the net electromagnetic interaction with the field by a factor of 4, but the doubled charge partially compensates.

2

, because the alpha particle carries twice the elementary charge of the proton, and force depends on charge alone, not mass.

4

, because the alpha particle has both twice the charge and twice the number of nucleons contributing to electromagnetic interactions.

1

, because both particles experience the same external electric field

E_0

, and the field exerts equal force on any charged particle placed within it.

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Physics 2 Quiz: Force On Charge In E Field

Practice Force On Charge In E Field in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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Question 1

What is the ratio of the magnitude of the force on the alpha particle to the magnitude of the force on the proton in this field?

$\frac{1}{2}$ , because the greater mass of the alpha particle reduces the net electromagnetic interaction with the field by a factor of 4, but the doubled charge partially compensates.
$2$ , because the alpha particle carries twice the elementary charge of the proton, and force depends on charge alone, not mass. (correct answer)
$4$ , because the alpha particle has both twice the charge and twice the number of nucleons contributing to electromagnetic interactions.
$1$ , because both particles experience the same external electric field $E_0$ , and the field exerts equal force on any charged particle placed within it.

Explanation: When a charged particle sits in an electric field, the force it experiences comes directly from Coulomb's law applied to a uniform field:

F = qE

. Notice what's in that equation — charge

q

and field strength

E

. Mass appears nowhere. This distinction is the heart of this question. The proton has charge

+e

, so it feels a force

F_p = eE_0

. The alpha particle has charge

+2e

, so it feels

F_\alpha = 2eE_0

. The ratio is simply

\frac{F_\alpha}{F_p} = \frac{2eE_0}{eE_0} = 2

, making B correct. A is a classic trap: it sneaks in the alpha particle's greater mass as though mass affects the electromagnetic force. Mass matters when you calculate acceleration (via

F = ma

), but the question asks about force, not acceleration. Don't conflate the two. C doubles down on a related misconception — "nucleons contributing to electromagnetic interactions" is physically wrong. Only net charge determines the electric force; neutrons are electrically neutral and contribute nothing to

F = qE

. D confuses "same field" with "same force." Yes, both particles are in the same field

E_0

, but the force each experiences scales with its own charge. Equal field ≠ equal force unless charges are also equal. A useful habit: whenever a question mixes charge, mass, and fields, consciously separate electromagnetic quantities (charge, field) from mechanical quantities (mass, inertia). Force in an electric field depends only on charge — save mass for when the question asks about acceleration or kinetic energy.

Question 2

A particle of charge $q$ and mass $m$ is released from rest in a uniform electric field $E$ . A second, identical particle is instead placed in a region where the electric field is $2E$ but where a uniform opposing drag force $F_d = qE$ also acts on it.

Compared to the net force on the first particle, the net force on the second particle is:

Equal in magnitude, because the electric force on the second particle is $2qE$ and the drag force is $qE$ , leaving a net force of $qE$ — identical to the net force on the first particle. (correct answer)
Twice as large, because the electric field acting on the second particle is doubled, and the drag force only becomes significant once the particle reaches a nonzero speed; at any given instant it is less than $qE$ .
Zero, because the drag force $qE$ and the doubled electric force $2qE$ together constitute an action–reaction pair that must sum to zero by Newton's third law.
Half as large, because the drag force $qE$ acts against the electric force $2qE$ , so the net force $qE$ must be further divided between the two particles sharing the same charge $q$ , yielding an effective force of $\frac{1}{2}qE$ .

Explanation: When analyzing net force problems, your job is always the same: identify every force acting on the object, note its direction, and sum them algebraically. Resist the urge to let extra information distract you from that simple process. For the first particle, the only force is the electric force:

F_1 = qE

. For the second particle, the electric force is

2qE

forward, and the drag force is

qE

backward. Summing these gives

F_2 = 2qE - qE = qE

. The net forces are equal — confirming that A is correct. Here's why each wrong answer fails. B claims the drag force "only becomes significant at nonzero speed," which misreads the problem entirely — the passage states

F_d = qE

as a constant, uniform force, not a velocity-dependent one like air resistance. It acts from the very first instant. C invokes Newton's third law, but that law describes force pairs between two different objects (e.g., particle pushing on field source, field source pushing on particle) — it never applies to two forces acting on the same object from different sources. The electric force and drag force are not an action–reaction pair. D introduces the idea of "dividing" the net force between two particles sharing charge

q

, which is physically meaningless. Each particle is an independent system; you never split a force across separate objects like that. The key strategy: always treat each object as its own isolated system and list forces acting on that object only. When a problem loads you with extra numbers, slow down and apply

F_{net} = \sum F

methodically — the algebra will cut through any distraction.

Question 3

A horizontal parallel-plate capacitor has plates separated by distance $d = 2.0\,\text{cm}$ and maintains a uniform electric field of magnitude $E = 3.0\times10^4\,\text{N/C}$ directed vertically upward (from the bottom plate to the top plate). A particle with mass $m = 2.0\times10^{-5}\,\text{kg}$ and unknown charge $q$ is observed to move with constant velocity horizontally through the capacitor.

What must be the sign and magnitude of the charge $q$ on the particle?

$q = +6.5\times10^{-9}\,\text{C}$ ; positive, so the downward electric force on the positive charge adds to gravity, and constant velocity is maintained by air resistance balancing both downward forces.
$q = -6.5\times10^{-9}\,\text{C}$ ; negative, so the downward electric force on the negative charge (opposite to the upward field) adds to gravity, and constant velocity is maintained by air resistance.
$q = +6.5\times10^{-9}\,\text{C}$ ; positive, so the upward electric force on the positive charge exactly balances the downward gravitational force on the particle. (correct answer)
$q = -6.5\times10^{-9}\,\text{C}$ ; negative, so the upward electric force on the negative charge (in the direction opposite to the upward field) exactly balances the downward gravitational force.

Explanation: When a charged particle moves with constant velocity, the net force on it must be zero — this is Newton's First Law. Since the particle moves horizontally, you need to focus on the vertical forces: gravity pulls it down, and the electric force can push it either up or down depending on the charge's sign. For vertical equilibrium, the electric force must point upward to cancel gravity. The electric field already points upward, and the electric force on a charge is

\vec{F} = q\vec{E}

. For this force to point upward (same direction as

\vec{E}

), the charge must be positive. Setting the magnitudes equal:

qE = mg

q = \frac{mg}{E} = \frac{(2.0\times10^{-5})(9.8)}{3.0\times10^4} \approx 6.5\times10^{-9}\,\text{C}

This confirms C — a positive charge whose upward electric force exactly balances gravity, requiring no other forces for constant velocity. A is wrong because a positive charge in an upward field experiences an upward force, not a downward one. Invoking air resistance is unnecessary and physically incorrect here. B is wrong on two counts: a negative charge does experience a downward electric force (opposing the upward field), but then gravity and the electric force both point down — nothing balances them, so constant velocity is impossible without an additional upward force. D is wrong because a negative charge in an upward field experiences a downward force (not upward), so it cannot balance gravity. Study tip: Whenever you see "constant velocity," immediately write

\sum F = 0

and identify every force direction before assigning signs. Getting force directions wrong on charged-particle problems is the most common trap on this topic.

Question 4

An electron is held stationary in a region of space where both a uniform gravitational field ( $g = 9.8\,\text{m/s}^2$ downward) and a uniform electric field are present. The electron has mass $m_e = 9.11\times10^{-31}\,\text{kg}$ and charge $-e = -1.60\times10^{-19}\,\text{C}$ .

For the electron to remain stationary, what must be the direction and approximate magnitude of the electric field?

The electric field must point downward with magnitude approximately $5.6\times10^{-11}\,\text{N/C}$ , because the force on a negative charge is opposite to $\vec{E}$ , producing an upward electric force that cancels gravity. (correct answer)
The electric field must point upward with magnitude approximately $5.6\times10^{-11}\,\text{N/C}$ , because the force on a negative charge is opposite to $\vec{E}$ , producing a downward electric force that doubles the gravitational force.
The electric field must point downward with magnitude approximately $5.6\times10^{-11}\,\text{N/C}$ , because the downward field exerts a downward force on the electron, and the reaction force from the field source balances gravity.
The electric field must point upward with magnitude approximately $5.6\times10^{-11}\,\text{N/C}$ , because a positive upward field exerts an upward force on any charge, directly canceling the downward gravitational force.

Explanation: When a charged particle is in equilibrium, the net force on it must be zero. Here, gravity pulls the electron downward with force

F_g = m_e g = (9.11\times10^{-31})(9.8) \approx 8.9\times10^{-30}\,\text{N}

. The electric force must therefore point upward with equal magnitude. The critical insight is how the sign of the charge affects force direction: for a negative charge, the electric force is opposite to

\vec{E}

, since

\vec{F} = q\vec{E}

and

q < 0

. So to produce an upward force on the electron,

\vec{E}

must point downward. Setting

eE = m_e g

gives

E = \frac{m_e g}{e} = \frac{8.9\times10^{-30}}{1.60\times10^{-19}} \approx 5.6\times10^{-11}\,\text{N/C}

. This confirms choice A is correct — a downward field of that magnitude produces an upward force on the negatively charged electron, exactly canceling gravity. Choice B describes an upward field, which would push the electron downward (opposite to

\vec{E}

for a negative charge), adding to gravity rather than canceling it — exactly the wrong direction. Choice C reaches the right field direction but through completely wrong reasoning. There is no "reaction force from the field source" acting here; Newton's third law pairs forces on different objects, not within the same system this way. Choice D incorrectly claims an upward electric field exerts an upward force on any charge — this ignores the sign of

q

entirely, which is the central concept being tested. Study tip: Whenever you see a negative charge in a field problem, immediately flag that

\vec{F}

and

\vec{E}

point in opposite directions. Writing out

\vec{F} = q\vec{E}

with the negative sign explicitly prevents this classic mistake.

Question 5

In a thought experiment, an electric field region contains two subregions. In subregion I ( $0 < x < a$ ), $\vec{E} = E_0\hat{x}$ . In subregion II ( $a < x < 2a$ ), $\vec{E} = -E_0\hat{x}$ . A particle of charge $+q$ and mass $m$ enters at $x=0$ moving in the $+x$ direction with initial speed $v_0$ . Ignore gravity.

As the particle moves from $x=0$ to $x=2a$ , which of the following correctly describes how the net impulse delivered by the electric force compares to the impulse delivered in subregion I alone?

The net impulse over the full path is zero, because the equal and opposite fields in the two regions deliver equal and opposite impulses — the particle's symmetric spatial path guarantees cancellation.
The net impulse over the full path is positive and greater than the impulse in region I alone, because the particle enters region II with higher speed than it had at $x=0$ , so region II's force acts on a faster-moving particle and contributes a larger impulse.
The net impulse over the full path equals the impulse in region I alone, because the opposing field in region II does negative work on the particle but does not transfer any net momentum — only energy is affected.
The net impulse over the full path is negative in sign (smaller than the impulse in region I), because the particle decelerates in region II and therefore spends more time there, giving region II a larger impulse magnitude than region I. (correct answer)

Explanation: Impulse is defined as

J = F \cdot \Delta t

, not

F \cdot \Delta x

. This distinction is the entire key to this problem. When you see a charged particle moving through oppositely directed fields, don't assume the impulses cancel just because the spatial regions are equal — you must think about time spent in each region. In subregion I, the force

F = qE_0

accelerates the particle from

v_0

to some higher speed

v_1

. Because it's speeding up, the particle crosses region I quickly. In subregion II, the force

-qE_0

decelerates the particle. It enters at

v_1 > v_0

and slows down, meaning it spends more time in region II than it did in region I. Since impulse equals force times time, and region II's force magnitude is identical to region I's but acts over a longer time interval, region II delivers a larger magnitude impulse — but in the negative direction. The net impulse over both regions is therefore negative (less than the impulse from region I alone), confirming D is correct. Choice A is wrong because equal spatial lengths do not imply equal time intervals. The cancellation argument fails precisely because the particle travels at different average speeds in each region. Choice B is wrong because higher speed means less time in region II, not more — and impulse depends on time, not speed directly. Choice C contains a subtle but serious error: force absolutely does transfer momentum (impulse = change in momentum), so claiming "only energy is affected" contradicts the impulse-momentum theorem. Remember: whenever force direction reverses over equal distances, ask yourself whether the time spent in each region is also equal — it usually isn't.

Question 6

Two point charges, $+Q$ and $-Q$ , are fixed in space separated by distance $d$ . A small test charge $+q$ (with $q \ll Q$ ) is placed at the midpoint between them.

Which of the following correctly describes the net electrostatic force on the test charge $+q$ at the midpoint?

The net force is zero, because the equal and opposite electric fields from $+Q$ and $-Q$ cancel at the midpoint, producing no net field and hence no force on $+q$ .
The net force is directed toward $-Q$ with magnitude $\frac{8kQq}{d^2}$ , because both individual forces on $+q$ point toward $-Q$ and add constructively. (correct answer)
The net force is directed toward $-Q$ with magnitude $\frac{2kQq}{d^2}$ , because both individual forces on $+q$ point toward $-Q$ and add constructively.
The net force is directed toward $+Q$ with magnitude $\frac{2kQq}{d^2}$ , because the repulsion from $+Q$ and the attraction toward $-Q$ act in opposite directions and cancel, leaving only the repulsive component.

Explanation: When analyzing forces on a charge placed between two opposite charges, resist the temptation to think about "field cancellation" — instead, carefully draw the force vectors on the test charge from each source charge separately. Place

+q

at the midpoint, so it sits a distance

r = d/2

from each charge. The force from

+Q

+q

is repulsive, pushing

+q

away from

+Q

— meaning it points toward $-Q$ . The force from

-Q

+q

is attractive, pulling

+q

toward

-Q

— it also points toward $-Q$ . Both forces act in the same direction, so they add. Each force has magnitude

F = \frac{kQq}{(d/2)^2} = \frac{4kQq}{d^2}

, giving a total of

\frac{8kQq}{d^2}

directed toward

-Q

. That confirms B is correct. A is wrong because it confuses the electric field at the midpoint (which does not cancel for a dipole — both charges contribute fields pointing in the same direction toward

-Q

) with a false cancellation. Even if fields did cancel somewhere, you'd need to analyze force directions correctly. C uses the right direction but makes an arithmetic error — it forgets to account for the

(d/2)^2

denominator, underestimating the force by a factor of 4. D incorrectly claims the repulsion from

+Q

and attraction from

-Q

oppose each other; since

+q

sits between them, both forces point the same way. A reliable strategy: always sketch the geometry and draw force arrows on the test charge before doing any math. Direction errors are the most common trap on Coulomb's Law problems.

Question 7

In a certain region, the electric field is given by $\vec{E}(x) = E_0 \left(1 + \frac{x}{L}\right)\hat{x}$ , where $E_0$ and $L$ are positive constants and $x$ is the position coordinate. A particle with charge $+q$ and mass $m$ is released from rest at $x = 0$ .

What is the magnitude of the electric force on the particle at the instant it has traveled a distance $L$ from its release point?

$qE_0$ , because at the release point $x=0$ the field equals $E_0$ and the force is $qE_0$ , which remains constant since the particle was released from rest.
$2qE_0$ , because substituting $x = L$ into the field expression gives $E(L) = 2E_0$ , and the force is $F = qE(L) = 2qE_0$ . (correct answer)
$qE_0\left(1 + \frac{v}{c}\right)$ , because the non-uniform field introduces a velocity-dependent correction to the force at the moment the particle reaches position $L$ .
$\frac{3}{2}qE_0$ , because the force is the average of the field values at $x=0$ and $x=L$ , giving $E_{avg} = \frac{E_0 + 2E_0}{2} = \frac{3E_0}{2}$ , and $F = q E_{avg}$ .

Explanation: When a question asks for the force on a charged particle at a specific position, your instinct should be to evaluate the electric field at that position and apply

F = qE

. The key word here is instantaneous — you need the field value where the particle currently is, not some average or earlier value. Since the particle travels a distance

L

from

x = 0

, it arrives at

x = L

. Substituting into the given field expression:

E(L) = E_0\left(1 + \frac{L}{L}\right) = E_0(1 + 1) = 2E_0

. The electric force at that instant is therefore

F = qE(L) = 2qE_0

, confirming B is correct. A commits a classic trap: assuming the force is constant just because the particle started from rest. Rest describes the initial velocity, not the field — the field is explicitly position-dependent, so the force changes as the particle moves. A freezes the field at

x = 0

and misapplies it to

x = L

. C introduces a relativistic-style velocity correction that has no physical basis here. Nothing in classical electrostatics makes the Coulomb force velocity-dependent; that idea belongs to a different context entirely. Treat it as a distractor designed to sound sophisticated. D confuses force at a point with work done over a displacement. Averaging field values is a technique sometimes used when calculating energy or displacement — not when finding the instantaneous force at a specific location. Study tip: Whenever you see "at the instant" or "at position

x

," that's your signal to simply evaluate the given field expression at that

x

-value and compute

F = qE

Question 8

A small test charge $q = -3\,\mu\text{C}$ is placed in a region where the electric field is described by $\vec{E} = (4\hat{x} - 3\hat{y})\times10^4\,\text{N/C}$ .

What is the magnitude of the electrostatic force on the test charge, and in which general direction does it point relative to the electric field vector?

Magnitude $0.15\,\text{N}$ ; the force is directed antiparallel to $\vec{E}$ and thus points in the direction $(-4\hat{x}+3\hat{y})$ . (correct answer)
Magnitude $0.15\,\text{N}$ ; the force is directed parallel to $\vec{E}$ and thus points in the direction $(4\hat{x}-3\hat{y})$ .
Magnitude $0.21\,\text{N}$ ; the force is directed antiparallel to $\vec{E}$ and thus points in the direction $(-4\hat{x}+3\hat{y})$ .
Magnitude $0.12\,\text{N}$ ; the force is directed antiparallel to $\vec{E}$ and thus points in the direction $(-4\hat{x}+3\hat{y})$ .

Explanation: When a charge sits in an electric field, the force on it is given by

\vec{F} = q\vec{E}

. The critical detail here is the sign of the charge — it controls both the magnitude and the direction of the force relative to

\vec{E}

. Start with the magnitude. The magnitude of the electric field is

|\vec{E}| = \sqrt{(4)^2 + (-3)^2} \times 10^4 = \sqrt{25} \times 10^4 = 5 \times 10^4\,\text{N/C}

. The magnitude of the force is then

|\vec{F}| = |q||\vec{E}| = (3 \times 10^{-6})(5 \times 10^4) = 0.15\,\text{N}

. Now for direction: since

q = -3\,\mu\text{C}

is negative, multiplying

\vec{E}

by a negative scalar flips the vector. So

\vec{F} = (-3\times10^{-6})(4\hat{x} - 3\hat{y})\times10^4

, which points in the

(-4\hat{x} + 3\hat{y})

direction — antiparallel to

\vec{E}

. This confirms A. Choice B has the correct magnitude but wrong direction — it ignores that the negative charge reverses the force vector relative to

\vec{E}

. This is the most tempting trap. Choice C arrives at the wrong magnitude (

0.21\,\text{N}

), likely from incorrectly computing

|\vec{E}| = (4+3)\times10^4

by adding components directly instead of using the Pythagorean theorem. Choice D uses yet another incorrect magnitude (

0.12\,\text{N}

), possibly from

|q| \times

a miscalculated field. Study tip: Always flag the sign of the charge first. Negative charge → force antiparallel to

\vec{E}

; positive charge → force parallel. This single step eliminates half the distractors on most force-field problems.

Question 9

A physicist reports that a small charged sphere experiences a force of $8.0\times10^{-4}\,\text{N}$ when placed in an electric field of $4.0\times10^3\,\text{N/C}$ . Later, the sphere is moved to a new location where the electric field has magnitude $1.0\times10^4\,\text{N/C}$ and the sphere carries a new charge of $5.0\times10^{-7}\,\text{C}$ .

What is the force on the sphere at the new location, and what was the charge on the sphere at the original location?

Force at new location: $2.0\times10^{-6}\,\text{N}$ ; original charge: $2.0\times10^{-7}\,\text{C}$ .
Force at new location: $5.0\times10^{-3}\,\text{N}$ ; original charge: $3.2\,\text{C}$ .
Force at new location: $5.0\times10^{-3}\,\text{N}$ ; original charge: $2.0\times10^{-7}\,\text{C}$ . (correct answer)
Force at new location: $1.25\times10^{-3}\,\text{N}$ ; original charge: $2.0\times10^{-7}\,\text{C}$ .

Explanation: When you see a question involving electric fields and forces, your anchor equation is

F = qE

, where

F

is the electric force,

q

is the charge, and

E

is the electric field magnitude. This question asks you to apply that relationship in two separate scenarios. For the original location, you're given

F = 8.0\times10^{-4}\,\text{N}

and

E = 4.0\times10^3\,\text{N/C}

. Solving for charge:

q = \frac{F}{E} = \frac{8.0\times10^{-4}}{4.0\times10^3} = 2.0\times10^{-7}\,\text{C}

. For the new location, you're given

q = 5.0\times10^{-7}\,\text{C}

and

E = 1.0\times10^4\,\text{N/C}

. So:

F = qE = (5.0\times10^{-7})(1.0\times10^4) = 5.0\times10^{-3}\,\text{N}

. This confirms C as the correct answer. A gets the original charge wrong (

2.0\times10^{-6}\,\text{C}

is off by a factor of 10) and also gives a completely incorrect new force — it looks like a decimal place error in both calculations. B gets the new force right but reports the original charge as

3.2\,\text{C}

, which is an enormous charge for a small sphere — a red flag. This likely comes from multiplying instead of dividing when solving

q = F/E

. D gets the original charge right but fumbles the new force, possibly by dividing instead of multiplying when applying

F = qE

. A quick study tip: always pause to identify which quantity is unknown before plugging into

F = qE

. The equation has three variables — if you confuse which one to solve for, you'll get a distractor answer every time.

Question 10

A particle with charge $q > 0$ experiences a force $\vec{F}_1$ when placed at point P in a region of space. The same particle is then replaced by a particle with charge $-2q$ at the same point P. The electric field at P is produced by a fixed external source and does not change.

If $\vec{F}_1 = F_0 \hat{x}$ , what is the force $\vec{F}_2$ on the $-2q$ charge at point P?

$\vec{F}_2 = +\frac{1}{2}F_0\hat{x}$ , because the $-2q$ charge alters the local electric field at P by attracting field lines, effectively halving the external field and reversing the sign of the force to give a net positive component.
$\vec{F}_2 = +2F_0\hat{x}$ , because the magnitude of the charge doubles, doubling the force, and the direction remains along $\hat{x}$ since the field at P is unchanged by the charge substitution.
$\vec{F}_2 = -F_0\hat{x}$ , because the negative charge simply reverses the direction of the force without changing its magnitude, since the magnitude of $-2q$ relative to $q$ is scaled only by the sign.
$\vec{F}_2 = -2F_0\hat{x}$ , because the charge magnitude doubles (giving factor of 2) and the negative sign reverses the direction, yielding a force antiparallel to $\vec{F}_1$ . (correct answer)

Explanation: When you see a question like this, reach for the fundamental relationship

\vec{F} = q\vec{E}

. The electric force on a charge depends on two things: the electric field at that point (set by the external source) and the charge placed there. Since the problem states the external field is fixed, your only job is to track how the charge changes. From the first scenario,

\vec{F}_1 = q\vec{E} = F_0\hat{x}

, so the field at P is

\vec{E} = \frac{F_0}{q}\hat{x}

. Now substitute the second charge:

\vec{F}_2 = (-2q)\vec{E} = (-2q)\left(\frac{F_0}{q}\hat{x}\right) = -2F_0\hat{x}

. The magnitude doubles because

|-2q| = 2|q|

, and the direction reverses because of the negative sign — making D the correct answer. A is wrong on two levels: it invents the idea that a test charge "attracts field lines" and alters the external field — test charges don't modify the field that's already there — and the arithmetic doesn't follow from any valid principle. B correctly identifies that the magnitude doubles, but ignores the negative sign entirely, flipping the direction of the force incorrectly. C accounts for the sign change (direction reversal) but ignores the factor of 2 from the larger magnitude, treating

|-2q|

as though it were just

|q|

. A useful habit: always separate magnitude and direction when analyzing force changes. Ask yourself, "Did the field change? Did the charge magnitude change? Did the sign change?" — then apply each effect independently. This prevents the kind of partial-credit thinking that traps students into answers B or C.