Physics 2 Quiz: Electric Field Lines

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Electric Field LinesQuestion 1 of 9

Two concentric spherical shells carry uniform surface charge densities. The inner shell of radius $r_1$ carries charge $+Q$ and the outer shell of radius $r_2$ carries charge $-3Q$ . An observer traces an electric field line starting just outside the inner shell and moving radially outward. Which of the following correctly describes the complete trajectory of this field line?

The field line originates on the inner shell, travels radially outward through the region between the shells, and terminates on the outer shell. No field lines from the inner shell extend beyond

r_2

, because the outer shell's charge magnitude of

3Q

exceeds the inner shell's

Q

, so the outer shell fully absorbs all outward flux from the inner shell.

The field line originates on the inner shell, travels radially outward through the region between the shells, and terminates on the outer shell only if it happens to be one of the

1/3

of total lines that are absorbed there. The remaining

2/3

of lines from the inner shell continue radially outward to infinity, since the outer shell only intercepts a fraction of the total flux proportional to the charge ratio.

The field line originates on the inner shell, travels radially outward through the region between the shells, and all lines terminate on the outer shell. This is because the net enclosed charge at any radius between

r_1

and

r_2

+Q

, which drives all flux outward into the outer shell, leaving no field lines to escape beyond

r_2

The field line originates on the inner shell and curves back inward in the region between the shells due to the strong attractive force exerted by the

-3Q

outer shell. As a result, the field line terminates back on the inner shell before ever reaching

r_2

, consistent with the net inward force on flux in that region.

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Physics 2 Quiz

Physics 2 Quiz: Electric Field Lines

Practice Electric Field Lines 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 Lines, 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

The field line originates on the inner shell, travels radially outward through the region between the shells, and terminates on the outer shell. No field lines from the inner shell extend beyond $r_2$ , because the outer shell's charge magnitude of $3Q$ exceeds the inner shell's $Q$ , so the outer shell fully absorbs all outward flux from the inner shell. (correct answer)
The field line originates on the inner shell, travels radially outward through the region between the shells, and terminates on the outer shell only if it happens to be one of the $1/3$ of total lines that are absorbed there. The remaining $2/3$ of lines from the inner shell continue radially outward to infinity, since the outer shell only intercepts a fraction of the total flux proportional to the charge ratio.
The field line originates on the inner shell, travels radially outward through the region between the shells, and all lines terminate on the outer shell. This is because the net enclosed charge at any radius between $r_1$ and $r_2$ is $+Q$ , which drives all flux outward into the outer shell, leaving no field lines to escape beyond $r_2$ .
The field line originates on the inner shell and curves back inward in the region between the shells due to the strong attractive force exerted by the $-3Q$ outer shell. As a result, the field line terminates back on the inner shell before ever reaching $r_2$ , consistent with the net inward force on flux in that region.

Explanation: Whenever you see a question involving concentric charged shells and field lines, your first tool should be Gauss's Law: the electric flux through any closed surface depends only on the enclosed charge. Consider a Gaussian sphere of radius

r

where

r_1 < r < r_2

. The enclosed charge is

+Q

, so the electric field points radially outward in that region. A field line starting just outside the inner shell travels outward through this region — that part is clear. Now ask: what happens at

r_2

? The inner shell contributes

+Q

of flux, but the outer shell carries

-3Q

, which means it needs to terminate

3Q

worth of flux. Since only

+Q

of flux arrives from inside, all of it gets absorbed. Every single field line from the inner shell terminates on the outer shell. This makes A correct. C is tempting but wrong in its reasoning. It correctly identifies that the net enclosed charge between the shells is

+Q

, but then incorrectly concludes this means all field lines are somehow "driven" into the outer shell by that enclosed charge. Gauss's Law tells you about flux magnitude, not whether lines escape — the reason no lines escape is that the outer shell has more than enough negative charge (

-3Q

) to absorb all incoming flux. B is wrong because field lines don't get "partially absorbed" based on a charge ratio. Every field line originating on the inner shell terminates on the outer shell, since

|-3Q| > +Q

. D is wrong because field lines between two concentric shells always travel radially outward — they never curve back inward. The geometry enforces radial symmetry. Study tip: Always apply Gauss's Law first. Whichever shell has greater charge magnitude controls whether flux escapes — if

|Q_{outer}| \geq |Q_{inner}|

, no field lines exit beyond the outer shell.

Question 2

A long, straight wire carries a uniform positive linear charge density $\lambda$ . A second long, straight wire, parallel to the first and separated by distance $d$ , carries a uniform negative linear charge density $-2\lambda$ . Which of the following correctly describes the electric field line pattern for this system?

All field lines from the positive wire terminate on the negative wire, and no field lines extend to or from infinity, because the two wires carry opposite signs of charge and opposite-sign charge systems always form self-contained field line patterns regardless of the ratio of their charge magnitudes.
All field lines from the positive wire terminate on the negative wire, and no field lines extend to infinity from either wire, because opposite charges always form closed field line systems regardless of the ratio of their charge magnitudes.
Half the field lines from the positive wire terminate on the negative wire and the other half extend to infinity, while the negative wire draws in additional lines from infinity to account for its excess charge, because the 2:1 charge ratio splits the positive wire's flux evenly between the two available sinks.
All field lines from the positive wire terminate on the negative wire. Additionally, the negative wire draws in extra field lines from infinity on all sides, because the net charge per unit length of the system is $-\lambda$ , so the far-field resembles that of a single negative wire with lines converging inward from infinity. (correct answer)

Explanation: When analyzing electric field line patterns for a multi-wire system, you need to think about two separate things: what happens locally between the wires, and what the system looks like far away. Gauss's Law governs both. Start with the far-field perspective. The net linear charge density of the system is

\lambda + (-2\lambda) = -\lambda

. At large distances, the two wires look like a single wire with charge density

-\lambda

. By Gauss's Law, a net negative line charge must have field lines converging inward from infinity — the system draws in lines from all directions far away. Now zoom in locally: every field line leaving the positive wire (

+\lambda

) must terminate somewhere negative, and the nearest available sink is the

-2\lambda

wire. So all field lines from the positive wire terminate on the negative wire. But the negative wire has twice the magnitude, meaning it needs additional flux — which it gets from infinity. This is exactly what D describes, making it correct. A and B share the same fundamental error: they claim opposite-sign charges always form self-contained, closed field line systems. This is only true when the magnitudes are equal. The ratio matters because it determines the net charge, which governs the far-field behavior. C is tempting but wrong on two counts. It incorrectly claims only half the positive wire's field lines reach the negative wire, and it invents a "2:1 split" rule that has no physical basis. All lines from the positive wire terminate on the negative wire — none escape to infinity. A useful rule of thumb: always compute the net charge of a system first. If it's nonzero, the system cannot be self-contained — field lines must extend to or come from infinity accordingly.

Question 3

Region I is described by electric field lines that are straight, horizontal, and become more densely packed as one moves to the right. Region II is described by electric field lines that are straight, horizontal, and uniformly spaced throughout.

At point P, field lines are converging (becoming closer together in the direction of the field). At point Q, the field lines are diverging (spreading apart in the direction of the field). Both P and Q are in regions free of any charge. What can be correctly inferred about the electric field magnitude $E$ and the electric potential $V$ at points P and Q?

The field magnitude $E$ is increasing at P and decreasing at Q, and since a stronger field corresponds to a steeper potential gradient, the potential $V$ must be lower at P than at Q, because stronger electric fields always point from higher to lower potential regions.
The field magnitude $E$ is increasing along the field direction at P and decreasing at Q, but no conclusion about the relative value of $V$ at P versus Q can be drawn from field line geometry alone, because determining potential requires integrating $\vec{E}$ along a path from a reference point — information not available from local geometry. (correct answer)
The potential $V$ is higher at P than at Q because converging field lines indicate field intensification, and by $\vec{E} = -\nabla V$ the potential surfaces are more compressed at P, which means the local potential value is necessarily greater there than at a point with more widely spaced equipotentials.
The field magnitude $E$ is decreasing at P and increasing at Q, because convergence of field lines means lines are being absorbed by the charge-free region, reducing net flux and therefore field strength, while divergence increases the local flux and field strength.

Explanation: When analyzing electric field line geometry, you need to carefully separate two distinct questions: (1) what does local field line spacing tell you about field magnitude, and (2) what does that tell you about electric potential? Field line density directly encodes field magnitude — where lines converge (pack together),

E

increases in that direction; where they diverge (spread apart),

E

decreases. So at P, the field is strengthening, and at Q, it is weakening. That part is straightforward. The trap this question sets is assuming you can compare the absolute values of

V

at P versus Q from local geometry alone. You cannot. Potential is determined by integrating

\vec{E}

along a path:

V(P) - V(Q) = -\int_Q^P \vec{E} \cdot d\vec{l}

. Without knowing the spatial relationship between P and Q, the path connecting them, and the full field distribution along that path, the local behavior of field lines at each point tells you nothing about their relative potentials. Answer B correctly captures this — it confirms the field magnitude behavior while honestly acknowledging the limit of local geometric reasoning. Answer A is seductive but flawed: it correctly identifies

E

behavior, then leaps to a conclusion about

V

at P versus Q that isn't justified. "Stronger field" tells you about the gradient of

V$ locally, not the *value* of

V$ compared to a distant point. Answer C makes the same error in reverse, incorrectly equating compressed equipotentials with a higher potential value. Answer D is simply wrong about the physics — converging field lines in a charge-free region do not mean flux is "absorbed"; Gauss's law requires zero net flux through any closed surface in a charge-free region, but local convergence just means the field is strengthening, not disappearing. Study tip: Always distinguish between local field behavior (readable from line spacing) and global potential comparisons (requiring integration along a path). Questions that mix these two are classic traps on Physics 2 exams.

Question 4

Two electric field line diagrams are described for different charge configurations. Configuration 1: Field lines form closed loops in a bounded region of space with no charges present. Configuration 2: Field lines begin and end on the same charge (a single isolated positive charge with field lines curving back and terminating on it). Which of the following statements correctly evaluates both configurations in the context of electrostatics?

Configuration 1 is impossible in electrostatics, but Configuration 2 is possible if the isolated positive charge is surrounded by a dielectric material. Polarization of the dielectric creates bound charge distributions that redirect field lines, potentially causing them to curve back and terminate on the original source charge.
Configuration 1 is impossible in electrostatics, but Configuration 2 is possible if the isolated positive charge is immersed in a strong external electric field oriented to oppose its own field, which could in principle redirect field lines back toward the source charge and cause them to terminate on it.
Both configurations are impossible in electrostatics. Closed field line loops violate $\nabla \times \vec{E} = 0$ , since a path along such a loop would yield a nonzero line integral, contradicting the irrotational nature of electrostatic fields. Field lines from a positive charge cannot terminate back on it, because Gauss's law requires lines to diverge from positive charges and terminate only on negative charges or at infinity. (correct answer)
Both configurations are possible under the right conditions: closed loops can form between equal and opposite charges when field lines reconnect along a neutral axis, and a charge can have field lines return to it if it is in a region where the net field from all other sources exactly cancels, creating a local field reversal.

Explanation: Whenever you see a question about electric field line behavior, anchor your thinking to two fundamental laws of electrostatics: Gauss's law (

\nabla \cdot \vec{E} = \rho/\epsilon_0

) and the irrotational condition (

\nabla \times \vec{E} = 0

). These two equations together tightly constrain what field lines can and cannot do. Configuration 1 fails because closed field line loops would mean a nonzero circulation of

\vec{E}

around that loop — but in electrostatics,

\oint \vec{E} \cdot d\vec{l} = 0

always. A closed loop directly violates this. Configuration 2 fails because Gauss's law dictates that field lines diverge from positive charges and can only terminate on negative charges or extend to infinity. A positive charge has no mechanism to "attract" its own field lines back; that would require a negative charge to act as a sink. Both configurations are therefore impossible, making C the correct answer. Choice A misunderstands dielectric polarization. While a dielectric does produce bound charges, those bound charges are negative on the surface nearest the positive source — they don't redirect field lines back onto the original charge; they simply weaken the field inside. Choice B is similarly flawed: an opposing external field can create a null point nearby, but it cannot cause the original charge's own field lines to loop back and terminate on it — that still violates Gauss's law. Choice D confuses superposition with field line topology; field lines never "reconnect" into loops between opposite charges, and a local field cancellation doesn't create termination on the source charge. Your study tip: memorize that

\nabla \times \vec{E} = 0

forbids loops, and

\nabla \cdot \vec{E} > 0

at positive charges forbids self-termination — these two rules eliminate a huge class of impossible configurations on exam questions.

Question 5

A student examines an electric field line diagram for a system of two point charges. In the diagram, all field lines originate from charge A and terminate on charge B. The density of field lines near charge A is exactly twice the density near charge B at equal distances from each respective charge.

Based solely on the field line diagram described, which of the following conclusions is best supported about the charges A and B?

Charge A is positive with magnitude twice that of charge B, and charge B is negative, because field lines emanate from positive charges and terminate on negative charges, and line density is proportional to charge magnitude. (correct answer)
Charge A is negative with magnitude twice that of charge B, and charge B is positive, because the higher density near A indicates a stronger sink, meaning more field lines are absorbed there per unit area.
Both charges are equal in magnitude but opposite in sign, because the total number of field lines entering and leaving any closed surface in the system must be conserved by Gauss's law regardless of local density differences.
Charge A is positive with magnitude twice that of charge B, and charge B is positive, because field lines only terminate on negative charges if the system is isolated, and the described geometry suggests both charges repel.

Explanation: When analyzing electric field line diagrams, two fundamental rules govern everything: field lines flow from positive to negative charges, and the number of field lines drawn from or to a charge is directly proportional to that charge's magnitude. Here, all field lines originate at charge A and terminate at charge B. Origination means A is the source — a positive charge. Termination means B is the sink — a negative charge. So far, both A and B differ in sign, with A positive and B negative. Now consider the density clue: the field line density near A is twice that near B at equal distances. Since density encodes magnitude, A carries twice the charge magnitude of B. This makes A confirm as answer A — positive, with magnitude

|q_A| = 2|q_B|

, and B negative. Choice B fails on the fundamental direction rule. Negative charges absorb field lines (they are sinks), but B is described as the termination point, making B negative — not A. Flipping the signs contradicts the origination-termination convention directly. Choice C misapplies Gauss's Law. While Gauss's Law is valid, it does not require equal charge magnitudes between two charges. A closed surface around A would show net outward flux proportional to

q_A

, and the charges can absolutely differ in magnitude. Choice D is internally contradicted — if field lines terminate on B, B must be negative. Field lines never terminate on positive charges regardless of system isolation. Study tip: Always memorize the two field line rules as a pair — direction (+ to −) and density (proportional to magnitude). Exam questions almost always test both simultaneously.

Question 6

A conducting sphere of radius $R$ carries a net charge $+Q$ . A thin, flat, infinite grounded conducting plane is brought close to (but not touching) the sphere. After electrostatic equilibrium is reached, which of the following best describes the electric field line configuration near the surface of the sphere?

The field lines near the sphere's surface are denser on the side facing away from the plane, because the plane repels positive charge on the sphere toward the far side by electrostatic induction, creating a higher surface charge density there.
The field lines near the sphere's surface remain perfectly symmetric and uniformly distributed, because the total charge on the sphere is fixed at $+Q$ and Gauss's law requires the field outside a conductor to depend only on the total enclosed charge, not on nearby conductors.
The field lines near the sphere's surface are no longer symmetric; they are denser on the side facing the plane than on the opposite side, because induced negative charges on the plane attract positive charges on the sphere toward the near side, redistributing the surface charge nonuniformly. (correct answer)
The field lines near the sphere's surface are uniformly distributed but globally weaker than in the absence of the plane, because the grounded plane absorbs some of the electric flux from the sphere, reducing the net field strength at the sphere's surface uniformly.

Explanation: Whenever you see a conducting sphere near a grounded plane, you need to think about electrostatic induction — not just Gauss's law in isolation. The key concept here is that nearby conductors fundamentally reshape charge distributions, even when total charge is conserved. Here's what actually happens: the grounded plane is connected to Earth, so charges can flow freely onto or off it. The positive charge

+Q

on the sphere attracts electrons from the ground onto the near side of the plane, creating a sheet of induced negative charge there. This induced negative charge then exerts an attractive Coulomb force on the positive charges of the sphere, pulling them toward the side facing the plane. The result is a nonuniform surface charge density — higher near the plane, lower on the far side — and therefore denser electric field lines on the side facing the plane. That's exactly what C describes, making it correct. A gets the physics backwards. The grounded plane doesn't repel positive charges — it develops induced negative charge that attracts the sphere's positive charges toward it, not away. B contains a subtle but critical misapplication of Gauss's law. Yes, the total charge is fixed at

+Q

, but Gauss's law tells you the total flux through a closed surface, not the local field distribution. Nearby conductors redistribute surface charge nonuniformly, breaking symmetry entirely. D is wrong because the plane doesn't uniformly weaken the field. It breaks the symmetry; some regions get stronger, others weaker, depending on geometry. Your study tip: when a conductor is grounded, always ask what charge it absorbs from or releases to Earth — this changes everything about the nearby field geometry.

Question 7

Comparing the two regions, which of the following statements about the electric field and charge distribution is correct?

Region I has greater electrostatic energy density than Region II where the field is strongest, but Region II must contain free charges to sustain a uniform field, while Region I is charge-free because Gauss's law only applies to regions where the field is nonuniform.
Both regions are charge-free, because in both cases the field lines are straight and do not curve. Curvature of field lines — not changes in spacing — is the indicator of local charge density according to Gauss's law, and neither region displays curved field lines.
Region II must contain a nonzero charge distribution to maintain the uniform field against the natural tendency of field lines to spread, while Region I is charge-free because the increasing density of field lines is a natural consequence of flux conservation as lines travel through space.
Region I has a nonzero volume charge density throughout, because increasing field line density means $\nabla \cdot \vec{E} \neq 0$ , which by Gauss's law implies $\rho \neq 0$ . Region II is consistent with being charge-free, because uniform spacing means $\partial E_x/\partial x = 0$ , so $\nabla \cdot \vec{E} = 0$ and Laplace's equation is satisfied. (correct answer)

Explanation: When you see a question involving electric field line spacing and charge distributions, your first instinct should be to reach for Gauss's law in differential form:

\nabla \cdot \vec{E} = \rho/\varepsilon_0

. This tells you that wherever the divergence of the electric field is nonzero, there must be a local charge density — and vice versa. In Region I, the field lines are horizontal but becoming more densely packed to the right. Denser field lines mean a stronger field magnitude in that direction, so

\partial E_x/\partial x > 0

. Since the field is purely horizontal,

\nabla \cdot \vec{E} = \partial E_x/\partial x \neq 0

, which by Gauss's law requires

\rho \neq 0

throughout that region. In Region II, the field lines are uniformly spaced, meaning constant field magnitude, so

\partial E_x/\partial x = 0

, giving

\nabla \cdot \vec{E} = 0

. Laplace's equation is satisfied, and no free charge is required. This makes D correct. Answer A is wrong because it invents a false rule — Gauss's law applies everywhere, not only where fields are nonuniform. Answer B is the most tempting trap: it conflates two different things. Curvature of field lines does indicate the presence of a transverse field component, but changes in spacing along the field direction independently signal nonzero divergence. Straight lines can still imply charge density. Answer C has the logic exactly backwards — field lines spreading or converging on their own (flux conservation in charge-free space) is precisely what happens in Region I's scenario, not Region II's. Your study tip: always translate field line descriptions into mathematical language. "Denser spacing" means

\partial E/\partial x \neq 0

; "uniform spacing" means

\partial E/\partial x = 0

. Then apply

\nabla \cdot \vec{E} = \rho/\varepsilon_0

directly.

Question 8

A student analyzes the electric field line pattern near a point labeled X in a charge-free region. At X, the field lines are curved, and the student notes that the curvature (bending) of the lines is directed perpendicular to the local field direction. The student claims: "The curvature of field lines at a point directly indicates the presence of an electric force component perpendicular to the field at that point, which means a charge at X would experience a net force in a direction other than along the field line."

Which of the following most accurately assesses the student's claim?

The claim is correct. A curved field line at X means the electric field vector itself has a component perpendicular to the line's local direction, which by $\vec{F} = q\vec{E}$ directly produces a perpendicular force component on any charge placed at X.
The claim is incorrect. The curvature of field lines reflects the spatial variation of the field direction (i.e., transverse gradients of the field), not a force component perpendicular to the field. The force on a point charge at X is always directed exactly along the local field line (parallel to $\vec{E}$ at X for a positive charge), regardless of the curvature of the field line through X. (correct answer)
The claim is correct only for negative charges. A negative charge at X experiences a force opposite to the field direction, and when field lines are curved, this reversal introduces an asymmetry that produces a net transverse force component on the negative charge.
The claim is incorrect specifically because in a charge-free region, field lines cannot be curved by definition — the absence of charges means $\nabla^2 \vec{E} = 0$ , which forces field lines to be straight, so the described scenario is physically impossible.

Explanation: When analyzing electric field lines, it's crucial to distinguish between what a field line tells you at a point versus what it tells you about the surrounding region. The electric field vector

\vec{E}

at any point is defined as the tangent to the field line at that point — it has no component perpendicular to the line by definition. Therefore, the force

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

on a charge at X is always directed exactly along the local field line (for positive charges), no matter how curved the line appears. B is correct because field line curvature encodes how the field direction changes across space — a geometric property of the field's spatial variation, not a local force component. A curved field line through X simply means the field direction rotates as you move along or across the line. This is related to transverse gradients like

\partial E_x/\partial y

, which describe the field's structure, not the force at a single point. A is wrong because it confuses the geometry of the field line pattern with the local field vector. The field vector at X is tangent to the line — it has zero perpendicular component by construction. C is wrong because reversing the charge sign reverses the force direction, but that reversed force is still perfectly antiparallel to

\vec{E}

— still along the field line. No transverse component is introduced for either sign of charge. D is wrong because field lines absolutely can be curved in charge-free regions. Maxwell's equations permit spatially varying field directions without requiring local charges — think of the field between two separated point charges in the intervening empty space. Remember: curvature of a field line is about the map, not the force at a point. Always evaluate force using the local

\vec{E}

vector (the tangent), not the line's overall shape.

Question 9

A physics student makes the following claim: "If electric field lines in a region are parallel and equally spaced, then the electric potential in that region must be zero everywhere."

Which of the following best evaluates the student's claim?

The claim is incorrect. Parallel, equally spaced field lines indicate a uniform electric field, which implies the potential varies linearly with position along the field direction — it need not be zero anywhere in the region. (correct answer)
The claim is correct. A uniform field (parallel, equally spaced lines) means the field is the same everywhere, and since $\vec{E} = -\nabla V$ , a spatially constant field requires that $V$ must also be spatially constant and therefore zero by symmetry.
The claim is incorrect only if the region contains free charges. In a charge-free region with uniform field lines, the divergence of $\vec{E}$ is zero, which by Poisson's equation requires $V = 0$ throughout the region.
The claim is correct in the specific case where the field lines are oriented perpendicular to the boundary of the region, because in that geometry the equipotential surfaces are parallel to the boundary where $V$ is typically defined as zero by convention.

Explanation: Whenever you see a question linking electric field geometry to electric potential, your anchor concept should be the relationship

\vec{E} = -\nabla V

. This equation tells you how changes in potential relate to the field — not what the potential's absolute value must be. Parallel, equally spaced field lines define a uniform electric field — constant in both magnitude and direction throughout the region. Integrating the relationship above along the field direction gives

V = -Ex + C

, where

C

is an arbitrary constant. This means potential varies linearly with position; it is not necessarily zero anywhere. Answer A captures this exactly: a uniform field implies linearly varying potential, with no requirement that it equal zero. Answer B contains a subtle but critical error. Yes, a spatially constant

\vec{E}

means

\nabla V

is constant — but a constant nonzero gradient means

V

changes with position. Saying

V

must therefore be spatially constant (let alone zero) is a logical contradiction. Symmetry arguments never force a potential to be zero unless a specific boundary condition or reference point is defined. Answer C confuses two separate ideas. Gauss's law in differential form,

\nabla \cdot \vec{E} = \rho/\epsilon_0

, being zero in a charge-free region says nothing about

V

equaling zero — that's a misapplication of Poisson's equation, which instead reads

\nabla^2 V = -\rho/\epsilon_0

. Answer D mistakes a convention for a physical requirement. Defining

V = 0

at a boundary is a choice, not a geometric consequence of field-line orientation. Your key takeaway:

\vec{E}

depends on the gradient of

V

, never its absolute value. Zero potential is always a reference choice — never forced by field geometry alone.