Physics 2 Quiz: Interpreting Field Lines And Equipotentials

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Interpreting Field Lines And EquipotentialsQuestion 1 of 5

Two infinite, parallel, oppositely charged conducting plates (a capacitor) are separated by distance $d$ . A thin dielectric slab of thickness $d/2$ and dielectric constant $\kappa = 3$ is inserted between the plates, touching the negative plate, with a vacuum gap of $d/2$ remaining adjacent to the positive plate. Compared to the uniform-field case with no dielectric (same charge $Q$ on the plates), which statement about the field lines and equipotentials in each region is correct?

The field line density (and hence

|\vec{E}|

) is the same in both the vacuum region and the dielectric region, because the surface charge density

\sigma

on the plates is unchanged and Gauss's law requires the same displacement field

D

throughout; the equipotentials are uniformly spaced throughout the entire gap.

The field line density is three times greater in the vacuum region than in the dielectric region because the dielectric reduces the field inside it by a factor of

\kappa = 3

; the equipotentials are more closely spaced in the vacuum region and more widely spaced in the dielectric region.

The field line density is three times greater in the dielectric region than in the vacuum region because bound surface charges on the dielectric add to the free charge, enhancing the field inside the dielectric; the equipotentials are more closely spaced in the dielectric region.

The field line density is the same in both regions because the boundary condition requires the normal component of

\vec{E}

to be continuous across any interface; the equipotentials are therefore uniformly spaced throughout, but the potential difference across the dielectric is three times that across the vacuum gap.

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Physics 2 Quiz

Physics 2 Quiz: Interpreting Field Lines And Equipotentials

Practice Interpreting Field Lines And Equipotentials 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 Interpreting Field Lines And Equipotentials, 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 density (and hence $|\vec{E}|$ ) is the same in both the vacuum region and the dielectric region, because the surface charge density $\sigma$ on the plates is unchanged and Gauss's law requires the same displacement field $D$ throughout; the equipotentials are uniformly spaced throughout the entire gap.
The field line density is three times greater in the vacuum region than in the dielectric region because the dielectric reduces the field inside it by a factor of $\kappa = 3$ ; the equipotentials are more closely spaced in the vacuum region and more widely spaced in the dielectric region. (correct answer)
The field line density is three times greater in the dielectric region than in the vacuum region because bound surface charges on the dielectric add to the free charge, enhancing the field inside the dielectric; the equipotentials are more closely spaced in the dielectric region.
The field line density is the same in both regions because the boundary condition requires the normal component of $\vec{E}$ to be continuous across any interface; the equipotentials are therefore uniformly spaced throughout, but the potential difference across the dielectric is three times that across the vacuum gap.

Explanation: Whenever you see a capacitor problem with a partial dielectric, your first tool should be Gauss's law in the form

\vec{D} = \epsilon_0 \kappa \vec{E}

. Since the plates carry fixed surface charge density

\sigma

, the displacement field

D = \sigma

is the same everywhere between the plates — but

\vec{E}

is not. In the vacuum region (adjacent to the positive plate),

E_{\text{vac}} = \frac{\sigma}{\epsilon_0}

. In the dielectric region (touching the negative plate),

E_{\text{diel}} = \frac{\sigma}{\kappa \epsilon_0} = \frac{\sigma}{3\epsilon_0}

. The electric field is three times weaker inside the dielectric. Since field line density is proportional to

|\vec{E}|

, the vacuum region has three times the field line density of the dielectric region. Equipotentials (surfaces of constant potential) are perpendicular to field lines, and their spacing is inversely proportional to field strength — so equipotentials are more closely packed in the vacuum and more spread out in the dielectric. This confirms B. Choice A confuses

D

with

E

. Yes,

D

is uniform throughout, but

E = D/(\kappa\epsilon_0)

varies with

\kappa

, so the field — and equipotential spacing — is not uniform. Choice C reverses the effect of the dielectric: bound surface charges on the dielectric actually oppose the applied field inside, reducing

E

, not enhancing it. Choice D incorrectly states that the normal component of

\vec{E}

is continuous across an interface; it is

D_\perp

that is continuous (for no free surface charge), not

E_\perp

. Study tip: Always distinguish

\vec{D}

(continuous across interfaces) from

\vec{E}

(reduced by

\kappa

inside a dielectric) — mixing them up is the most common trap in dielectric capacitor problems.

Question 2

A conducting sphere of radius $R$ carries a total charge $+Q$ . A thin, uncharged conducting spherical shell of inner radius $2R$ and outer radius $3R$ surrounds it concentrically. The system is in electrostatic equilibrium.

Which of the following correctly describes the electric field line pattern and equipotential structure in the region $R < r < 2R$ between the inner sphere and the shell?

Field lines extend radially outward from the inner sphere and terminate on the inner surface of the shell; the equipotentials are concentric spherical shells with spacing that decreases with increasing $r$ , and the potential at $r = 2R^-$ (just inside the shell) equals the potential at $r = 3R^+$ (just outside the shell).
Field lines extend radially outward from the inner sphere and terminate on the inner surface of the shell; the equipotentials are concentric spherical shells, but the potential difference between any two equipotentials in this region is zero because the induced charges on the shell exactly cancel the field of the inner sphere.
Field lines extend radially outward from the inner sphere through the region $R < r < 2R$ , pass through the conducting shell, and continue outward beyond $r = 3R$ ; the equipotentials are concentric spherical shells with spacing that increases with $r$ , and the potential inside the conducting shell equals the potential on its outer surface.
Field lines extend radially outward from the inner sphere and terminate on the inner surface of the shell; the equipotentials are concentric spherical shells, and the potential at $r = 2R$ equals the potential at $r = 3R$ because the shell is a conductor. (correct answer)

Explanation: When analyzing a conducting shell problem, focus on two key principles: (1) electric field lines cannot exist inside a conductor, and (2) the entire volume of a conductor is a single equipotential. In the region

R < r < 2R

, Gauss's Law tells you the field behaves like a point charge:

E = \frac{kQ}{r^2}

directed radially outward. Field lines originate on the inner sphere's surface charge

+Q

and terminate on

-Q

induced on the shell's inner surface (

r = 2R

). Since the shell is uncharged overall,

+Q

appears on the outer surface (

r = 3R

). The equipotentials are concentric spheres, and because

E \propto 1/r^2

, the potential varies as

V \propto 1/r

— spacing between equipotentials actually decreases as you move outward (more potential drop per unit length closer to the sphere). Crucially, the entire conductor — from

r = 2R

r = 3R

— sits at one potential, so

V(2R) = V(3R)

. This confirms D is correct. A is wrong because it claims equipotential spacing decreases with increasing

r

, which is backwards — spacing increases as the field weakens. It also incorrectly states that

V(2R^-) = V(3R^+)

; the potential just outside the shell (

3R^+

) is lower than at

2R

because the field does work in that outer region. B is wrong because the induced charges don't cancel the field in the gap — they redirect field lines, not eliminate them. A non-zero field exists throughout

R < r < 2R

. C is wrong on two counts: field lines cannot pass through a conductor, and equipotential spacing increases (not decreases) with

r

in this region. Remember: a conductor in equilibrium is always one equipotential. Any question involving nested conductors will hinge on this fact — use it to quickly eliminate answers that assign different potentials to different surfaces of the same conductor.

Question 3

An electric dipole consists of charges $+q$ and $-q$ separated by distance $2a$ . Consider a point P located on the perpendicular bisector of the dipole axis at distance $r \gg a$ from the center. A student claims: 'At point P, the equipotential surface passing through P is perpendicular to the dipole axis, because the electric field at P points radially outward from the dipole center.' Which of the following correctly evaluates this claim?

The claim is entirely correct: the equipotential through P is the perpendicular bisector plane ( $V = 0$ ), and the field at P points radially outward from the center, so both the conclusion and the reasoning are valid.
The claim is partially correct in its conclusion but wrong in its reasoning: the equipotential through P is approximately perpendicular to the dipole-to-P line only far from the dipole, and the field at P actually points radially outward but weakens as $1/r^3$ , not $1/r^2$ .
The claim has a correct conclusion but wrong reasoning: the equipotential through P is indeed the perpendicular bisector plane ( $V = 0$ ), which is perpendicular to the dipole axis. However, the field at P points antiparallel to the dipole moment (from $+q$ toward $-q$ ), not radially outward — and it is this axial field direction that is perpendicular to the bisector plane. (correct answer)
The claim is entirely incorrect: the field at P points radially inward toward the dipole center, and the equipotential surface through P is a curved, non-planar surface because the dipole field has no exact flat equipotential at any finite distance.

Explanation: When analyzing electric dipoles, you need to separately evaluate two things: the potential and the field direction at a given point — they're related but distinct, and confusing them is exactly the trap this question sets. On the perpendicular bisector of a dipole, every point is equidistant from

+q

and

-q

. Since the potentials from each charge are equal in magnitude but opposite in sign, they cancel exactly:

V = 0

everywhere on that plane. This makes the entire perpendicular bisector plane a flat equipotential surface — and since it's perpendicular to the dipole axis by geometry, the conclusion about the equipotential is correct. The reasoning, however, is wrong. The electric field on the perpendicular bisector does not point radially outward from the center. The field components pointing toward/away from the center (the radial components) cancel by symmetry. What remains are the components pointing antiparallel to the dipole moment — that is, from

+q

toward

-q

, along the dipole axis direction. This field is parallel to the dipole axis and therefore perpendicular to the bisector plane — which is why the field is perpendicular to the equipotential, as it should be (

\vec{E} \perp

equipotential surfaces is always true). So C is correct: right conclusion, wrong reasoning. A fails because it accepts the flawed "radially outward" reasoning as valid. B introduces a false comparison to

1/r^3

vs.

1/r^2

falloff, which is irrelevant to the directional error. D is doubly wrong — the field isn't radially inward, and the bisector plane is a flat equipotential. Study tip: Always verify field direction and potential independently for dipoles. The field on the bisector axis is antiparallel to

\vec{p}

— memorize this as a key dipole fact.

Question 4

A student draws a proposed electric field line diagram for a region of space and claims it represents a valid electrostatic field. The diagram shows field lines that (i) are closed loops (beginning and ending at the same point), (ii) are uniformly spaced throughout a region where the student also draws closely spaced equipotential lines in one sub-region, and (iii) never cross each other. Which of the following correctly identifies all the violations of valid electrostatic field line rules present in this diagram?

Only feature (i) violates the rules: electrostatic field lines cannot form closed loops because $\nabla \times \vec{E} = 0$ implies the field is conservative. Features (ii) and (iii) are both consistent with valid electrostatic field diagrams.
Features (i) and (ii) both represent violations: closed loops violate the conservative nature of electrostatic fields, and uniformly spaced field lines (implying uniform $|\vec{E}|$ everywhere) are mutually inconsistent with closely spaced equipotentials in only one sub-region (which would imply stronger $|\vec{E}|$ there). The two parts of the diagram contradict each other. (correct answer)
Features (i) and (iii) both violate the rules: closed loops are forbidden, and field lines in a valid diagram must cross each other at source points to represent superposition of multiple charge distributions.
Only feature (i) is a violation, and it automatically makes feature (ii) invalid as well, since a closed-loop field would force non-uniform potential spacing everywhere; feature (iii) is a correctly satisfied rule because field lines never cross in any valid diagram.

Explanation: When analyzing electric field line diagrams, you need to check each feature against the fundamental rules derived from Maxwell's equations for electrostatics: (1) field lines cannot form closed loops, (2) field line spacing reflects field strength, and (3) field lines never cross. The closed loops in feature (i) directly violate

\nabla \times \vec{E} = 0

, which means the electrostatic field is conservative — no closed path can have a net change in potential, so field lines (which point in the direction of decreasing potential) cannot return to their starting point. This makes (i) a clear violation. Feature (ii) contains a subtler but equally real contradiction. Uniformly spaced field lines imply a uniform field magnitude

|\vec{E}|

throughout the entire region. However, the spacing between equipotential surfaces is related to field strength by

|\vec{E}| = -\frac{\Delta V}{\Delta d}

, meaning closely spaced equipotentials indicate a stronger field in that sub-region. These two claims cannot coexist — the diagram contradicts itself internally. This makes B correct: both (i) and (ii) are violations. Feature (iii) — field lines never crossing — is actually a correctly followed rule, not a violation. Field lines crossing would imply two different field directions at one point, which is physically impossible. Choice A misses the contradiction in (ii), treating uniform spacing and non-uniform equipotentials as compatible when they are not. Choice C wrongly claims field lines must cross at source points — they never cross, period. Choice D incorrectly ties (i) and (ii) together causally and misrepresents the nature of the (ii) violation. Study tip: Always check field line spacing and equipotential spacing together — they must tell a consistent story about

|\vec{E}|

in every sub-region.

Question 5

A student examines an electric field line diagram showing two regions: Region I, where field lines are straight, parallel, and uniformly spaced, and Region II (adjacent to Region I), where field lines curve and converge toward a point. The boundary between the two regions is a flat surface perpendicular to the field lines in Region I.

Which of the following statements about the equipotential surfaces in this configuration is correct?

In Region I, equipotential surfaces are flat planes parallel to the boundary, and in Region II, equipotential surfaces are spherical shells centered on the convergence point, with the spacing between successive equipotentials decreasing as one moves toward the convergence point. (correct answer)
In Region I, equipotential surfaces are flat planes parallel to the boundary, and in Region II, equipotential surfaces are spherical shells centered on the convergence point, with the spacing between successive equipotentials increasing as one moves toward the convergence point.
In both Region I and Region II, equipotential surfaces are flat planes, but their orientation rotates continuously from parallel to the boundary in Region I to parallel to the field lines in Region II.
In Region I, equipotential surfaces are flat planes perpendicular to the field lines, and in Region II, the equipotential surfaces are also flat planes but tilted at an angle to the field lines because the field is non-uniform, causing the potential to vary along any given plane.

Explanation: Whenever you see a question pairing electric field diagrams with equipotential surfaces, anchor yourself to the fundamental rule: equipotential surfaces are always perpendicular to electric field lines, and the spacing between equipotentials tells you about field strength — closer spacing means stronger field. In Region I, the field lines are straight, parallel, and uniformly spaced, which is the signature of a uniform electric field — like that between parallel plates. Because all field lines point the same direction and are evenly spaced, the equipotential surfaces must be flat planes oriented perpendicular to those lines (parallel to the boundary). In Region II, field lines converge toward a point, which mimics a point charge geometry. The equipotentials around a point charge are spherical shells centered on that charge. Crucially, because the field lines converge (field gets stronger closer to the point), the same potential difference

\Delta V = -\int \vec{E} \cdot d\vec{r}

is covered over a shorter distance near the convergence point — so successive equipotential shells are more closely spaced as you approach it. This makes A correct. B is wrong because it reverses the spacing logic — spacing decreases toward the convergence point, not increases. C is wrong because equipotential surfaces never rotate to become parallel to field lines; they are always perpendicular to them by definition. D is wrong on two counts: in a non-uniform field, equipotentials are curved (not flat tilted planes), and field lines are never parallel to equipotentials. Your go-to memory anchor: E-field lines ⊥ equipotentials, always — and denser field lines mean denser (closer-spaced) equipotentials.