Electric Field Lines - Physics 2

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?
Answer: 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. 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.
Based solely on the field line diagram described, which of the following conclusions is best supported about the charges A and B?
Answer: 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. 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.
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?
Answer: 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. 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.
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?
Answer: 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. 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.
Which of the following most accurately assesses the student's claim?
Answer: 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. 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.
Which of the following best evaluates the student's claim?
Answer: 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. 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.
Comparing the two regions, which of the following statements about the electric field and charge distribution is correct?
Answer: 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. 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.
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?
Answer: 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. 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.
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?
Answer: 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. 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.

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