Gauss's Law for Electric Fields - Physics 2

Two infinite nonconducting sheets have $+\sigma$ and $-\sigma$ . Field between them:
Answer: $\sigma/\varepsilon_0$ Each sheet alone creates field sigma/(2 epsilon0). Between the sheets the positive sheet points away from it and the negative sheet points toward it, so the two contributions point the same way inside and add to sigma/epsilon0. The tempting single-sheet value sigma/(2 epsilon0) ignores that both sheets contribute.
A spherical conducting shell of inner radius $a$ and outer radius $b$ carries a net charge of $+3Q$ . A point charge of $-Q$ is placed at the center of the shell. Using Gauss's law, what is the electric field magnitude at a distance $r$ where $a < r < b$ (inside the conducting material), and what is the surface charge density on the outer surface of the shell?
Answer: $E = 0$ inside the conductor; outer surface charge density $\sigma_{outer} = \frac{2Q}{4\pi b^2}$ , because the $-Q$ point charge induces $+Q$ on the inner surface, leaving $+2Q$ on the outer surface from the shell's net charge. Whenever you see a problem involving a conducting shell with an interior charge, your first instinct should be to apply two foundational principles: Gauss's law and the electrostatic property that conductors have zero internal electric field. Inside a conductor in electrostatic equilibrium, free charges redistribute instantly to cancel any internal field, so $E = 0$ for $a < r < b$ — no exceptions. This immediately eliminates choice D, which incorrectly claims the point charge creates a nonzero field inside the conducting material. That would only be true in empty space, not within a conductor. Now for the surface charges. Because $E = 0$ everywhere inside the conductor, any Gaussian surface drawn within the material encloses zero net charge. The $-Q$ point charge at the center must therefore be exactly compensated — meaning $+Q$ is induced on the inner surface (at radius $a$ ). Since the shell's total net charge is $+3Q$ and $+Q$ of it is pulled to the inner surface, charge conservation requires the remaining $+2Q$ to reside on the outer surface (at radius $b$ ). The outer surface charge density is then $\sigma_{outer} = \frac{2Q}{4\pi b^2}$ . This is choice A, and it's correct. Choice B incorrectly adds the full $+3Q$ shell charge to the outer surface along with an extra $+Q$ induction, double-counting the induced charge. Choice C ignores the induction process entirely, wrongly assuming the shell's $+3Q$ sits untouched on the outer surface regardless of the interior charge. Study tip: Always track charge using conservation — identify what's induced on the inner surface first, then subtract from the shell's net charge to find the outer surface charge. This two-step method prevents the double-counting trap that makes B so tempting.
What is the ratio of the electric field magnitude at $r = R$ (the surface) to the electric field magnitude at $r = 2R$ (outside the cylinder)?
Answer: $\frac{E(R)}{E(2R)} = 2$ , because $E(R) = \rho_0 R/(3\epsilon_0)$ inside and $E(2R) = \rho_0 R/(6\epsilon_0)$ outside, giving a ratio of 2. Whenever you see a non-uniform charge distribution in a cylindrical geometry, your first instinct should be Gauss's Law — but you must carefully compute the enclosed charge by integrating $\rho(r)$ over the actual volume, not assuming uniformity. For a cylindrical Gaussian surface of radius $r$ and length $L$ , Gauss's Law gives $E(2\pi r L) = Q_{enc}/\epsilon_0$ . Inside ( $r \leq R$ ), the enclosed charge is: $Q_{enc} = \int_0^r \rho_0 \frac{r'}{R}(2\pi r' L)\,dr' = \frac{2\pi \rho_0 L}{R}\cdot\frac{r^3}{3}$ So $E(r) = \frac{\rho_0 r^2}{3\epsilon_0 R}$ , giving $E(R) = \frac{\rho_0 R}{3\epsilon_0}$ . Outside ( $r > R$ ), all charge is enclosed: $Q_{enc} = \frac{2\pi \rho_0 L R^2}{3}$ , so $E(r) = \frac{\rho_0 R^2}{3\epsilon_0 r}$ , giving $E(2R) = \frac{\rho_0 R}{6\epsilon_0}$ . The ratio is $E(R)/E(2R) = 2$ , confirming choice B. Choice A is wrong because continuity of the field at the surface doesn't mean the field is equal at $r = R$ and $r = 2R$ — those are two different points, and the field outside falls as $1/r$ . Choice C uses a flawed "uniform-density approximation" that misapplies the interior formula; the non-uniform density changes the enclosed charge calculation entirely. Choice D correctly notes the $1/r$ exterior behavior but incorrectly concludes the two values are equal — doubling $r$ halves the outside field, but $E(R)$ from the interior formula is not the same starting point. As a study tip: always re-derive $Q_{enc}$ from scratch when $\rho$ is non-uniform — shortcut formulas assuming constant density will mislead you every time.
Point charge $+q$ at center of uncharged conducting shell. Field outside at distance $r$ is
Answer: outward $kq/r^2$ The uncharged shell has -q induced on its inner surface and +q on its outer surface. A Gaussian surface outside the shell encloses the central +q, the inner -q, and the outer +q, so the net enclosed charge is +q. Thus the field is outward kq/ $r^2$ . The tempting zero-field answer confuses shielding of the interior from outside fields with the net charge that still produces an external field.
Nonconducting sphere radius $R$ has charge density $\rho=ar^2$ . For $r<R$ , $E$ is proportional to
Answer: $r^3$ Gauss's law at radius r<R: the enclosed charge is the integral of a r'^2 over volume, proportional to $r^5$ . Dividing by the Gaussian surface area 4 pi $r^2$ leaves E proportional to $r^3$ . The tempting wrong choice is $r^2$ because that is rho, but you must integrate the charge first.
An infinitely long cylindrical shell of radius $R$ carries a uniform surface charge density $\sigma > 0$ . A second coaxial infinitely long solid cylinder of radius $R/2$ is placed inside, carrying a uniform volume charge density $\rho < 0$ such that the total charge per unit length of the entire system is zero. A student constructs a Gaussian cylinder of radius $r$ with $R/2 < r < R$ and length $L$ . Which of the following correctly describes the electric field in this region?
Answer: $E = \frac{\rho R^2}{8\epsilon_0 r}$ directed radially inward, because only the inner solid cylinder contributes to the enclosed charge and $\rho < 0$ , making the field point inward with magnitude that decreases as $1/r$ . When applying Gauss's Law to cylindrical symmetry, your first job is always to identify what charge is enclosed by your Gaussian surface — not the total charge of the entire system. For a Gaussian cylinder of radius $r$ (where $R/2 < r < R$ ) and length $L$ , only the inner solid cylinder of radius $R/2$ lies inside the surface. Its enclosed charge is $Q_{enc} = \rho \cdot \pi (R/2)^2 \cdot L = \frac{\rho \pi R^2 L}{4}$ . Applying Gauss's Law: $E(2\pi r L) = \frac{Q_{enc}}{\epsilon_0}$ , which gives $E = \frac{\rho R^2}{8\epsilon_0 r}$ . Since $\rho < 0$ , this value is negative, meaning the field points radially inward — confirming B is correct. A is wrong because the zero total charge condition applies to the whole system, not to any sub-region. Gauss's Law depends only on enclosed charge, not global charge balance. The shell at radius $R$ contributes nothing to the enclosed charge here, so the field is nonzero. C is tempting but subtly wrong. The field from a negative enclosed charge genuinely points inward — you cannot simply relabel it as pointing outward with the same magnitude. Direction is physically meaningful; flipping it misrepresents the field. D contains a calculation error: it drops a factor of 2 in the denominator. The area of the Gaussian surface is $2\pi r L$ , so dividing correctly yields $8\epsilon_0 r$ in the denominator, not $4\epsilon_0 r$ . Study tip: In Gauss's Law problems, always sketch which charges fall inside your surface before calculating — the global charge distribution is a deliberate distraction.
Insulating sphere radius $R$ has uniform charge. At $r=R/2$ , $E$ is how many times $E$ at $r=2R$ ?
Answer: 2 times Inside a uniformly charged sphere, E grows linearly with r: E = kQr/ $R^3$ . At r = R/2, this gives E = kQ/( $2R^2$ ). Outside, the sphere acts like a point charge, so at r = 2R, E = kQ/ $(2R)^2$ = kQ/( $4R^2$ ). Dividing gives 2. The tempting wrong answer is 16 times from using the inverse-square law at R/2, but inside the sphere only the enclosed charge matters and the field is not point-like.
A Gaussian surface is a cube of side length $L$ centered at the origin. The electric field throughout all of space is given by $\mathbf{E} = E_0 x \hat{x}$ , where $E_0$ is a positive constant and $x$ is the $x$ -coordinate. What is the total charge enclosed within the Gaussian cube?
Answer: $Q_{enc} = \epsilon_0 E_0 L^3$ , found by computing the net flux through the two faces perpendicular to $\hat{x}$ (the flux through the other four faces is zero since $\mathbf{E} \cdot \hat{n} = 0$ on them) and applying Gauss's law. When you see a question involving a non-uniform electric field and a symmetric surface, your first instinct should be Gauss's law: $\Phi_{net} = Q_{enc}/\epsilon_0$ . The key is carefully computing the flux through each face. The cube spans from $-L/2$ to $+L/2$ along each axis. Since $\mathbf{E} = E_0 x\hat{x}$ has no $y$ or $z$ components, the dot product $\mathbf{E} \cdot \hat{n} = 0$ on all four faces parallel to the $x$ -axis — those faces contribute nothing. Only the two faces perpendicular to $\hat{x}$ matter. On the right face ( $x = +L/2$ , outward normal $+\hat{x}$ ): flux $= E_0(L/2) \cdot L^2$ . On the left face ( $x = -L/2$ , outward normal $-\hat{x}$ ): flux $= -E_0(-L/2) \cdot L^2 = +E_0(L/2)L^2$ . Both faces contribute positively, giving net flux $\Phi = E_0 L^3$ . Applying Gauss's law: $Q_{enc} = \epsilon_0 E_0 L^3$ , confirming D. Choice A drops the factor of $L$ from the face separation, essentially treating the cube as a 2D surface and ignoring how the field varies over the cube's depth. Choice B incorrectly multiplies the correct result by an extra factor of 2 — both faces already contribute equally and are already summed; there's no additional doubling. Choice C is a tempting symmetry trap: yes, $\mathbf{E}$ is antisymmetric in sign at the two faces, but the outward normals are also opposite, so both flux contributions are positive, not canceling. Your study tip: always track the outward normal direction on each face separately — antisymmetry in $\mathbf{E}$ combined with antisymmetry in $\hat{n}$ produces constructive, not destructive, contributions to net flux.
A student attempts to apply Gauss's law to find the electric field of an electric dipole (two equal and opposite charges $+q$ and $-q$ separated by distance $d$ ) by drawing a spherical Gaussian surface centered midway between the charges. The student argues: "Since the total enclosed charge is zero, Gauss's law tells us $E = 0$ everywhere on this surface." Which of the following is the most precise and complete critique of this reasoning?
Answer: The student correctly applies Gauss's law to conclude the net flux is zero, but incorrectly equates zero net flux with zero field. Gauss's law relates the total flux through a closed surface to enclosed charge; it does not imply $E = 0$ pointwise unless $\mathbf{E}$ is uniform in magnitude and normal to the surface, allowing it to factor out of the integral. Whenever you see a question invoking Gauss's law, your first instinct should be to recall exactly what the law says: $\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$ . This is a statement about the total flux — a single integrated quantity — not about the field at any individual point on the surface. The student's reasoning commits a critical logical error. Yes, the enclosed charge is zero, so the net flux through the surface is indeed zero. But "net flux = 0" means the field lines entering the surface exactly cancel those exiting it. It says absolutely nothing about whether $\mathbf{E}$ vanishes at any specific point. You can only factor $E$ out of the integral — and thereby conclude $E = 0$ pointwise — when symmetry guarantees that $|\mathbf{E}|$ is constant and $\mathbf{E}$ is everywhere perpendicular (or parallel) to the surface. A dipole has no such symmetry; the field varies wildly in both magnitude and direction across the spherical surface. Answer B captures this distinction precisely and completely. Answer A is partially true but overstated. Gauss's law is always mathematically valid — it's a fundamental theorem. The issue isn't validity; it's that without symmetry, you can't extract $E$ from the integral to solve for it usefully. Answer C is confused and self-contradicting; it correctly notes $Q_{\text{enc}} = 0$ but then invents a false requirement about surface centering that doesn't exist in Gauss's law. Answer D is entirely fabricated. Gauss's law places no restriction on where you "want" the field — there's no such requirement in the law. Study tip: Always ask two questions when applying Gauss's law — "What is $Q_{\text{enc}}$ ?" and "Does symmetry let me pull $E$ outside the integral?" Both conditions must be satisfied to find $E$ pointwise.
A long coaxial cable consists of a solid inner conductor of radius $a$ carrying free charge per unit length $+\lambda$ , surrounded by a thin conducting cylindrical shell of radius $b > a$ carrying free charge per unit length $+\lambda$ (same sign and magnitude). In electrostatic equilibrium, what is the electric field magnitude at $r = (a+b)/2$ (midway between the conductors)?
Answer: $E = \frac{\lambda}{2\pi\epsilon_0\left(\frac{a+b}{2}\right)}$ , because only the inner conductor's charge contributes within the Gaussian surface; the outer shell's charges reside on its inner and outer surfaces but the net charge enclosed by a surface between the conductors is only $+\lambda$ . Whenever you see a coaxial cable problem in electrostatics, your instinct should be to reach for Gauss's Law with a cylindrical Gaussian surface. The electric field between the conductors depends entirely on the net charge enclosed by your chosen surface — not on charges sitting outside it. Draw a cylindrical Gaussian surface of radius $r = \frac{a+b}{2}$ and length $\ell$ . This surface sits in the gap between the two conductors. The only charge it encloses is the free charge on the inner conductor: $+\lambda\ell$ . Because the outer shell is a conductor in electrostatic equilibrium, any charge it carries must reside on its surfaces — specifically, $-\lambda$ on its inner surface and $+2\lambda$ on its outer surface (to keep the shell's net charge $+\lambda$ ). But both of these surface charges lie outside your Gaussian surface and contribute nothing to the enclosed charge. Applying Gauss's Law: $E(2\pi r \ell) = \frac{\lambda \ell}{\epsilon_0} \implies E = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{\lambda}{2\pi\epsilon_0\!\left(\frac{a+b}{2}\right)}$ This confirms C is correct. A is wrong because a grounded shell would carry $-\lambda$ on its inner surface, canceling the inner field — but here the outer shell carries $+\lambda$ total, so it cannot cancel anything inside. B incorrectly claims $-\lambda$ induced on the inner surface adds to the enclosed charge; induced charges are on the outer shell, outside the Gaussian surface, and don't count. D uses an incorrect formula — the standard cylindrical result always carries $2\pi\epsilon_0$ , not $\pi\epsilon_0$ . Study tip: Always identify where charges reside on conductors before counting enclosed charge — induced surface charges can fool you if you forget which surface they sit on relative to your Gaussian surface.
Using the inverse of Gauss's law (inferring charge distribution from the field), what is the volume charge density $\rho(r)$ for $r < R$ , and is there a surface charge at $r = R$ ?
Answer: $\rho(r) = \frac{4\epsilon_0 E_0 r}{R^2}$ for $r < R$ , with no surface charge at $r = R$ because the field is continuous across the boundary. Whenever you see a problem asking you to infer charge distribution from an electric field, your go-to tool is the differential form of Gauss's law: $\nabla \cdot \mathbf{E} = \rho/\epsilon_0$ . In spherical coordinates with purely radial fields, the divergence is $\frac{1}{r^2}\frac{d}{dr}(r^2 E)$ , not simply $dE/dr$ . That extra geometric factor is the heart of this problem. For $r < R$ , apply the full spherical divergence to $E = E_0 r^2/R^2$ : $\rho = \epsilon_0 \cdot \frac{1}{r^2}\frac{d}{dr}\!\left(r^2 \cdot \frac{E_0 r^2}{R^2}\right) = \epsilon_0 \cdot \frac{1}{r^2}\frac{d}{dr}\!\left(\frac{E_0 r^4}{R^2}\right) = \epsilon_0 \cdot \frac{4E_0 r}{R^2} = \frac{4\epsilon_0 E_0 r}{R^2}$ For the surface charge, check whether the field is continuous at $r = R$ . Inside: $E(R^-) = E_0 R^2/R^2 = E_0$ . Outside: $E(R^+) = E_0 R/R = E_0$ . The field matches exactly, so there is no surface charge — confirming answer D. Answer A uses only $dE/dr$ without the $1/r^2 \frac{d}{dr}(r^2 E)$ structure, missing the full divergence entirely. Answer B gets the divergence formula right but then incorrectly concludes there is a negative surface charge — the field is actually continuous at $r = R$ , so no surface charge exists. Answer C computes $\frac{1}{r^2}\frac{d}{dr}(r^2 E)$ but makes an algebra error, dropping the factor of 4 and arriving at a constant density instead of one proportional to $r$ . Study tip: Always use the full spherical divergence $\frac{1}{r^2}\frac{d}{dr}(r^2 E)$ for radial fields — surface charges appear only when $E$ is discontinuous at a boundary, so always check continuity explicitly.
A hollow non-conducting spherical shell of inner radius $R_1$ and outer radius $R_2$ carries a total charge $Q > 0$ distributed non-uniformly such that the volume charge density varies only with $r$ . A student claims that Gauss's law can be used to find $E(r)$ for $R_1 < r < R_2$ without knowing the exact functional form of $\rho(r)$ , provided only that the total charge $Q$ is known. Which statement best evaluates this claim?
Answer: The claim is partially correct: Gauss's law can determine $E(r)$ for $r > R_2$ from the total charge $Q$ , but inside the shell the enclosed charge at radius $r$ depends on the distribution, so $E(r)$ for $R_1 < r < R_2$ cannot be found without $\rho(r)$ . Whenever you see a Gauss's law question involving a non-uniform charge distribution, your first instinct should be to check two things independently: Is there spherical symmetry? and What is the enclosed charge at the radius of interest? Here, because $\rho$ depends only on $r$ (not on angle), the system does retain full spherical symmetry. This means the electric field must be purely radial, and you can confidently write $\oint \mathbf{E}\cdot d\mathbf{A} = E(r)\cdot 4\pi r^2$ . Gauss's law is absolutely applicable. The critical issue, however, is what goes on the right side of that equation. For a Gaussian surface at radius $r$ where $R_1 < r < R_2$ , you need $Q_{enc}(r)$ , which is only the charge residing between $R_1$ and $r$ — a fraction of $Q$ that depends entirely on how $\rho$ is distributed radially. Without knowing $\rho(r)$ , you cannot compute that integral. Only for $r > R_2$ does the enclosed charge equal the full $Q$ , making $E(r)$ determinable from $Q$ alone. This makes C the correct answer. A is wrong because it misidentifies the problem — symmetry is preserved, so the flux integral can be simplified; the real obstacle is computing $Q_{enc}$ , not setting up the surface integral. B is wrong because spherical symmetry is not broken when $\rho = \rho(r)$ ; non-uniformity in $r$ alone preserves the angular symmetry Gauss's law requires. D is wrong because it falsely claims $Q_{enc}$ is determined by $Q$ and geometry alone — inside the shell, geometry alone is insufficient without the radial distribution. Study tip: Always ask "what is my enclosed charge?" separately from "can I simplify the flux integral?" — these are two independent steps in any Gauss's law problem.
Long nonconducting cylinder radius $R$ has uniform volume charge density. For $r<R$ , $E$ is proportional to
Answer: $r$ For a Gaussian surface inside the cylinder at radius r, the enclosed charge is proportional to $r^2$ while the surface area is proportional to r. Gauss's law gives E times area proportional to enclosed charge, so E is proportional to $r^2$ / r = r. The tempting 1/r answer applies outside a line charge or outside the cylinder, not inside the uniform cylinder.
Using superposition with Gauss's law results for infinite planes, what is the magnitude and direction of the electric field in the region between the two planes and in the region to the right of Plane 2 (taking rightward as positive)?
Answer: Between the planes: $E = \frac{3\sigma}{2\epsilon_0}$ directed rightward; to the right of Plane 2: $E = \frac{\sigma}{2\epsilon_0}$ directed leftward. When you encounter superposition problems with infinite charged planes, your go-to tool is the result from Gauss's law: each infinite plane independently produces a uniform field of magnitude $\frac{|\sigma_{surface}|}{2\epsilon_0}$ pointing away from a positive plane and toward a negative plane, everywhere in space. Here, Plane 1 carries $+\sigma$ and Plane 2 carries $-2\sigma$ . Plane 1 contributes $\frac{\sigma}{2\epsilon_0}$ rightward everywhere to its right (including the between region and beyond Plane 2). Plane 2 contributes $\frac{2\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$ directed toward it — meaning leftward to its right, and rightward to its left (i.e., in the between region). Between the planes: Both contributions point rightward, so they add: $\frac{\sigma}{2\epsilon_0} + \frac{\sigma}{\epsilon_0} = \frac{\sigma}{2\epsilon_0} + \frac{2\sigma}{2\epsilon_0} = \frac{3\sigma}{2\epsilon_0}$ rightward. To the right of Plane 2: Plane 1 still points rightward $\left(\frac{\sigma}{2\epsilon_0}\right)$ , but Plane 2 now points leftward $\left(\frac{\sigma}{\epsilon_0}\right)$ , giving a net $\frac{\sigma}{\epsilon_0} - \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{2\epsilon_0}$ leftward. This confirms A. Choice B incorrectly adds the between-region fields as if one cancels the other. Choice C flips the direction between the planes, confusing which way each plane's field points in that region. Choice D doubles the fields without the factor of 2 in the denominator from Gauss's law, using $\frac{\sigma}{\epsilon_0}$ instead of $\frac{\sigma}{2\epsilon_0}$ per plane. Your strategy: always draw each plane separately, assign field arrows based on sign, then vector-add region by region. Forgetting the factor of $\frac{1}{2}$ per plane is the most common error here.

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