Electric Flux - Physics 2

1. What is the electric flux through a Gaussian spherical surface of radius $r$ where $R_1 < r < R_2$ (i.e., inside the conducting material of the shell)?
   Answer: $\Phi = 0$, because the Gaussian surface lies entirely within the conducting material, where the electric field is zero in electrostatic equilibrium; therefore the flux integral $\oint \vec{E} \cdot d\vec{A} = 0$ by direct evaluation. Whenever you see a Gaussian surface drawn inside a conductor, your first instinct should be: what does electrostatics say about the electric field there? In electrostatic equilibrium, free charges in a conductor redistribute until the internal field is exactly zero — this is a foundational property of conductors. Because the Gaussian surface at $R_1 < r < R_2$ lies entirely within the conducting material, the electric field is zero at every point on that surface. Gauss's law states $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$, and since $\vec{E} = 0$ everywhere on the surface, the flux integral evaluates directly to zero — making B correct. Now here's the subtle but critical follow-up: if the flux is zero, then $Q_{\text{enc}} = 0$ as well. This forces a $+Q$ charge to appear on the inner surface (to cancel the central $-Q$), and the remaining $+2Q$ moves to the outer surface — a useful result, but not what the question asks. A and D both correctly identify the charge distribution (inner surface $+Q$, outer surface $+2Q$), and D's reasoning is even self-consistent with Gauss's law — but both answers make a fatal error: they assume $\vec{E} \neq 0$ inside the conductor, which violates electrostatic equilibrium. The zero flux comes from the zero field directly, not from a charge-counting argument. C incorrectly treats the entire $+3Q$ shell charge as enclosed, ignoring that charge resides on surfaces, not uniformly throughout. Study tip: When a Gaussian surface sits inside a conductor, evaluate $\vec{E}$ first — the field argument gives flux = 0 immediately, before you ever count charges.
2. What is the electric flux through one of the three faces of the cube that share the corner where the charge is located?
   Answer: $\Phi = 0$, because each of the three faces adjacent to the charge lies in a plane that contains the charge itself; the electric field from $+Q$ is directed radially outward from the charge and therefore lies parallel to each of these faces, so the field has no component along the face normal and contributes zero flux through each adjacent face. When dealing with Gauss's Law and flux through specific surfaces, always ask yourself: what is the geometry of the field relative to each face's normal vector? Here, the charge $+Q$ sits exactly at the corner of the cube. Consider any one of the three faces that share that corner. The charge lies in the plane of each of those faces — right at the face's corner. Since the electric field radiates outward from $+Q$ in all directions, and every field line from a charge lying within a plane either stays parallel to that plane or points away from it along the surface, the field has zero component perpendicular to each adjacent face. Flux is defined as $\Phi = \int \vec{E} \cdot d\vec{A}$, and since $\vec{E}$ is everywhere parallel to each adjacent face ($\vec{E} \perp \hat{n}$ at all points), the dot product is zero — making $\Phi = 0$ for each of the three adjacent faces. D is correct. Choice A correctly identifies that the cube encloses $Q/8$ of the total flux (by the eight-cube symmetry argument), but then incorrectly assumes that flux is distributed equally across all six faces — it isn't, because the geometry is not symmetric about the charge's position. Choice B uses the right total flux but invents a solid-angle argument that contradicts the actual geometry; adjacent faces subtend zero solid angle from a charge lying in their own plane. Choice C makes the opposite error from A and B — it correctly rejects equal distribution but wrongly concludes all flux exits through the adjacent faces, when in fact it exits through the opposite three faces. Study tip: Whenever a charge lies on a surface (face, edge, or corner), the flux through that surface is always zero — the field never "punches through" a plane it originates from.
3. What is the electric flux through this square surface?
   Answer: $\Phi = 0$, because the surface lies in the $xz$-plane and its outward normal is $\hat{y}$; the electric field from the line charge at any point on this surface points in the $\hat{x}$-direction (radially away from the $z$-axis, since $y = 0$ on this surface), so $\vec{E} \cdot \hat{y} = 0$ everywhere on the surface. When calculating electric flux, the single most important thing to check before doing any integration is the geometric relationship between the electric field direction and the surface's outward normal. Flux is defined as $\Phi = \iint \vec{E} \cdot \hat{n} \, dA$, so if $\vec{E} \perp \hat{n}$ everywhere on a surface, the flux is immediately zero — no integration needed. Here's why D is correct: the square lies in the $xz$-plane, meaning every point on it has $y = 0$. The outward normal to this surface is $\hat{y}$. The line charge runs along the $z$-axis, and by symmetry, its electric field points radially outward in the $xy$-plane. At any point where $y = 0$, that radial direction is purely $\hat{x}$. Therefore $\vec{E} = E(x)\hat{x}$ on the entire surface, and $\vec{E} \cdot \hat{y} = 0$ everywhere. The flux is exactly zero. Choice A applies the correct field magnitude $E = \lambda/(2\pi\varepsilon_0 d)$ but ignores that this field is perpendicular to the surface normal — multiplying by area gives nothing physically meaningful here. Choice B correctly identifies that the field varies with $x$ and even performs the right integral for a surface facing the line charge, but the surface here faces sideways (normal is $\hat{y}$, not $\hat{x}$), so the setup doesn't apply. Choice C misapplies Gauss's law — that law relates flux through a closed surface to enclosed charge, not through arbitrary open surfaces. Your takeaway: always identify $\hat{n}$ and $\vec{E}$'s direction before calculating anything. A zero dot product means zero flux, instantly.
4. What is the net electric flux through this rectangular Gaussian surface?
   Answer: $\Phi = \frac{\rho \pi R^2 L}{\varepsilon_0}$, because Gauss's law gives the net flux as $Q_{\text{enc}}/\varepsilon_0$, and the enclosed charge is determined by the volume of the cylindrical charge distribution ($\pi R^2 L$) that fits inside the Gaussian surface, regardless of the shape of the Gaussian surface itself. Whenever you see a Gauss's law problem, anchor yourself to one foundational principle: **the net electric flux through any closed surface equals the total enclosed charge divided by $\varepsilon_0$**, regardless of the surface's shape. The shape of the Gaussian surface never changes how much charge is inside it. Here, the cylindrical charge distribution has radius $R$ and length $L$, giving a volume of $\pi R^2 L$. The enclosed charge is therefore $Q_{\text{enc}} = \rho \pi R^2 L$. Plugging into Gauss's law: $\Phi = \frac{Q_{\text{enc}}}{\varepsilon_0} = \frac{\rho \pi R^2 L}{\varepsilon_0}$ This is answer C, and it's correct because only the actual charge sitting inside the Gaussian surface contributes to net flux — the geometry of the box is irrelevant. A is wrong because it invents a fictional concept called "charge redistribution at boundaries." No such correction exists in Gauss's law — uncharged corner regions contribute zero charge and therefore zero net flux. B is wrong because it confuses the volume of the Gaussian surface with the volume of the charge distribution. Gauss's law integrates over the enclosed charge, not the enclosed space. The $4R^2 L$ rectangular volume is partially empty (the four corners contain no charge). D is wrong for a similar reason — it fabricates a geometric correction factor. Gauss's law requires no such scaling; the integral automatically accounts for only the charge present. Your study tip: whenever a problem uses an unconventional Gaussian surface shape, it's almost always testing whether you remember that $\Phi = Q_{\text{enc}}/\varepsilon_0$ depends purely on enclosed charge, never on surface geometry.
5. What is the net electric flux through the entire closed surface of the cube?
   Answer: $\Phi = E_0 L^2$, contributed entirely by the two faces perpendicular to $\hat{x}$, since the $\hat{z}$ component of the field is uniform and contributes zero net flux through the top and bottom faces. When tackling electric flux problems with non-uniform fields, your most powerful tools are direct surface integration and Gauss's Law together — use them to check each other. The net flux through a closed surface equals $\oint \vec{E} \cdot d\vec{A}$. For this cube, you can evaluate each pair of faces separately. For the two faces perpendicular to $\hat{x}$ (at $x = 0$ and $x = L$), the outward normals are $-\hat{x}$ and $+\hat{x}$ respectively. The $x$-component of $\vec{E}$ is $E_0(1 + x/L)$. At $x = 0$, the flux is $-E_0 L^2$; at $x = L$, it is $+2E_0 L^2$. The net contribution is $E_0 L^2$. For the faces perpendicular to $\hat{z}$ (top and bottom), the $\hat{z}$ component $E_0$ is constant, so the outward flux through the top equals $+E_0 L^2$ and through the bottom equals $-E_0 L^2$, canceling exactly. The faces perpendicular to $\hat{y}$ have no $\hat{y}$ component, contributing nothing. The total is $\Phi = E_0 L^2$, confirming A is correct. Choice B is wrong because it double-counts: the uniform $\hat{z}$ component produces zero net flux — equal amounts enter and exit. Choice C contains a subtle error: the divergence of $\vec{E}$ is $\partial E_x/\partial x = E_0/L$, which is not zero, so by Gauss's Law there is enclosed charge and nonzero flux. Choice D incorrectly averages the field rather than computing the flux difference between the two faces properly. As a strategy, always check each component of $\vec{E}$ against its corresponding pair of faces independently. A uniform component along any axis always contributes zero net flux — that's your quick filter for eliminating terms before calculating.
6. What is the electric flux through the curved hemispherical surface only (not including the flat base)?
   Answer: $\Phi_{\text{curved}} = \pi R^2 E_0$, because the flux through the curved surface equals the flux through the flat circular base by symmetry, and the flat base has area $\pi R^2$ with normal $\hat{z}$, giving $\Phi = E_0 \pi R^2$; the sign is positive since the field exits through the curved top. When you encounter electric flux problems involving curved surfaces, your most powerful tool is often Gauss's Law combined with the superposition of surfaces — not a brute-force integral. Here's the key insight: the hemisphere (curved surface) plus the flat circular base together form a closed surface enclosing no charge. By Gauss's Law, the total flux through any closed surface with no enclosed charge is zero: $\Phi_{\text{curved}} + \Phi_{\text{flat}} = 0$ The flat base lies in the $xy$-plane with its outward normal pointing downward ($-\hat{z}$, outward from the enclosed volume). The flux through it is: $\Phi_{\text{flat}} = \vec{E} \cdot (-\hat{z}) \cdot \pi R^2 = -E_0 \pi R^2$ Therefore: $\Phi_{\text{curved}} = -\Phi_{\text{flat}} = +\pi R^2 E_0$ This confirms answer A is correct — the curved surface flux equals $\pi R^2 E_0$, positive because the field genuinely exits through the curved top. Answer B is a trap: integrating $E_0\cos\theta$ over the hemisphere does not yield $2\pi R^2 E_0$. The $\cos\theta$ factor and the area element combine to give exactly $\pi R^2 E_0$, not twice that. Answer C misapplies Gauss's Law. That law gives zero flux only through a closed surface with no enclosed charge — the curved hemisphere alone is an open surface, so its flux need not be zero. Answer D gets the magnitude right but the sign wrong, confusing the orientation of the outward normal on the curved surface. Study tip: When you see flux through an open curved surface, mentally "close" it with a simple flat cap, apply Gauss's Law, then subtract the easy flat-surface flux. This avoids messy integrals entirely.
7. What is the electric flux through the disk?
   Answer: $\Phi = \frac{\pi a^2 E_0}{2}$, because integrating $E_0(r/a)^2$ over the disk using the area element $dA = 2\pi r\, dr$ gives $\int_0^a E_0\frac{r^2}{a^2} \cdot 2\pi r\, dr = \frac{2\pi E_0}{a^2}\cdot\frac{a^4}{4} = \frac{\pi a^2 E_0}{2}$. When a field varies across a surface, you can't just multiply a single field value by the area — you must integrate. Electric flux is defined as $\Phi = \int \vec{E} \cdot d\vec{A}$, and for a non-uniform field, this integral must account for how the field strength changes across the surface. Since the field $\vec{E} = E_0(r/a)^2\hat{z}$ is perpendicular to the disk (which lies in the $xy$-plane), the dot product simply gives $E_0(r/a)^2$. Using circular symmetry, the area element is $dA = 2\pi r\, dr$, so the flux becomes: $\Phi = \int_0^a E_0\frac{r^2}{a^2} \cdot 2\pi r\, dr = \frac{2\pi E_0}{a^2}\int_0^a r^3\, dr = \frac{2\pi E_0}{a^2}\cdot\frac{a^4}{4} = \frac{\pi a^2 E_0}{2}$ This confirms B is correct. Choice A makes an error in averaging. For a circular geometry, the area-weighted average of $(r/a)^2$ is not $1$ — that claim is fabricated with no mathematical support. The actual weighted average is $1/2$, which is exactly what the correct integral produces. Choice C uses a one-dimensional integral $\int_0^a E_0(r/a)^2\,dr$ and multiplies by the circumference $2\pi a$. This treats the disk like a rectangle rather than using the proper polar area element $2\pi r\, dr$, missing the crucial factor of $r$ inside the integral. Choice D claims the area-weighted average of $(r/a)^2$ is $1/4$, but the correct calculation gives $1/2$, as shown above. Your takeaway: whenever a field depends on position, always set up the full integral with the correct area element — in polar coordinates, that's $dA = 2\pi r\, dr$, not just $dr$.
8. What is the electric flux through the spherical shell due to both charges combined?
   Answer: $\Phi = \frac{Q}{\varepsilon_0}$, because only the charge enclosed by the shell matters for net flux through a closed surface; the external $-Q/2$ charge contributes zero net flux regardless of its location outside the shell. Whenever you see a question involving electric flux through a closed surface, your first instinct should be to reach for Gauss's Law: $\Phi = \frac{Q_{\text{enc}}}{\varepsilon_0}$. The entire power of this law rests on one critical fact — only the charge inside the closed surface determines the net flux through it. External charges are irrelevant to the net result. Here, the spherical shell encloses only $+Q$ at its center. Applying Gauss's Law directly gives $\Phi = \frac{Q}{\varepsilon_0}$, making B the correct answer. Yes, the external $-Q/2$ charge creates field lines that pierce the shell — but every field line that enters the shell from outside must also exit it somewhere else. The inward and outward contributions from any external charge cancel exactly, contributing zero net flux. A makes a fundamental error by treating Gauss's Law as depending on the total charge in the problem. It doesn't — only enclosed charge matters. The $-Q/2$ charge is outside the surface and is simply irrelevant to net flux. C incorrectly tries to calculate a partial flux contribution based on solid angle geometry. While solid angles are useful in other contexts, Gauss's Law already accounts for geometry automatically — any external charge contributes exactly zero net flux, not "half" or any other fraction. D constructs a physically creative but ultimately false narrative about field lines terminating before crossing the shell. Field lines from $+Q$ radiate outward through the shell in all directions; the external charge doesn't "intercept" them before they cross the surface. Study tip: When you see an external charge in a Gauss's Law problem, it's almost always a deliberate distractor. Train yourself to immediately identify $Q_{\text{enc}}$ and ignore everything outside the surface.
9. What is the net electric flux through the entire closed surface $S$ (treating it as a single closed surface enclosing the region between $r = R$ and $r = 2R$)?
   Answer: $\Phi = 0$, because the field $\vec{E} = A/r^2 \hat{r}$ has zero divergence for $r \neq 0$, meaning no charge is enclosed in the region between the two spheres, and by Gauss's law the net flux through any closed surface in a charge-free region is zero. Whenever you see a question combining Gauss's law with a field like $\vec{E} = A/r^2\,\hat{r}$, your first instinct should be to check the divergence and identify what charge is enclosed by the closed surface. Gauss's law states that the net electric flux through any closed surface equals $Q_\text{enc}/\varepsilon_0$. The field $\vec{E} = A/r^2\,\hat{r}$ is mathematically identical in form to a Coulomb field, and its divergence is exactly zero for all $r \neq 0$. Since the closed surface $S$ encloses only the shell region between $r = R$ and $r = 2R$ — with no charge at the origin inside that volume — $Q_\text{enc} = 0$, and therefore $\Phi = 0$. Answer C is correct. You can verify this directly: the outward flux through the outer sphere is $\frac{A}{(2R)^2} \cdot 4\pi(2R)^2 = 4\pi A$, and the flux entering through the inner sphere (inward, so negative for a closed surface) is $-\frac{A}{R^2} \cdot 4\pi R^2 = -4\pi A$. These cancel exactly, giving zero net flux. A is wrong because the formula presented is fabricated — there is no law relating flux to a "geometric mean" of areas and field differences. B makes a critical sign error: the inner sphere's outward normal points inward (toward the origin), so its contribution is negative, not additive. D correctly notes that no charge is enclosed but then contradicts itself by claiming a nonzero net flux — this directly violates Gauss's law. The key strategy: always assign normal directions carefully for closed surfaces. The outward normal on the inner shell points toward the origin, making its flux contribution negative, which is precisely why the $1/r^2$ field produces perfect cancellation.

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