Physics 2 Quiz: Electric Flux

What this quiz covers

This quiz focuses on Electric Flux, 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.

Question 1

1. $\Phi_{\text{curved}} = \pi R^2 E_0$, 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. (correct answer)
2. $\Phi_{\text{curved}} = 2\pi R^2 E_0$, because integrating $E_0 \cos\theta$ over the full hemispherical surface area $2\pi R^2$ yields a factor that, after integration from $\theta = 0$ to $\pi/2$, gives twice the flat-base area times the field strength.
3. $\Phi_{\text{curved}} = 0$, because the hemisphere is an open surface with no enclosed charge, and Gauss's law requires the flux through any surface in a charge-free region to be zero regardless of the field configuration.
4. $\Phi_{\text{curved}} = -\pi R^2 E_0$, because the outward normal on the curved surface has a component that opposes $\hat{z}$ on the sides of the hemisphere, and a careful integration shows the net contribution is equal in magnitude but opposite in sign to the flux through the flat base.

Question 2

1. $\Phi = \pi a^2 E_0$, because the field ranges from $0$ at the center to $E_0$ at the rim; taking the simple arithmetic mean of these boundary values gives an average field of $E_0/2$, but since the field is quadratic rather than linear, the correct average is $E_0$, and multiplying by the disk area $\pi a^2$ yields this result.
2. $\Phi = \frac{\pi a^2 E_0}{2}$, because integrating $E_0(r/a)^2$ over the disk using the area element $dA = 2\pi r\, dr$ gives $\int_0^a E_0\frac{r^2}{a^2} \cdot 2\pi r\, dr = \frac{2\pi E_0}{a^2}\cdot\frac{a^4}{4} = \frac{\pi a^2 E_0}{2}$. (correct answer)
3. $\Phi = \frac{2\pi a^2 E_0}{3}$, because integrating the field over the disk requires only $\int_0^a E_0(r/a)^2\, dr = E_0 a/3$, and multiplying by the circumference $2\pi a$ gives $2\pi a^2 E_0/3$.
4. $\Phi = \pi a^2 E_0 / 4$, because the quadratic field profile weights the central low-field region more heavily than the high-field rim in a circular geometry; the area-weighted average of $(r/a)^2$ over a disk works out to $1/4$, so the effective flux is $E_0/4$ times the disk area.

Question 3

1. $\Phi = \frac{Q}{24\varepsilon_0}$, because by symmetry the charge $+Q$ at the corner is shared equally among 8 identical cubes that tile around that corner, giving a total flux of $Q/(8\varepsilon_0)$ through this cube, which is then divided equally among all 6 faces.
2. $\Phi = \frac{Q}{48\varepsilon_0}$, because the charge at the corner contributes $Q/(8\varepsilon_0)$ total flux through the cube, and although the flux is not distributed uniformly, the three faces adjacent to the charge each subtend a smaller solid angle as seen from $+Q$ than the opposite faces do, with each adjacent face receiving $Q/(48\varepsilon_0)$.
3. $\Phi = \frac{Q}{8\varepsilon_0}$, because the charge $+Q$ at the corner is shared among 8 surrounding cubes, so the total flux through this cube is $Q/(8\varepsilon_0)$; since the charge sits at the corner of each adjacent face, all of this flux exits through the three adjacent faces rather than the three opposite ones.
4. $\Phi = 0$, 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. (correct answer)

Question 4

1. $\Phi = \frac{\lambda a^2}{2\pi \varepsilon_0 d}$, because the electric field at distance $d$ from the line charge has magnitude $\lambda/(2\pi\varepsilon_0 d)$, and multiplying by the surface area $a^2$ gives the flux through the square.
2. $\Phi = \frac{\lambda a}{2\pi \varepsilon_0}\ln\!\left(\frac{d+a/2}{d-a/2}\right)$, because the flux must be computed by integrating the radial field component across the surface, and the $x$-variation of the field over the width $a$ produces a logarithmic result.
3. $\Phi = \frac{\lambda a^2}{2\pi \varepsilon_0 d}$, because by Gauss's law, any surface near a line charge intercepts a fraction of the total flux proportional to the area of the surface divided by the distance, regardless of the surface orientation.
4. $\Phi = 0$, 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. (correct answer)

Question 5

1. $\Phi = \frac{(4 - \pi)\rho R^2 L}{\varepsilon_0}$, because the rectangular box encloses both the cylindrical charge (volume $\pi R^2 L$) and the uncharged corner regions (volume $(4-\pi)R^2 L$), and the net flux accounts for the charge redistribution at the boundaries between the charged and uncharged regions.
2. $\Phi = \frac{4\rho R^2 L}{\varepsilon_0}$, because the rectangular Gaussian surface encloses a square cross-section of area $(2R)^2 = 4R^2$ times length $L$, and Gauss's law uses the volume of the Gaussian surface to determine the enclosed charge.
3. $\Phi = \frac{\rho \pi R^2 L}{\varepsilon_0}$, 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. (correct answer)
4. $\Phi = \frac{2\rho R^2 L}{\varepsilon_0}$, because the circular cross-section of the charge distribution does not fill the square cross-section of the Gaussian surface; the flux must be corrected by the ratio of the inscribed circle area to the square area, which gives an effective factor of $\pi/4 \approx 0.785$, approximately $1/2$ after rounding.

Question 6

1. $\Phi = \frac{-Q}{\varepsilon_0}$, because the only charge that contributes to the flux is the central $-Q$ charge; the induced charge on the inner surface of the shell lies at $r = R_1$, which is outside this Gaussian surface.
2. $\Phi = 0$, 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. (correct answer)
3. $\Phi = \frac{2Q}{\varepsilon_0}$, because the net charge enclosed by the Gaussian surface includes $-Q$ at the center plus the $+3Q$ net charge on the shell, giving a total of $+2Q$ enclosed, and the flux follows from Gauss's law.
4. $\Phi = \frac{-Q}{\varepsilon_0}$, because the conducting shell redistributes its charge so that $+Q$ appears on the inner surface and $+2Q$ on the outer surface; since the outer surface charge at $R_2 > r$ is not enclosed, only the central $-Q$ contributes, giving $-Q/\varepsilon_0$.

Question 7

1. $\Phi = 4\pi A(4R^2 - R^2)/(R^2 \cdot 4R^2) = 3\pi A/R^2$, because the net flux is determined by the difference in field strengths at the two boundaries multiplied by a geometric mean of their surface areas.
2. $\Phi = 4\pi A$, because the flux through the outer sphere of radius $2R$ is $A/(2R)^2 \cdot 4\pi(2R)^2 = 4\pi A$, while the flux through the inner sphere is $A/R^2 \cdot 4\pi R^2 = 4\pi A$, but since both faces have outward normals in the same radial direction, their contributions add.
3. $\Phi = 0$, 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. (correct answer)
4. $\Phi = 4\pi A$, because even though no charge is enclosed between the shells, the $1/r^2$ dependence of the field means the outward flux through the outer sphere exceeds the inward flux entering through the inner sphere by $4\pi A$, reflecting the difference in their surface areas.

Question 8

1. $\Phi = \frac{Q}{2\varepsilon_0}$, because the two charges together have a net charge of $+Q/2$, and Gauss's law uses the total charge in the problem rather than just the charge inside the surface.
2. $\Phi = \frac{Q}{\varepsilon_0}$, 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. (correct answer)
3. $\Phi = \frac{3Q}{4\varepsilon_0}$, because the external $-Q/2$ charge at distance $2R$ subtends a solid angle that allows roughly one-half of its inward flux to penetrate the shell, reducing the total outward flux by $Q/(4\varepsilon_0)$.
4. $\Phi = 0$, because the field lines from $+Q$ that travel toward the $-Q/2$ charge terminate on it without crossing the shell surface on the far side, while field lines on the near side re-enter the shell, causing complete cancellation of outward and inward contributions.

Question 9

1. $\Phi = E_0 L^2$, 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. (correct answer)
2. $\Phi = 2E_0 L^2$, because both the $\hat{x}$-varying component and the $\hat{z}$ component each contribute $E_0 L^2$ to the net outward flux through their respective pairs of faces.
3. $\Phi = 0$, because the divergence of $\vec{E}$ is zero everywhere inside the cube, implying by Gauss's law that no net charge is enclosed and thus no net flux passes through the surface.
4. $\Phi = E_0 L^2 / 2$, because the average value of the $x$-component of the field over the two perpendicular faces reduces the net contribution by a factor of one-half compared to a uniform field of magnitude $E_0$.