Physics 2 Quiz: Common Eandm Pitfalls

What this quiz covers

This quiz focuses on Common Eandm Pitfalls, 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. The first student is correct. Because the battery has been disconnected, charge Q is constant, and inserting the dielectric increases both capacitance and voltage proportionally, raising the energy by a factor of $\kappa$.
2. Neither student is correct. After the battery is disconnected and the dielectric inserted, the voltage across the capacitor remains $V_0$ because charge is conserved, so the energy $U = \frac{1}{2}CV^2$ does increase by $\kappa$ as the first student claims, but only if the capacitor is reconnected to the battery first.
3. The second student is correct, but the energy decreases by a factor of $\kappa^2$ rather than $\kappa$, because both the capacitance and the voltage change simultaneously when the dielectric is inserted with the battery disconnected.
4. The first student's formula is right but the wrong version was used. With constant charge, the energy is $U = Q^2/(2C)$; as C increases by $\kappa$, the energy decreases by a factor of $\kappa$. The second student is correct, and the energy becomes $U_0/\kappa$. (correct answer)

Question 2

1. The original student is correct. When traversing a battery from − to + terminal (gaining potential), the term is positive; when traversing a resistor in the direction of current flow (losing potential), the term is negative. Both objections are wrong. (correct answer)
2. The first classmate is correct. A battery always represents a voltage loss for the traversing loop, so its contribution to KVL is always negative regardless of traversal direction through the battery.
3. The second classmate is correct. Because current flows through the resistors in the traversal direction, the resistors are adding energy to the circuit, so their KVL contributions should be positive, making the equation $+12 + IR_1 + IR_2 = 0$.
4. Both classmates are partially correct: the battery sign depends on orientation relative to traversal, giving $+12$, but resistors traversed in the current direction also contribute positive terms, yielding $+12 + IR_1 + IR_2 = 0$.

Question 3

1. The Gaussian surface shape is wrong: an infinite plane has spherical symmetry, so a spherical Gaussian surface must be used rather than a cylinder. Using the correct surface eliminates the $2\pi$ factor and yields $E = \sigma/\epsilon_0$.
2. The analogy fails because Gauss's law requires a spherical surface whenever the source is planar. Using a cylindrical surface incorrectly retains the factor of $2\pi r$ from the line-charge geometry in the denominator, preventing the correct cancellation that would give the plane-charge result.
3. The student incorrectly evaluated the flux through the cylindrical surface. Because the field is perpendicular to the plane, it is parallel to the curved side of the cylinder (zero flux there) and perpendicular to only the two flat end-caps. The total flux is $2EA$, not $2\pi r L \cdot E$, so Gauss's law gives $E = \sigma/(2\epsilon_0)$, not $\sigma/(2\pi\epsilon_0)$. (correct answer)
4. The error lies in the enclosed charge: for an infinite plane, only the charge on one side of the Gaussian surface should be counted, so the enclosed charge is $\sigma A/2$ rather than $\sigma A$. Correcting this and summing both end-caps gives $E = \sigma/(4\epsilon_0)$.

Question 4

1. The student's setup and result are both correct. In the Biot–Savart law, $\hat{r}$ points from the source to the field point ($+\hat{y}$ here), the cross product $\hat{x} \times \hat{y} = +\hat{z}$ is correct by the right-hand rule, and wrapping the right hand around the wire confirms the field is in $+\hat{z}$ above the wire. (correct answer)
2. The student identified $\hat{r}$ correctly as $+\hat{y}$, but the cross product $\hat{x} \times \hat{y}$ actually equals $-\hat{z}$ by the right-hand rule in a standard right-handed coordinate system, so the field at P points in $-\hat{z}$, not $+\hat{z}$.
3. The cross product result $\hat{x} \times \hat{y} = +\hat{z}$ is correct, but the student has the direction of $\hat{r}$ reversed. In the Biot–Savart law, $\hat{r}$ points from the field point back to the source element, so $\hat{r} = -\hat{y}$. This gives $\hat{x} \times (-\hat{y}) = -\hat{z}$, meaning the field actually points in $-\hat{z}$.
4. Both the direction of $\hat{r}$ and the cross product are wrong. The Biot–Savart $\hat{r}$ points from P to the source ($-\hat{y}$), and $\hat{x} \times (-\hat{y}) = +\hat{z}$ by the right-hand rule, so the field is in $+\hat{z}$ — the right answer arrived at through two compensating errors.

Question 5

1. The classmate is correct. The resistor current must be computed as $I = (\mathcal{E} - V_C)/R$, which at $t=0$ gives $\mathcal{E}/R$ and decays exponentially, whereas differentiating $V_C$ artificially introduces a factor of C that distorts the units and the physical meaning.
2. The student's method contains a conceptual error because $I = C(dV_C/dt)$ applies only when the capacitor is discharging; during charging, the correct relation is $I = -C(dV_C/dt)$, which introduces a sign flip and changes the direction of the current in the final expression.
3. The student's method is valid in principle, but there is a sign error: differentiating $(1 - e^{-t/RC})$ gives $+e^{-t/RC}/RC$, so the current should be $I = \mathcal{E}/(RC^2)\,e^{-t/RC}$, which has incorrect units of A/F rather than amperes.
4. The student's method is valid and the result is correct. The relation $I = C(dV_C/dt)$ is the fundamental capacitor current–voltage relationship, and differentiating $\mathcal{E}(1-e^{-t/RC})$ correctly yields $I = (\mathcal{E}/R)e^{-t/RC}$, which agrees with $I = (\mathcal{E} - V_C)/R$ at all times. (correct answer)

Question 6

1. There is no error; in a parallel combination, the voltage across C₃ equals the voltage across the series branch C₁₂, so substituting that voltage is correct and gives the right energy for C₃.
2. The student has confused series and parallel voltage rules: in a parallel combination, every branch shares the same terminal voltage V, so $V_3 = V$, not the voltage across the series branch. Using a reduced voltage underestimates $U_3$. (correct answer)
3. The student has the topology backward: C₃ is actually in series with the parallel combination of C₁ and C₂, so the correct voltage across C₃ is determined by the charge-division rule, not by setting $V_3 = V$.
4. The error is in the equivalent-capacitance formula: adding C₁₂ and C₃ in parallel should use the reciprocal sum $\frac{1}{C_{eq}} = \frac{1}{C_{12}} + \frac{1}{C_3}$, which then propagates to give an incorrect $V_3$.

Question 7

1. The current-divider formula is applied incorrectly: for resistors in parallel the correct form is $I_1 = I_{tot}\frac{R_1}{R_1+R_2}$, not $\frac{R_2}{R_1+R_2}$. Using this formula gives $I_1 = 2\,\text{A}$ and $P_1 = 16\,\text{W}$.
2. The reasoning contains no errors. The current-divider formula $I_1 = I_{tot}\frac{R_2}{R_1+R_2}$ correctly reflects that the branch with smaller resistance carries more current, yielding $I_1 = 6\,\text{A}$ and $P_1 = 144\,\text{W}$. (correct answer)
3. The equivalent resistance formula is wrong: parallel resistors obey $R_{eq} = R_1 + R_2 = 16\,\Omega$, not the product-over-sum formula, so the total current and all subsequent calculations are incorrect.
4. The power formula $P_1 = I_1^2 R_1$ is invalid for parallel circuits. Because every branch shares the same voltage, only $P_1 = V^2/R_1$ may be used; a current-based formula is inapplicable here.

Question 8

1. The original student is correct: when $\theta = \omega t$ is the angle the normal makes with $\vec{B}$, the flux is $\Phi = B_0 A \sin(\omega t)$ because the component of $\vec{B}$ along $\hat{n}$ is $B_0\sin\theta$. Faraday's law then correctly yields a negative cosine EMF.
2. The second classmate is correct: the original flux $\Phi = B_0 A\sin(\omega t)$ is right, but Faraday's law should not include the negative sign for an externally driven rotating loop, because the minus sign applies only to loops responding to a changing external field, not to mechanically rotated loops. The EMF is therefore $+B_0 A\omega\cos(\omega t)$.
3. The third student is correct: $\Phi = \vec{B}\cdot\hat{n}\,A = B_0 A\cos(\omega t)$, since the dot product depends on $\cos\theta$, not $\sin\theta$. At $t = 0$ the loop lies in the xy-plane so $\hat{n} = \hat{z}$ and flux is at its maximum — consistent only with the cosine formula. Differentiating gives $\mathcal{E} = B_0 A\omega\sin(\omega t)$. (correct answer)
4. The original student is correct about the flux but should have used $\hat{n} = -\hat{z}$ as the outward normal at $t = 0$, which introduces a compensating sign change. With this convention, $\Phi = -B_0 A\sin(\omega t)$ and Faraday's law gives $\mathcal{E} = +B_0 A\omega\cos(\omega t)$, agreeing with the second classmate.