Physics 2 Quiz: Units And Sign Conventions In Eandm

What this quiz covers

This quiz focuses on Units And Sign Conventions In Eandm, 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 integral setup contains a sign error: the limits of integration should run from $a$ to $b$, not $b$ to $a$, to match the displacement direction. With corrected limits, the result is $\frac{\lambda}{2\pi\epsilon_0}\ln(b/a)$, which is still positive for $b > a$, so the final answer and physical reasoning are correct despite the sign error in setup.
2. The setup, evaluation, and reasoning are all correct. The formula $V(a) - V(b) = -\int_b^a E_s \, ds$ is valid because reversing the integration limits introduces a compensating sign flip, yielding $+\frac{\lambda}{2\pi\epsilon_0}\ln(b/a) > 0$, confirming that $a$ (closer to the line) is at higher potential. (correct answer)
3. The evaluation is wrong because the correct antiderivative of $1/s$ is $\ln|s| + C$, which must be evaluated as $[\ln s]_b^a = \ln(a) - \ln(b) = \ln(a/b)$, giving a negative result that contradicts the stated sign. The student's final answer has the wrong sign.
4. The setup is wrong because $V(a) - V(b) = -\int_a^b \mathbf{E} \cdot d\mathbf{l}$, not $-\int_b^a$. The corrected integral gives $-\frac{\lambda}{2\pi\epsilon_0}\ln(b/a) < 0$, implying $V(a) < V(b)$, which is inconsistent with a positive line charge, revealing an error in the physical premise.

Question 2

1. The flux calculation and zero-EMF result are correct, and the student's final claim is also correct: a negative flux that becomes more negative as $B_0$ increases gives $d\Phi_B/dt < 0$, so $\mathcal{E} = -d\Phi_B/dt > 0$ regardless of the rate of change.
2. The flux calculation is correct and the EMF for constant $B$ is correctly zero, but the student's area vector choice is physically invalid. By convention, the area vector of a loop in the $xy$-plane must point in the $+z$ direction, so choosing $-\hat{z}$ violates the standard sign convention for flux.
3. The flux calculation is wrong because magnetic flux must always be positive. The correct flux is $+1.57 \times 10^{-3} \text{ Wb}$, and the resulting EMF from Faraday's law would be negative for an increasing $B_0$.
4. The flux calculation and the zero-EMF result for constant $B$ are both correct. However, the student's final claim is wrong: the sign of the induced EMF depends on whether $B_0$ increases or decreases, not on the sign of $\Phi_B$ alone. The student's statement is only valid for increasing $B_0$. (correct answer)

Question 3

1. Both the unit analysis and the interpretation are correct. The units of $k q_1 q_2 / r$ reduce to joules, and a negative potential energy means an attractive configuration requires positive external work equal to $|U|$ to separate the charges. (correct answer)
2. The unit analysis is correct, but the interpretation is wrong. A negative $U$ means the system is in a repulsive configuration, so the external agent receives energy rather than supplies it when separating the charges.
3. The unit analysis is correct, but the interpretation is wrong. A negative $U$ means the system is bound, so energy must be removed from the system—not added—to separate the charges, meaning the external agent does negative work.
4. The unit analysis is wrong because $\mu\text{C}$ values must be converted to nanocoulombs before the formula applies, and the interpretation is wrong because the sign of $U$ alone cannot determine the direction of work without knowing the sign of the force.

Question 4

1. The cross product is wrong: $\hat{x} \times \hat{z} = -\hat{y}$ is correct, but the final force is $-0.24\hat{y}\text{ N}$, not $+0.24\hat{y}\text{ N}$. The student made a sign error by applying the negative charge's sign twice—once in the magnitude factor and once in the 'reversal' claim.
2. The cross product is correct ($\hat{x} \times \hat{z} = -\hat{y}$), the sign handling is correct (negative $q$ reverses direction), and the final result $+0.24\hat{y}\text{ N}$ is correct. The student's full calculation contains no errors. (correct answer)
3. The cross product is wrong: $\hat{x} \times \hat{z} = +\hat{y}$, not $-\hat{y}$. With the correct cross product, the force on the negative charge becomes $-0.24\hat{y}\text{ N}$, pointing in the $-\hat{y}$ direction.
4. The cross product is wrong: $\hat{x} \times \hat{z} = +\hat{y}$, not $-\hat{y}$. With the correct cross product and the sign of the negative charge, the force is $+0.24\hat{y}\text{ N}$, which agrees with the student's final answer even though the intermediate step was wrong.

Question 5

1. The loop equation is correct, $I = 1 \text{ A}$ is correct, and the terminal-voltage claim is correct: conventional current entering a resistor from one side means that side is at higher potential, consistent with $V = IR = 8 \text{ V}$. (correct answer)
2. The loop equation is correct and $I = 1 \text{ A}$ is correct, but the terminal-voltage claim is wrong: the side where current exits a resistor is at higher potential, so the right terminal of $R_2$ is at higher potential.
3. The loop equation has a sign error because the EMF term should be negative when traversed in the direction of conventional current inside the source, giving $I = -1 \text{ A}$, which means the actual current is counterclockwise and the terminal-voltage claim is reversed.
4. The loop equation is correct and $I = 1 \text{ A}$ is correct, but the terminal-voltage claim is ambiguous without knowing the internal resistance of the source, because the terminal voltage of $R_2$ depends on the voltage drop across the source's internal resistance as well.

Question 6

1. The first student is correct. Since $4\pi$ is dimensionless, it does not contribute units, so $\mu_0$ itself carries T·m/A. Dimensional analysis confirms: $(\text{T}\cdot\text{m/A})(\text{A})(\text{m})/(\text{m}^2) = \text{T}$. (correct answer)
2. The second student is correct. Because $4\pi$ appears in the denominator, dimensional analysis must treat $\mu_0/(4\pi)$ as a single combined constant; assigning units to $\mu_0$ alone is physically meaningless in SI.
3. Neither student is correct. The Biot–Savart formula yields units of Wb/m (webers per meter), not tesla, because the cross product $d\mathbf{l} \times \hat{r}$ introduces an extra factor of meters that prevents the result from reducing to tesla.
4. The first student is correct in the unit assignment but wrong about the equivalence: the SI unit T·m/A cannot be rewritten as N/A² or H/m, because those alternative forms apply only in Gaussian units and introduce a factor of $4\pi$ when converted to SI.

Question 7

1. Her analysis is wrong in the intermediate step: $\text{C}^2/(\text{N}\cdot\text{m}^2)$ times $\text{N}\cdot\text{m/(C}\cdot\text{m}\cdot\text{s)}$ yields $\text{C/(N}\cdot\text{m}\cdot\text{s)}$, not $\text{C/(m}^2\cdot\text{s)}$, because the Newton units do not fully cancel, and the result is not dimensionally equivalent to current density.
2. Her analysis is wrong because she should use $[\epsilon_0] = \text{F/m}$ and $[E] = \text{V/m}$, giving $\frac{\text{F}}{\text{m}} \cdot \frac{\text{V}}{\text{m}\cdot\text{s}} = \frac{\text{F}\cdot\text{V}}{\text{m}^2\cdot\text{s}} = \frac{\text{C}}{\text{m}^2\cdot\text{s}}$, which simplifies to $\text{A/m}^2$. Her method gives the right answer but uses a non-standard form of $[\epsilon_0]$ that introduces an error of a factor of $4\pi$ in SI.
3. Her analysis is wrong because she identified the final unit as $\text{A/m}^2$ when the correct unit of current density is $\text{A/m}$ (amperes per meter), which describes surface current density, not volume current density, and the displacement current is inherently a surface phenomenon.
4. Her analysis is correct. Substituting $\text{V} = \text{N}\cdot\text{m/C}$ gives $\frac{\text{C}^2}{\text{N}\cdot\text{m}^2} \cdot \frac{\text{N}\cdot\text{m}}{\text{C}\cdot\text{m}\cdot\text{s}} = \frac{\text{C}}{\text{m}^2 \cdot \text{s}} = \text{A/m}^2$, which is indeed the SI unit of current density. (correct answer)

Question 8

1. The student correctly identifies that $U$ increases with $d$ and correctly concludes the force is attractive, since a system naturally moves to minimize energy, and attraction would decrease $d$, reducing $U$. The error is only in stating that the 'electric field does work'—it is the external agent, not the field, that transfers energy.
2. The student's energy formula is correct, but the conclusion that $U$ increases with $d$ is wrong at fixed $Q$. Because $C$ decreases with $d$ and $U = Q^2/(2C)$, increasing $d$ actually decreases $U$, so the electric force must be repulsive to push the plates apart and do negative work on the system.
3. The student's energy formula and the conclusion that $U$ increases with $d$ are correct, but the sign reasoning is inverted. If an external agent must do positive work to separate the plates against an attractive force, the energy input from the external agent accounts for the increase in $U$. The electric force is attractive, not repulsive, consistent with opposite charges on the plates. (correct answer)
4. The student's energy formula contains a unit error: $Q^2 d/(2\epsilon_0 A)$ has units of $\text{C}^2 \cdot \text{m} / (\text{C}^2/(\text{N}\cdot\text{m}^2) \cdot \text{m}^2) = \text{N}\cdot\text{m}^3/\text{m} = \text{N}\cdot\text{m}^2$, not joules, so the entire analysis is dimensionally invalid before any physical reasoning is applied.

Question 9

1. The student applied the divergence operator incorrectly. The correct result is $[\nabla \cdot \mathbf{J}] = \text{A/m}^2$ (not A/m³), because $\nabla \cdot$ acts on the vector components without reducing the overall dimension by one power of length.
2. The student's error is in the divergence step: $\nabla \cdot$ does not divide by meters but by seconds, because spatial derivatives in Maxwell's equations carry implicit time dependence. Correcting this gives $[\nabla \cdot \mathbf{J}] = \text{C/(m}^2\cdot\text{s}^2)$, which still differs from $[\partial\rho/\partial t]$.
3. The student overlooks that $1 \text{ A} = 1 \text{ C/s}$ by definition, so $\text{A/m}^3 = \text{C/(m}^3\cdot\text{s)}$. Both sides carry identical dimensions, and the continuity equation is dimensionally consistent without any conversion factor. (correct answer)
4. The student's definition of $\rho$ is wrong. In the continuity equation, $\rho$ represents charge per unit volume per unit time, with units C/(m³·s), so $[\partial\rho/\partial t] = \text{C/(m}^3\cdot\text{s}^2)$. This corrected form makes both sides dimensionally consistent.