Physics 2 Quiz: Coulombs Law

What this quiz covers

This quiz focuses on Coulombs Law, 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. $\frac{F_0}{8}$, because after contact the charge on each sphere is $+Q/2$, and after grounding one sphere to zero, the remaining charge of $Q/2$ on the other sphere interacts with zero charge, giving a force reduced by the factor of $1/2 \times 1/2 = 1/4$ relative to the original $Q/2$ state, so $F_0/4 \times 1/2 = F_0/8$.
2. $\frac{F_0}{4}$, because after contact each sphere holds $+Q/2$, and after one sphere is grounded its charge becomes zero, so the force is proportional to $(Q/2)(0) = 0$; but residual induction creates a force of $F_0/4$ due to image-charge effects on the grounded sphere.
3. $0$, because one sphere carries no charge after grounding, and by Coulomb's law the force between a charged object and an uncharged object is identically zero for point charges, regardless of the charge on the other sphere. (correct answer)
4. $\frac{F_0}{2}$, because only one of the two original charges $+Q$ is removed by grounding while the other retains its post-contact charge of $+Q/2$, and Coulomb's law scales linearly with each charge, yielding half the force of the post-contact configuration.

Question 2

1. $0.20\,\text{m}$ from $q_1$, placing $q_3$ between the two charges, where the $1/r^2$ dependence compensates for the 4:1 charge magnitude ratio. (correct answer)
2. $0.30\,\text{m}$ from $q_1$, placing $q_3$ at the midpoint between the charges, where equal distances from both charges combine with the charge ratio to produce zero net force.
3. $0.60\,\text{m}$ from $q_1$, placing $q_3$ on the far side of $q_2$, where the forces from the two opposite-sign charges are directed oppositely and can balance despite the charge magnitude difference.
4. $0.20\,\text{m}$ from $q_1$ on the far side of $q_1$ from $q_2$, placing $q_3$ in the exterior region where the $1/r^2$ dependence can offset the charge magnitude ratio for opposite-sign charges.

Question 3

1. $\frac{kq^2}{a^2}\left(2\sqrt{2} - \frac{1}{2}\right)$, because the two nearest-neighbor attractive forces (each of magnitude $kq^2/a^2$) add directly along the diagonal to give $2\sqrt{2}\,kq^2/a^2$, and the repulsive diagonal force of $kq^2/(2a^2)$ is then subtracted.
2. $\frac{kq^2}{a^2}\left(\sqrt{2} + \frac{1}{2}\right)$, because the two side-neighbor attractive forces have a vector resultant of $\sqrt{2}\,kq^2/a^2$ directed toward the opposite corner, and the repulsive diagonal force of $kq^2/(2a^2)$ acts in the same direction, so the two contributions add.
3. $\frac{kq^2}{a^2}\left(\sqrt{2} - \frac{1}{2}\right)$, because the two adjacent $-q$ charges each exert an attractive force of magnitude $kq^2/a^2$ along the sides; their vector resultant points diagonally with magnitude $\sqrt{2}\,kq^2/a^2$, and the repulsive force from the $+q$ at the opposite corner has magnitude $kq^2/(2a^2)$ and opposes this resultant. (correct answer)
4. $\frac{kq^2}{a^2}\left(1 - \frac{1}{2}\right) = \frac{kq^2}{2a^2}$, because each side-neighbor force contributes a diagonal component of $kq^2/a^2 \cdot \cos 45° = kq^2/(\sqrt{2}\,a^2)$, these two components sum to $\sqrt{2}\,kq^2/a^2$, and the repulsive diagonal force equals $kq^2/(2a^2)$, so the net result simplifies to $kq^2/(2a^2)$.

Question 4

1. $F$, because the doubling of $q_B$ and the addition of $q_C$ are exactly offset by the factor-of-2 reduction from the increased separation, and the net effect returns the force to its original value.
2. $2F$, because the total charge at the new location is $3q_B$ (doubled $q_B$ plus original $q_C = q_B$), giving a factor of 3 in the numerator, while the distance factor $(\sqrt{2})^2 = 2$ goes in the denominator, yielding $3F/2$... but rounding and the discrete nature of the problem makes this approximately $2F$.
3. $\frac{3\sqrt{2}}{2}F$, because the force must be computed separately for $2q_B$ and $q_C$ at the new location and then added using the superposition principle, with the $\sqrt{2}$ factor entering through the vector nature of the distance scaling.
4. $\frac{3F}{2}$, because the total charge at the new location is $2q_B + q_B = 3q_B$, the new distance is $r\sqrt{2}$ so the distance-squared factor becomes $2r^2$, and thus the new force is $\frac{k q_A (3q_B)}{2r^2} = \frac{3}{2} \cdot \frac{kq_Aq_B}{r^2} = \frac{3F}{2}$. (correct answer)

Question 5

1. Magnitude $F_0$, attractive, because tripling both charges gives a factor of $9$ in the numerator, tripling the distance gives a factor of $9$ in the denominator, and the sign reversal changes the interaction from repulsive to attractive with no effect on magnitude. (correct answer)
2. Magnitude $F_0$, repulsive, because the magnitude calculation gives the same factor of $9/9 = 1$, but the sign reversal of one charge does not affect the force magnitude or character since Coulomb's law is symmetric in the charge signs for magnitude calculations.
3. Magnitude $3F_0$, attractive, because tripling each charge triples the force independently for each charge (a total factor of $6$ increase), tripling the distance reduces force by $9$, giving a net factor of $6/9 = 2/3$... and with the sign reversal the force becomes attractive with magnitude $3F_0$.
4. Magnitude $9F_0$, attractive, because tripling each charge multiplies the force by $3 \times 3 = 9$ while tripling the distance multiplies by $1/9$, so the magnitudes cancel; however, the sign reversal converts the original repulsive force to attractive and introduces an additional factor of 9 from the change in field orientation.

Question 6

1. The particle at $(d, 0)$ has a greater initial acceleration because it lies along the x-axis directly toward $+Q$, whereas the particle at $(0, d)$ is displaced perpendicularly, so its Coulomb force has a smaller effective component along its direction of motion.
2. The particle at $(0, d)$ has a greater initial acceleration because it lies along the y-axis, and its Coulomb force toward the origin is therefore entirely along one coordinate direction, making it more efficient at accelerating the particle than the force on the particle at $(d, 0)$.
3. The initial accelerations differ by a factor of $\sqrt{2}$, because the Coulomb force on the particle at $(0, d)$ must be resolved into components along both coordinate axes while the force on the particle at $(d, 0)$ acts purely along one axis, introducing a geometric factor of $\sqrt{2}$ in the net acceleration.
4. Both particles have identical initial accelerations of magnitude $\frac{kQq}{md^2}$, because both are at the same distance $d$ from $+Q$, carry the same charge magnitude $q$, and have the same mass $m$, so Coulomb's law gives the same force magnitude and Newton's second law gives the same acceleration magnitude. (correct answer)

Question 7

1. $4F_0$, because removing the dielectric multiplies the force by $\kappa = 4$, and halving the distance multiplies it by $4$ as well, giving a net factor of $16$; but the two effects partially offset each other to yield $4F_0$.
2. $16F_0$, because the dielectric constant of 4 suppressed the original force by a factor of 4, and halving the separation increases the force by a factor of 4 via the inverse-square law, so the two multiplicative factors of 4 combine to give a total factor of 16. (correct answer)
3. $8F_0$, because the dielectric suppresses force by $\kappa$, so removing it multiplies the force by 4; halving the distance multiplies by $2^2 = 4$; but one must subtract the correction for the change in medium polarization, yielding a net factor of 8.
4. $2F_0$, because the inverse-square law gives a factor of $4$ for halving the distance, but this gain is partially offset by the factor of $\kappa = 4$ that is lost when moving to vacuum, and the net change is only a doubling of $F_0$.