📝 Inverse square vector fields (16 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 16 questions available

←Previous Topic:Vector field notation and components📝 Inverse square vector fieldsGradient vector fields:Next Topic→

What is Inverse square vector fields?

Inverse square vector fields:
Inverse square fields vary as $1/r^2$ from a source, given by $\mathbf{F}(\mathbf{r}) = c \frac{\mathbf{r}}{|\mathbf{r}|^3}$ in 3D, where $\mathbf{r} = \langle x,y,z \rangle$ .

Example:
Gravitational field $\mathbf{F} = -GmM \frac{\mathbf{r}}{r^3}$ attracts masses toward the origin.

Reason:
These fields model fundamental forces (gravity, electrostatics) and have zero divergence except at the source, satisfying Gauss's law.

Easy

Medium

Hard

📝 All Inverse square vector fields MCQs

Q1. A field has magnitude $F(r)=12/r^2$ . If a particle moves from $r=2$ to $r=6$ , what happens to the field magnitude?

A.It becomes one-third as large

B.It becomes one-ninth as large ✅

C.It becomes three times as large

D.It becomes nine times as large

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Because the field follows an inverse-square relationship, its magnitude is proportional to $1/r^2$ . Increasing the distance by a factor of $3$ therefore decreases the magnitude by $3^2=9$ . Thus the new magnitude is $1/9$ of the original magnitude.

Q2. Which statement best distinguishes an inverse-square field from a field that decreases linearly with distance?

A.Doubling distance always halves the inverse-square field

B.Doubling distance reduces an inverse-square field to one-fourth its original value ✅

C.The inverse-square field is constant along every radial direction

D.The inverse-square field increases as distance increases

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For an inverse-square field, $F\propto 1/r^2$ . Therefore replacing $r$ by $2r$ gives $F'=F/4$ , not $F/2$ . A linear decrease would follow a different mathematical relationship, so confusing these behaviors is a common modeling error.

Q3. Two sensors are placed at distances $r$ and $2r$ from the same isolated source. The first sensor measures field magnitude $E$ . Assuming the source is unchanged, what should the second sensor measure?

2E

E/2

E/4

✅

E/8

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The field magnitude is proportional to $1/r^2$ . At the second sensor the distance is doubled, so the denominator becomes $(2r)^2=4r^2$ . Consequently, the second measurement is $E/4$ . The result follows from the scaling law rather than direct numerical substitution.

Q4. A student claims that moving from $r=3$ to $r=9$ causes an inverse-square field to become three times weaker because the distance becomes three times larger. What is the flaw?

A.The field should become three times stronger

B.The field should become six times weaker

C.The field should become nine times weaker ✅

D.The field should remain unchanged

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The student correctly notices that the distance increases by a factor of $3$ , but incorrectly applies a linear scaling rule. Because the field depends on $1/r^2$ , the factor must also be squared. Therefore the field becomes $1/9$ of its original value, meaning it is nine times weaker.

Q5. A spherical surface is centered on a point source. If the radius of the surface is doubled, which combination correctly describes the change in surface area and field magnitude?

A.Area doubles and field doubles

B.Area becomes four times larger and field becomes one-fourth as large ✅

C.Area becomes twice as large and field becomes one-fourth as large

D.Area becomes four times larger and field remains constant

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The surface area of a sphere is proportional to $r^2$ , so doubling the radius makes its area four times larger. The inverse-square field simultaneously decreases by a factor of four. This linked behavior explains why total outward influence can remain consistent across spherical surfaces.

Q6. A radial field has magnitude $F(r)=K/r^2$ . A researcher doubles the source strength represented by $K$ while moving the observation point to twice its original distance. What happens to the measured field?

A.It doubles

B.It becomes four times larger

C.It remains unchanged

D.It becomes half as large ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The source-strength parameter changes the field by a factor of $2$ , while doubling the distance changes the field by a factor of $1/4$ . Combining both effects gives $2\times1/4=1/2$ . Therefore the measured field becomes half its original magnitude.

Q7. A field produced by a point source is measured as $20$ units at $r=5$ . A model predicts $5$ units at $r=10$ . A second model predicts $10$ units. Which model is consistent with inverse-square behavior?

A.The first model only ✅

B.The second model only

C.Both models

D.Neither model

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Doubling the distance from $5$ to $10$ must reduce an inverse-square field by a factor of $4$ . Thus $20$ units should become $5$ units. The prediction of $10$ units corresponds to inverse-first-power behavior instead, so only the first model is consistent.

Q8. A graph of field magnitude versus distance starts very large near the source and decreases rapidly, then gradually flattens as distance increases. Which mathematical model best matches this qualitative graph?

F(r)=Kr

F(r)=K/r

F(r)=K/r^2

✅

F(r)=K-r^2

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: An inverse-square graph falls steeply near small $r$ and becomes progressively flatter as $r$ increases. The model $K/r^2$ has exactly this behavior. A $K/r$ curve also decreases, but less rapidly, while the other choices do not reproduce the observed positive decaying shape.

Q9. A graph shows two radial field curves from the same source. Curve A is consistently four times Curve B at every displayed distance. What is the most reasonable conclusion if both curves follow inverse-square behavior?

A.Their distances differ by a factor of four

B.Their source strengths differ by a factor of four ✅

C.Their distances differ by a factor of two

D.Their source strengths differ by a factor of sixteen

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For an inverse-square field, $F=K/r^2$ . If two measurements are made at the same distance and one field is four times the other, their proportionality constants must differ by a factor of four. A distance factor of two would instead produce a field ratio of four only when comparing different observation distances.

Q10. A student uses $F=K/r^2$ but forgets that $r$ represents distance from the source and substitutes the horizontal coordinate $x$ for $r$ . Why can this produce an incorrect field map?

A.Because

x

is always negative

B.Because radial distance generally depends on both

x

and

y

in a plane ✅

C.Because inverse-square fields cannot be represented graphically

D.Because

r

must always equal

x+y

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a point source at the origin, radial distance in a plane is $r=\sqrt{x^2+y^2}$ , not simply $x$ . Replacing $r$ with one coordinate ignores points that have different distances but identical $x$ -coordinates. This can distort both field magnitude and spatial symmetry.

Q11. At a location, two identical sources produce field contributions in opposite directions along the same line. If their magnitudes at that location are equal, what is the resultant field?

A.It is twice either contribution

B.It is equal to one contribution

C.It is zero ✅

D.It depends only on the distance squared

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Vector fields must be combined using both magnitude and direction. Equal vectors pointing in opposite directions cancel exactly, giving a resultant of zero. Simply adding their magnitudes would ignore direction and is therefore inappropriate for this situation.

Q12. A simulation shows that when distance changes from $r$ to $3r$ , the displayed field decreases by a factor of $9$ . A programmer argues that the simulation is wrong because the distance increased only threefold. Which response is correct?

A.The programmer is correct because the field must decrease threefold

B.The simulation is correct because the distance factor is squared ✅

C.The simulation is wrong because inverse-square fields decrease by sixfold

D.The programmer is correct because the field should increase threefold

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: An inverse-square law squares the distance scaling factor. Replacing $r$ by $3r$ changes $1/r^2$ into $1/(3r)^2=1/(9r^2)$ . Therefore the field becomes one-ninth of its original magnitude, so a ninefold decrease is exactly what the model predicts.

Q13. A field magnitude is modeled by $F(r)=K/r^2$ . An engineer wants the field to be no more than $1/16$ of its current value without changing $K$ . By what factor should the distance be increased?

A.2

B.4 ✅

C.8

D.16

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: We require $F'/F=1/16$ . Since $F'/F=(r/r')^2$ , we need $(r/r')^2=1/16$ . Taking the positive square root gives $r'/r=4$ . Therefore the distance must be increased by a factor of four, not sixteen.

Q14. A field diagram contains arrows pointing radially outward. Near the source the arrows are long and widely separated farther away. Which interpretation is most consistent with an inverse-square field?

A.The field direction changes randomly with distance

B.The decreasing arrow length reflects decreasing field magnitude while radial symmetry reflects direction ✅

C.The increasing spacing proves that the field magnitude increases

D.The arrows must all have identical magnitudes because they point outward

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: In a radial inverse-square field, direction is determined by the line connecting the source and observation point, while magnitude decreases as $1/r^2$ . A graphical representation therefore commonly uses shorter arrows farther from the source. The changing arrow length communicates magnitude, not a change in radial direction.

Q15. Two sources have strengths $K$ and $4K$ . At distances $r$ and $2r$ , respectively, their inverse-square field magnitudes are compared. What is the ratio of the first field to the second field?

A.0.04236111111111107 ✅

B.0.043055555555555625

C.0.08402777777777781

D.0.16736111111111107

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The first magnitude is $K/r^2$ . The second is $4K/(2r)^2=4K/(4r^2)=K/r^2$ . Thus the two magnitudes are equal, giving a ratio of $1:1$ . This illustrates how source strength and distance effects can exactly compensate each other.

Q16. Consider several identical sources arranged symmetrically around a point. Their individual fields at the center have equal magnitudes and directions that cancel in pairs. What can be concluded about the net field at the center?

A.It must equal the field of one source

B.It must be zero because the vector contributions cancel ✅

C.It must be four times the field of one source

D.It must depend only on the inverse-square distance law

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The inverse-square law determines each individual contribution, but the net field requires vector addition. In a symmetric arrangement, equal contributions can occur in opposite directions and cancel pairwise. When every contribution has a matching opposite vector, the resultant field at the center is exactly zero.

🔗 Related Topics (MCQs)

📝 Vector Fields in Calculus MCQs 📝 Graphical Representations of Vector Fields MCQs 📝 Vector field notation and components MCQs 📝 Gradient vector fields MCQs 📝 Conservative vector fields and potential functions MCQs

←Previous Topic:Vector field notation and components📝 Inverse square vector fieldsGradient vector fields:Next Topic→