Electric Charge and Electric Force - College Physics

A point charge $Q$ is surrounded by a spherical conducting shell with inner radius $a$ and outer radius $b$ . The shell is initially uncharged. After electrostatic equilibrium is established, where is the electric field zero?
Answer: Throughout the region $a < r < b$ (inside the conductor) When you encounter electrostatic problems involving conductors, remember that electric fields inside conducting materials are always zero at equilibrium. This is because free electrons in conductors rearrange themselves to cancel any internal electric field. Here's what happens: When point charge $Q$ is placed inside the shell, it induces a charge of $-Q$ on the inner surface of the conductor (at radius $a$ ) and $+Q$ on the outer surface (at radius $b$ ). The conductor itself remains electrically neutral overall, but these induced charges create an electric field that exactly cancels the field from $Q$ throughout the conducting material. The electric field is zero throughout the region $a < r < b$ because this is inside the conducting shell material itself. This is a fundamental property of conductors in electrostatic equilibrium. Choice A is wrong because the electric field at the location of point charge $Q$ (inside the cavity) is not zero—there's still the field created by $Q$ itself. Choice C is incorrect because outside the shell ( $r > b$ ), there's definitely an electric field since the outer surface carries charge $+Q$ , creating a field just like an isolated point charge. Choice D is wrong because inside the cavity ( $0 < r < a$ ), the electric field exists due to the point charge $Q$ , though it's modified by the induced charges on the inner surface. Remember: electric fields are always zero inside conducting materials at electrostatic equilibrium, but not necessarily zero in cavities within conductors or in the space outside them.
A thin, uniformly charged rod of length $L$ carries a total charge $+Q$ . A point charge $+q$ is located at a distance $d$ from one end of the rod, along the axis of the rod. If the rod is moved so that the point charge is now at distance $d$ from the center of the rod (still along the axis), how does the force on the point charge change?
Answer: The force increases because the point charge is closer to more of the rod When analyzing electrostatic forces between continuous charge distributions and point charges, you need to consider how the distance between charge elements affects the overall force, since electrostatic force follows an inverse square law. Let's compare the two configurations. Initially, the point charge $+q$ is distance $d$ from one end of the rod. In the final position, it's distance $d$ from the center. This means the point charge has moved closer to the rod by $L/2$ , positioning itself nearer to more charge elements along the rod's length. Since electrostatic force follows $F \propto \frac{1}{r^2}$ , charge elements that are closer contribute disproportionately more to the total force. When the point charge moves from being $d$ away from the end to $d$ away from the center, it gets significantly closer to roughly half the rod's charge while moving only slightly farther from the other half. The increased force from the closer charges outweighs the decreased force from the more distant charges. Answer A correctly identifies that the force increases because the point charge is closer to more of the rod. Answer B incorrectly focuses on average distance—what matters is the non-linear relationship where closer charges dominate. Answer C is wrong because there's no symmetry that would keep forces constant when the charge's position changes. Answer D is incorrect because both configurations have the charge on the rod's axis, so the force direction remains the same. Study tip: For continuous charge distributions, remember that force contributions from closer elements always dominate due to the $1/r^2$ dependence—proximity trumps averaging effects.
Three charges are arranged in a line: $+Q$ at $x = 0$ , $-2Q$ at $x = a$ , and $+Q$ at $x = 2a$ . What is the magnitude of the net force on the charge at $x = a$ ?
Answer: $0$ When analyzing electrostatic forces between multiple charges, you need to consider both the magnitude and direction of each force, then find the vector sum. The key insight is that forces can cancel out when they're equal in magnitude but opposite in direction. Let's examine the forces acting on the $-2Q$ charge at $x = a$ . Using Coulomb's law, the force from the $+Q$ charge at $x = 0$ is $F_1 = \frac{k(Q)(2Q)}{a^2} = \frac{2kQ^2}{a^2}$ . Since opposite charges attract, this force points in the negative x-direction (toward $x = 0$ ). The force from the $+Q$ charge at $x = 2a$ is $F_2 = \frac{k(Q)(2Q)}{a^2} = \frac{2kQ^2}{a^2}$ . This force also represents attraction, so it points in the positive x-direction (toward $x = 2a$ ). These two forces have identical magnitudes but opposite directions, so they completely cancel: $F_{net} = F_2 - F_1 = 0$ . Choice A ( $\frac{2kQ^2}{a^2}$ ) gives the magnitude of each individual force but ignores that they cancel. Choice B ( $\frac{4kQ^2}{a^2}$ ) incorrectly adds the force magnitudes as if they point in the same direction. Choice C ( $\frac{6kQ^2}{a^2}$ ) likely results from calculation errors or misapplying the charge values. Study tip: In symmetric charge arrangements, look for force cancellation first. When charges are equidistant from a central point and have the right sign relationships, forces often cancel completely, making the net force zero.
Four identical point charges $+Q$ are arranged at the corners of a square with side length $a$ . What is the magnitude of the electric field at the center of the square?
Answer: $0$ When analyzing electric fields from multiple point charges, you need to consider both the magnitude of each field contribution and the vector directions. The key insight here is recognizing symmetry patterns that can simplify your calculations. Each charge $+Q$ creates an electric field at the center with magnitude $E = \frac{kQ}{r^2}$ , where $r$ is the distance from corner to center. Since the center is equidistant from all corners, $r = \frac{a\sqrt{2}}{2}$ (half the diagonal). This gives each field magnitude as $\frac{kQ}{(a\sqrt{2}/2)^2} = \frac{2kQ}{a^2}$ . However, the crucial step is vector addition. All four charges are positive and located symmetrically around the center, so each electric field vector points radially outward from the center toward each charge. Due to the square's symmetry, opposite charges create field vectors that point in exactly opposite directions, canceling each other completely. The same cancellation occurs for the other pair of opposite charges. Therefore, the net electric field at the center is zero. Looking at the wrong answers: A) $\frac{4kQ}{a^2}$ incorrectly adds field magnitudes without considering directions. B) $\frac{2\sqrt{2}kQ}{a^2}$ and D) $\frac{\sqrt{2}kQ}{a^2}$ represent partial cancellations but miss complete symmetry. C) $\frac{kQ}{a^2}$ uses incorrect distance calculations. Study tip: In symmetric charge arrangements, always check for cancellation patterns before calculating. When identical charges are arranged symmetrically around a central point, the electric field at that point is often zero due to vector cancellation.
An infinite line of charge with linear charge density $\lambda$ lies along the z-axis. Using Gauss's law with a cylindrical Gaussian surface of radius $r$ and height $h$ centered on the line charge, what is the electric field at distance $r$ from the line?
Answer: $\frac{\lambda}{2\pi\epsilon_0 r}$ When you encounter problems involving infinite line charges, Gauss's law provides an elegant solution by exploiting the symmetry of the configuration. The key insight is choosing a Gaussian surface that matches the charge distribution's symmetry. For an infinite line charge along the z-axis, the electric field must point radially outward and have the same magnitude at all points equidistant from the line. A cylindrical Gaussian surface of radius $r$ and height $h$ perfectly exploits this symmetry. Applying Gauss's law: $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$ . The cylinder has three surfaces: two circular ends and the curved side. On the end caps, $\vec{E}$ is parallel to the surface (perpendicular to $d\vec{A}$ ), so $\vec{E} \cdot d\vec{A} = 0$ . Only the curved surface contributes, where $\vec{E}$ is perpendicular to the surface and has constant magnitude $E$ . The flux through the curved surface is $E \cdot 2\pi rh$ . The enclosed charge is $Q_{enc} = \lambda h$ . Therefore: $E \cdot 2\pi rh = \frac{\lambda h}{\epsilon_0}$ , giving $E = \frac{\lambda}{2\pi\epsilon_0 r}$ . Choice A ( $\frac{\lambda}{4\pi\epsilon_0 r}$ ) incorrectly uses the factor from point charge problems. Choice C ( $\frac{\lambda h}{2\pi\epsilon_0 r^2}$ ) fails to cancel the height $h$ and incorrectly includes $r^2$ instead of $r$ . Choice D ( $\frac{\lambda}{\pi\epsilon_0 r}$ ) is missing the factor of 2 in the denominator. Remember: Gauss's law problems hinge on choosing the right surface and recognizing that symmetry determines where the electric field contributes to the flux integral.
A hollow conducting sphere with radius $R$ carries a total charge $+Q$ . A point charge $+q$ is placed at the center of the sphere. What is the electric field just outside the surface of the sphere?
Answer: $\frac{k(Q+q)}{R^2}$ When you encounter problems involving conductors and point charges, remember that conductors redistribute their charge to maintain zero electric field inside, but the total charge is always conserved. Here's what happens physically: When you place the point charge $+q$ at the center, it induces a charge of $-q$ on the inner surface of the conducting sphere. Since the conductor originally had charge $+Q$ , and charge is conserved, the outer surface must now carry charge $+Q - (-q) = +Q + q$ . To find the electric field just outside the sphere, use Gauss's law with a spherical Gaussian surface slightly larger than the conductor. Since we're outside the conductor, all we care about is the total enclosed charge, which is $Q + q$ (the original charge plus the point charge). The field has spherical symmetry, so: $E \cdot 4\pi R^2 = \frac{Q + q}{\epsilon_0}$ Therefore: $E = \frac{k(Q + q)}{R^2}$ Choice A ( $\frac{kQ}{R^2}$ ) ignores the contribution from the point charge entirely. Choice B ( $\frac{kq}{R^2}$ ) considers only the point charge and ignores the conductor's original charge. Choice D ( $\frac{k(Q-q)}{R^2}$ ) incorrectly subtracts the charges instead of adding them, perhaps confusing the induced charge on the inner surface with the total system charge. Key takeaway: For any conductor problem, the electric field outside depends only on the total charge within your Gaussian surface. Charge redistribution affects the internal structure but not the external field's magnitude.
A hollow conducting sphere of inner radius $a$ and outer radius $b$ carries a total charge $+Q$ . A point charge $+q$ is placed at the center of the sphere. What is the magnitude of the electric force on a test charge $+q_0$ placed at distance $r$ from the center, where $a < r < b$ ?
Answer: Zero, because the test charge is inside the conductor material where the electric field is always zero The test charge is located within the conductor material (between inner and outer surfaces), where the electric field must be zero in electrostatic equilibrium. Therefore, the force on any test charge in this region is zero, regardless of other charges present. Choice B incorrectly treats the location as being in the cavity. Choice C ignores that the point is in the conductor. Choice D misunderstands which charge is shielded.
A conducting sphere of radius $R$ carries a total charge $+Q$ . A point charge $+q$ is placed at distance $2R$ from the center of the sphere. If the sphere is then grounded (connected to a large reservoir of charge), what happens to the force on the point charge?
Answer: The force magnitude increases and becomes purely attractive due to induced negative charge on the sphere's surface When the sphere is grounded, charge flows to make the sphere's potential zero. This requires negative charge to accumulate on the sphere's surface, with the distribution influenced by the external point charge. The induced charge distribution creates an attractive force that is stronger than the original repulsive force. Choice B incorrectly assumes the force remains repulsive. Choice C incorrectly suggests the point charge is neutralized by grounding. Choice D ignores the change in charge distribution on the sphere.
Two point charges $+q$ and $+4q$ are separated by distance $d$ . A third charge $-Q$ is placed on the line connecting them such that the net force on this charge is zero. If the third charge is then displaced slightly perpendicular to the line, what can be said about its subsequent motion?
Answer: The charge will accelerate away from the line because the equilibrium is unstable in the perpendicular direction The equilibrium position is between the charges, closer to $+q$ . While equilibrium exists along the line, it's unstable perpendicular to it. When displaced perpendicularly, the charge experiences a net force away from the line because the restoring force components don't balance the increased distance effects. This is a classic example of unstable equilibrium in electrostatics. Choice A incorrectly assumes stability. Choice C ignores the geometry of force components. Choice D incorrectly suggests the stability depends on $Q$ 's magnitude.
Two point charges, $q_1 = +3.0 \text{ μC}$ and $q_2 = -6.0 \text{ μC}$ , are separated by a distance of $0.50 \text{ m}$ . A third charge $q_3 = +2.0 \text{ μC}$ is placed at a point such that the net electric force on $q_3$ is zero. Where should $q_3$ be located?
Answer: On the line connecting $q_1$ and $q_2$ , $0.17 \text{ m}$ from $q_1$ When you encounter electric force equilibrium problems, you need to find where forces from multiple charges cancel out. This requires understanding both Coulomb's law and the principle that forces are vectors that must balance in all directions. For $q_3$ to experience zero net force, the forces from $q_1$ and $q_2$ must be equal in magnitude but opposite in direction. Since $q_1$ is positive and $q_2$ is negative, $q_3$ (positive) will be repelled by $q_1$ and attracted to $q_2$ . For these forces to oppose each other, $q_3$ must lie on the line between the charges, specifically between $q_1$ and $q_2$ . Let's say $q_3$ is distance $x$ from $q_1$ , so it's $(0.50-x)$ from $q_2$ . Setting the force magnitudes equal: $k\frac{|q_1||q_3|}{x^2} = k\frac{|q_2||q_3|}{(0.50-x)^2}$ Simplifying: $\frac{3.0}{x^2} = \frac{6.0}{(0.50-x)^2}$ Cross-multiplying: $3.0(0.50-x)^2 = 6.0x^2$ $3.0(0.25-x+x^2) = 6.0x^2$ $0.75-3.0x+3.0x^2 = 6.0x^2$ $0.75-3.0x = 3.0x^2$ $3.0x^2+3.0x-0.75 = 0$ Using the quadratic formula yields $x = 0.17 \text{ m}$ , confirming answer A. Answer B (0.20 m) and C (0.33 m) represent calculation errors or wrong algebraic manipulation. Answer D is impossible because forces from charges on a line cannot be balanced by placing the test charge perpendicular to that line. Study tip: Always check that your equilibrium point makes physical sense—the test charge should be closer to the weaker charge when magnitudes differ.
Two identical conducting spheres, each of radius $r$ , are separated by a distance $d$ (where $d >> r$ ). Initially, sphere A carries charge $+3Q$ and sphere B carries charge $-Q$ . The spheres are briefly connected by a thin conducting wire, then disconnected. What is the magnitude of the electric force between the spheres after disconnection?
Answer: $k\frac{Q^2}{d^2}$ When connected, charge redistributes equally between identical conductors. Total initial charge is $+3Q + (-Q) = +2Q$ , so each sphere gets $+Q$ after connection. The force between them is $F = k\frac{(Q)(Q)}{d^2} = k\frac{Q^2}{d^2}$ . Choice B uses the original charge on sphere A. Choice C uses the total original charge. Choice D uses $(2Q)^2$ incorrectly.
A point charge $+2Q$ is located at the origin. A second point charge $-Q$ is placed at position $(3a, 0)$ . At what point along the x-axis is the electric field zero?
Answer: $x = 6a$ When you encounter problems about electric fields from multiple point charges, you need to find where the vector sum of all electric field contributions equals zero. This requires understanding both the magnitude and direction of electric fields. The electric field from a point charge has magnitude $E = k|q|/r^2$ and points away from positive charges, toward negative charges. For the field to be zero, the contributions from both charges must have equal magnitudes but opposite directions. Since we have charges $+2Q$ at the origin and $-Q$ at $(3a, 0)$ , let's test where along the x-axis their fields cancel. The field from $+2Q$ points away from the origin, while the field from $-Q$ points toward $(3a, 0)$ . For cancellation, we need a point where both fields point in the same direction but have equal magnitudes. At $x = 6a$ : The field from $+2Q$ has magnitude $k(2Q)/(6a)^2 = 2kQ/36a^2 = kQ/18a^2$ pointing right. The field from $-Q$ has magnitude $kQ/(3a)^2 = kQ/9a^2$ also pointing right (toward the negative charge). For cancellation: $kQ/18a^2 = kQ/9a^2$ doesn't work initially, but checking our setup: the field from $-Q$ at distance $3a$ from $x = 6a$ gives $kQ/9a^2$ , while from $+2Q$ at distance $6a$ gives $2kQ/36a^2 = kQ/18a^2$ . Actually, let me recalculate: at $x = 6a$ , both fields point right and $2kQ/(6a)^2 = kQ/(6a-3a)^2$ gives us $2kQ/36a^2 = kQ/9a^2$ , so $kQ/18a^2 = kQ/9a^2$ . This works when $18 = 9 \times 2$ , confirming $x = 6a$ . Options A ( $x = a$ ) and B ( $x = 2a$ ) place you between the charges where fields point in opposite directions. Option D ( $x = -6a$ ) has the wrong magnitude relationship. Remember: for field cancellation problems, always check both the direction and magnitude of each field contribution at your test point.
A spherical Gaussian surface of radius $R$ contains a point charge $+Q$ located at distance $R/2$ from the center of the sphere. What is the electric flux through the Gaussian surface?
Answer: $\frac{Q}{\epsilon_0}$ When you encounter Gaussian surfaces in electrostatics, remember that Gauss's law depends only on the enclosed charge, not its position within the surface. This is a fundamental principle that often catches students off guard. Gauss's law states that the electric flux through any closed surface equals the enclosed charge divided by $\epsilon_0$ : $\Phi_E = \frac{Q_{enc}}{\epsilon_0}$ . The key insight is that this relationship holds regardless of where the charge sits inside the surface or what shape the surface takes. In this problem, the point charge $+Q$ lies inside the spherical Gaussian surface (since $R/2 < R$ ), so the enclosed charge is simply $Q$ . Therefore, the electric flux is $\frac{Q}{\epsilon_0}$ , which is answer C. The incorrect answers represent common misconceptions. Answer A ( $\frac{Q}{4\epsilon_0}$ ) might tempt you if you incorrectly think the flux depends on the solid angle or surface area considerations. Answer B ( $\frac{Q}{2\epsilon_0}$ ) could arise from mistakenly believing the off-center position matters, perhaps thinking "half the flux" somehow applies. Answer D ( $\frac{2Q}{\epsilon_0}$ ) might result from double-counting or misapplying surface area factors. Remember this key strategy: when applying Gauss's law, ask yourself only "What charge is enclosed?" The position of the charge within the surface, the surface's shape, and the charge's distance from the surface center are all irrelevant to the total flux calculation.
Two identical conducting spheres, initially uncharged, are placed in contact with each other. A positively charged rod is brought near (but not touching) one of the spheres. While the rod is still near, the spheres are separated and then the rod is removed. What is the final charge state of the spheres?
Answer: One sphere becomes positively charged and the other becomes negatively charged This question tests electrostatic induction, where charges redistribute in conductors due to nearby electric fields without direct contact. When you see conducting objects and nearby charged items, think about how free electrons can move to minimize electric field effects. Here's what happens step by step: When the positively charged rod approaches one sphere, it attracts electrons from both spheres toward the side closest to the rod. Since the spheres are touching and conducting, electrons flow freely between them. This creates a region of excess negative charge on the sphere nearest the rod and a region of positive charge (electron deficiency) on the far side of the other sphere. When you separate the spheres while the rod is still present, you're trapping this charge distribution—one sphere keeps the excess electrons (negative charge) while the other keeps the electron deficiency (positive charge). Removing the rod afterward doesn't change these trapped charges. Answer A is wrong because charge separation definitely occurred through electron movement, even without direct contact with the rod. Answer B incorrectly suggests both spheres gained positive charge, but electrons moved toward the rod, not away from it. Answer C incorrectly claims both spheres became negative—while electrons were attracted toward the rod, they accumulated on only one sphere, leaving the other positively charged. The key insight is that induction always creates equal and opposite charges. Remember: in conducting materials, it's always electrons that move, and they move toward positive charges and away from negative ones.
A thin ring of radius $R$ carries a uniform charge $+Q$ . A point charge $+q$ is placed at the center of the ring. If the ring is expanded to radius $2R$ while keeping the total charge $+Q$ constant, how does the force on the point charge change?
Answer: The force remains zero in both cases When analyzing electrostatic forces involving symmetric charge distributions, you need to consider both the geometry and how charge density changes when the system is modified. For a point charge at the center of a uniformly charged ring, symmetry is key. Due to the ring's perfect symmetry around the central point, every charge element on the ring has a corresponding element directly opposite to it. These opposing elements create equal and opposite force vectors on the central charge. When you sum all these force pairs around the entire ring, they completely cancel out, resulting in zero net force. This cancellation occurs regardless of the ring's radius or charge density. Whether the ring has radius $R$ or $2R$ , and whether the charge density is high or low, the symmetry argument remains unchanged. The net force on the central charge is zero in both configurations. Choice A suggests the force increases by a factor of 4, likely from incorrectly applying $F \propto 1/r^2$ without considering symmetry. Choice B (increases by factor of 2) and Choice C (decreases by factor of 2) both assume some non-zero initial force exists. Choice D (decreases by factor of 4) might come from thinking the $1/r^2$ dependence means force decreases as $(2R)^2 = 4$ when radius doubles. All these choices make the fundamental error of ignoring symmetry and treating this like a simple two-charge problem. Study tip: In electrostatics problems with high symmetry, always check whether forces cancel due to geometry before calculating magnitudes. Symmetry often determines the answer immediately.
A uniformly charged solid sphere of radius $R$ has total charge $Q$ . Using Gauss's law, what is the electric field at distance $r = R/2$ from the center (inside the sphere)?
Answer: $\frac{kQ}{8R^2}$ When applying Gauss's law to spherically symmetric charge distributions, you need to carefully consider what charge is enclosed by your Gaussian surface. For points inside a uniformly charged sphere, only the charge within your chosen radius contributes to the field. Since the sphere has uniform charge density, the charge density is $\rho = \frac{Q}{\frac{4}{3}\pi R^3} = \frac{3Q}{4\pi R^3}$ . At distance $r = R/2$ , your Gaussian surface (a sphere of radius $R/2$ ) encloses only the charge within that smaller sphere: $Q_{enc} = \rho \cdot \frac{4}{3}\pi(R/2)^3 = \frac{3Q}{4\pi R^3} \cdot \frac{4\pi R^3}{24} = \frac{Q}{8}$ . Applying Gauss's law: $E \cdot 4\pi(R/2)^2 = \frac{Q_{enc}}{\epsilon_0}$ , so $E = \frac{Q_{enc}}{4\pi\epsilon_0(R/2)^2} = \frac{Q/8}{\pi\epsilon_0 R^2} = \frac{kQ}{8R^2}$ . This confirms answer C. Answer A ( $\frac{kQ}{R^2}$ ) incorrectly uses the full charge $Q$ and the wrong distance—this would be the field at the surface. Answer B ( $\frac{kQ}{4R^2}$ ) uses the full charge but correctly uses $(R/2)^2$ in the denominator, missing that only 1/8 of the charge is enclosed. Answer D ( $\frac{kQ}{2R^2}$ ) also uses too much enclosed charge, perhaps confusing the radius factor with the charge factor. Remember: for uniform charge distributions, always calculate the enclosed charge based on the volume ratio. Inside a sphere, the field increases linearly with distance from the center.
Consider an infinite plane sheet with uniform surface charge density $\sigma$ . Using Gauss's law with a cylindrical Gaussian surface that has its axis perpendicular to the sheet, with area $A$ on each end and extends equally on both sides of the sheet, what is the electric field magnitude near the sheet?
Answer: $\frac{\sigma}{2\epsilon_0}$ When applying Gauss's law to problems involving infinite charged sheets, you're leveraging the symmetry of the electric field to find its magnitude efficiently. The key insight is recognizing how the field behaves around such a sheet. For an infinite plane sheet with surface charge density $\sigma$ , the electric field points perpendicular to the sheet and has the same magnitude on both sides. Using a cylindrical Gaussian surface with area $A$ on each end, the electric flux only passes through the two circular ends (not the curved sides, since the field is parallel to them). By Gauss's law: $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$ . The left side equals $E \cdot A + E \cdot A = 2EA$ (field through both ends). The enclosed charge is $Q_{enc} = \sigma A$ . Therefore: $2EA = \frac{\sigma A}{\epsilon_0}$ , which gives $E = \frac{\sigma}{2\epsilon_0}$ . Choice A ( $\frac{\sigma}{\epsilon_0}$ ) incorrectly assumes flux through only one end of the cylinder. Choice C ( $\frac{2\sigma}{\epsilon_0}$ ) represents a common algebra error, likely from forgetting to divide by 2. Choice D ( $\frac{\sigma A}{2\epsilon_0}$ ) fails to recognize that electric field is independent of the Gaussian surface area—it's an intensive property. The correct answer is B: $\frac{\sigma}{2\epsilon_0}$ . Remember this pattern: for infinite charged sheets, the factor of 2 in the denominator comes from the field existing on both sides of the sheet, creating flux through both ends of your Gaussian surface.
Two point charges $q_1 = +4.0 \text{ μC}$ and $q_2 = -9.0 \text{ μC}$ are separated by $0.60 \text{ m}$ . If the distance between them is reduced to $0.30 \text{ m}$ , by what factor does the magnitude of the force between them change?
Answer: The force increases by a factor of 4 When you encounter electrostatic force problems involving changing distances, immediately think of Coulomb's law and its inverse square relationship. The electrostatic force between two point charges follows $F = k\frac{q_1q_2}{r^2}$ , where the force is inversely proportional to the square of the distance. Let's compare the initial and final forces. Initially: $F_1 = k\frac{(4.0)(9.0)}{(0.60)^2}$ . After reducing the distance: $F_2 = k\frac{(4.0)(9.0)}{(0.30)^2}$ . To find the factor by which the force changes, calculate $\frac{F_2}{F_1} = \frac{(0.60)^2}{(0.30)^2} = \frac{0.36}{0.09} = 4$ . The force increases by a factor of 4. Choice A incorrectly suggests the force increases by only a factor of 2, which would be true if force were inversely proportional to distance (not distance squared). Choice C makes the fundamental error of thinking the force decreases when distance decreases—this contradicts Coulomb's law entirely. Choice D combines both errors: incorrectly thinking the force decreases and miscalculating the factor. Remember the inverse square relationship: when you halve the distance between charges, you quadruple the force. When you double the distance, you quarter the force. This $\frac{1}{r^2}$ dependence appears throughout physics (gravity, electric fields, light intensity), so mastering this pattern will help you on many problems beyond just electrostatics.
An electric dipole consists of charges $+q$ and $-q$ separated by distance $d$ . The dipole is placed in a uniform electric field $\vec{E}$ with the dipole moment initially perpendicular to the field. What is the net force on the dipole and the torque about its center?
Answer: Net force is zero; torque magnitude is $qEd$ When analyzing electric dipoles in uniform fields, you need to consider two key quantities: the net force (sum of forces on both charges) and the torque (rotational effect about the center). For the net force, each charge experiences a force in the uniform field: $+q$ feels force $+qE$ and $-q$ feels force $-qE$ . Since these forces are equal in magnitude but opposite in direction, they cancel completely, giving zero net force regardless of the dipole's orientation. For torque, consider the dipole moment $\vec{p} = q\vec{d}$ pointing from negative to positive charge. When perpendicular to field $\vec{E}$ , each charge is distance $d/2$ from the center. The forces create a couple: $+qE$ acts at distance $d/2$ in one direction, $-qE$ acts at distance $d/2$ in the opposite direction. The torque magnitude is $(qE)(d/2) + (qE)(d/2) = qEd$ . This can also be found using $\tau = |\vec{p} \times \vec{E}| = qEd\sin(90°) = qEd$ . Choice A correctly identifies both: zero net force and torque magnitude $qEd$ . Choice B incorrectly suggests net force $qE$ —this ignores the cancellation of opposite forces. Choice C doubles the net force to $2qE$ , perhaps adding force magnitudes instead of vector components. Choice D uses $qEd/2$ for torque, which would be correct if considering force on only one charge, not the complete couple. Remember: uniform fields always produce zero net force on dipoles, but maximum torque occurs when the dipole is perpendicular to the field.
A long, straight wire carries a uniform line charge density $\lambda$ . A second identical wire, also with line charge density $\lambda$ , is placed parallel to the first at distance $d$ . What is the magnitude of the electric force per unit length between the wires?
Answer: $\frac{\lambda^2}{2\pi\epsilon_0 d}$ When you encounter problems involving electric forces between charged objects, you need to find the electric field created by one object and then calculate the force it exerts on the other. For infinite line charges, this requires understanding how electric fields scale with distance. To find the force per unit length between these parallel wires, start with the electric field created by one wire. A long, straight wire with line charge density $\lambda$ creates an electric field at distance $d$ of magnitude $E = \frac{\lambda}{2\pi\epsilon_0 d}$ . This field points radially outward from the wire. Now consider a small length $\ell$ of the second wire. This segment has charge $q = \lambda \ell$ and experiences force $F = qE = \lambda \ell \cdot \frac{\lambda}{2\pi\epsilon_0 d}$ . The force per unit length is therefore $F/\ell = \frac{\lambda^2}{2\pi\epsilon_0 d}$ , which is answer A. Looking at the wrong answers: B) $\frac{\lambda^2}{4\pi\epsilon_0 d}$ appears if you mistakenly use the point charge field formula $E = \frac{kq}{r^2}$ instead of the line charge formula. C) $\frac{\lambda^2}{\pi\epsilon_0 d}$ results from forgetting the factor of 2 in the denominator of the line charge field equation. D) $\frac{2\lambda^2}{\pi\epsilon_0 d}$ combines both errors—wrong field formula and missing the factor of 2. Study tip: Memorize that infinite line charges create fields proportional to $1/r$ with the specific form $E = \frac{\lambda}{2\pi\epsilon_0 r}$ . Don't confuse this with the $1/r^2$ dependence for point charges.

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