Capacitors - College Physics

A capacitor consists of two conducting spheres of radii $R_1$ and $R_2$ (where $R_2 > R_1$ ) separated by a large distance $d \gg R_2$ . The capacitance of this system can be approximated as two isolated spheres, each with capacitance $4\pi\epsilon_0 R$ . If both spheres initially have zero charge and are then connected to opposite terminals of a battery with voltage $V$ , what is the approximate charge on the smaller sphere?
Answer: $Q_1 = 4\pi\epsilon_0 V \frac{R_1 R_2}{R_1 + R_2}$ , using the equivalent capacitance formula for the series combination When you encounter capacitors with complex geometries, the key is identifying whether they're in series or parallel and applying the appropriate combination rules. Here, two isolated conducting spheres connected to opposite battery terminals form a series capacitor system. For capacitors in series, you use the reciprocal formula: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$ . With each sphere having capacitance $C_1 = 4\pi\epsilon_0 R_1$ and $C_2 = 4\pi\epsilon_0 R_2$ , this gives: $\frac{1}{C_{eq}} = \frac{1}{4\pi\epsilon_0 R_1} + \frac{1}{4\pi\epsilon_0 R_2} = \frac{1}{4\pi\epsilon_0}\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{1}{4\pi\epsilon_0}\left(\frac{R_1 + R_2}{R_1 R_2}\right)$ Therefore: $C_{eq} = 4\pi\epsilon_0 \frac{R_1 R_2}{R_1 + R_2}$ In series capacitors, all capacitors carry the same charge, so $Q_1 = C_{eq} \times V = 4\pi\epsilon_0 V \frac{R_1 R_2}{R_1 + R_2}$ . Answer A incorrectly assumes charge distributes proportionally to radius, which would apply to parallel capacitors, not series. Answer B has the wrong proportionality factor and misunderstands how sphere size affects charge distribution. Answer D incorrectly uses $R_2$ as the leading coefficient and applies faulty reasoning about "equal dependence" on both radii. Remember: when capacitors connect to opposite battery terminals, they're in series regardless of their physical arrangement. Always use the series combination formula and recognize that all series capacitors carry identical charge.
A capacitor is formed by two concentric conducting spheres with inner radius $a$ and outer radius $b$ . If the capacitance of this configuration is $C = 4\pi\epsilon_0\frac{ab}{b-a}$ , and the inner sphere is charged to potential $V_0$ while the outer sphere is grounded, what is the electric field at a distance $r$ where $a < r < b$ ?
Answer: $E = \frac{V_0 ab}{r^2(b-a)}$ , pointing radially outward when the inner sphere is positively charged When you encounter concentric spherical capacitors, you're dealing with spherical symmetry where the electric field points radially and depends only on distance from the center. The key insight is connecting the given capacitance formula to find the charge, then applying Gauss's law. Start by finding the charge on the inner sphere using $Q = CV_0 = 4\pi\epsilon_0\frac{ab}{b-a} \cdot V_0$ . Between the spheres, by Gauss's law, the electric field at distance $r$ is $E = \frac{Q}{4\pi\epsilon_0 r^2}$ . Substituting the charge: $E = \frac{4\pi\epsilon_0\frac{ab}{b-a} \cdot V_0}{4\pi\epsilon_0 r^2} = \frac{V_0 ab}{r^2(b-a)}$ . The field points radially outward when the inner sphere carries positive charge. Choice A incorrectly includes an extra factor of $\frac{ab}{4\pi\epsilon_0}$ and has the nonsensical term $\frac{b-a}{b-a}$ in the numerator, suggesting dimensional confusion. Choice C places $ab$ in the wrong position and has an incorrect $r^2$ dependence - it should be $\frac{ab}{r^2(b-a)}$ , not $\frac{ab}{r^2} \cdot \frac{1}{b-a}$ . Choice D has the capacitance terms inverted ( $\frac{b-a}{ab}$ instead of $\frac{ab}{b-a}$ ) and incorrectly states the field points inward, which would only be true for negative charge. Remember: for spherical capacitors, always use the given capacitance to find charge, then apply Gauss's law. The $\frac{1}{r^2}$ dependence is fundamental, but pay careful attention to how the geometric factors $a$ , $b$ , and $(b-a)$ are arranged.
A parallel-plate capacitor has plate area $A$ and separation $d$ . A dielectric slab of thickness $d/2$ , area $A$ , and dielectric constant $\kappa$ is inserted so that it fills exactly half the space between the plates. What is the capacitance of this configuration?
Answer: $C = \epsilon_0 A \frac{2\kappa}{d(\kappa + 1)}$ , accounting for the series combination of air and dielectric regions When a dielectric partially fills a parallel-plate capacitor, you must analyze the electric field configuration carefully. The dielectric slab and air gap create two distinct regions with different electric properties, but they share the same electric field strength due to boundary conditions. The key insight is recognizing this as a series combination of two capacitors. The dielectric region (thickness $d/2$ ) has capacitance $C_1 = \frac{\kappa \epsilon_0 A}{d/2} = \frac{2\kappa \epsilon_0 A}{d}$ , while the air gap (thickness $d/2$ ) has capacitance $C_2 = \frac{\epsilon_0 A}{d/2} = \frac{2\epsilon_0 A}{d}$ . For capacitors in series: $\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{2\kappa \epsilon_0 A} + \frac{d}{2\epsilon_0 A} = \frac{d(\kappa + 1)}{2\kappa \epsilon_0 A}$ Therefore: $C = \frac{2\kappa \epsilon_0 A}{d(\kappa + 1)}$ , confirming answer B. A incorrectly treats this as a simple averaging problem, ignoring the series nature of the configuration. C makes a conceptual error about "effective separation" - the physical separation doesn't change, but the electric field distribution does. D mistakenly assumes a parallel arrangement, which would apply if the dielectric covered the full gap but only part of the plate area. Study tip: When dielectrics partially fill capacitors, always identify whether you have series (different regions along field lines) or parallel (different regions across field lines) combinations. The geometry determines the circuit model you should use.
A parallel-plate capacitor with plate area $A$ and plate separation $d$ is connected to a battery of voltage $V$ . After the capacitor is fully charged, the battery is disconnected, and then the plate separation is doubled while keeping the plate area constant. What happens to the energy stored in the capacitor?
Answer: The energy is doubled because the capacitance is halved and the charge remains constant When analyzing capacitor problems, you need to identify what remains constant after changes are made. Here, the key insight is that once the battery is disconnected, the charge on the capacitor plates cannot change - it's trapped. Let's work through what happens step by step. Initially, the capacitance is $C_0 = \epsilon_0 A/d$ and the stored energy is $U_0 = \frac{1}{2}CV^2$ . When you double the plate separation to $2d$ , the new capacitance becomes $C = \epsilon_0 A/(2d) = C_0/2$ - half the original value. Since charge is conserved after disconnection, we use the energy formula $U = \frac{Q^2}{2C}$ . With the same charge $Q$ but half the capacitance, the new energy becomes $U = \frac{Q^2}{2(C_0/2)} = \frac{Q^2}{C_0} = 2U_0$ . The energy doubles. Let's examine why the other choices fail: Choice B incorrectly assumes voltage stays constant, but voltage actually increases when you separate the plates while keeping charge fixed. Choice C suggests energy stays the same, missing that the $Q^2/C$ relationship means energy increases when capacitance decreases at constant charge. Choice D gets the direction right but calculates the wrong factor - energy doubles, not quadruples. Study tip: In capacitor problems, always first determine what stays constant after the change (charge if battery disconnected, voltage if battery connected). Then choose the appropriate energy formula: $U = \frac{1}{2}CV^2$ for constant voltage or $U = \frac{Q^2}{2C}$ for constant charge.
A cylindrical capacitor consists of two coaxial conducting cylinders with inner radius $a$ , outer radius $b$ , and length $L$ . The capacitance per unit length is $\frac{2\pi\epsilon_0}{\ln(b/a)}$ . If the inner cylinder is charged to $+Q$ and the outer to $-Q$ , what is the energy density at radius $r$ where $a < r < b$ ?
Answer: $u = \frac{Q^2}{8\pi^2\epsilon_0 L^2 r^2}$ , independent of the outer radius and proportional to charge squared When analyzing energy density in electrostatic systems, you need to use the relationship $u = \frac{1}{2}\epsilon_0 E^2$ , where the electric field must be found first using the charge distribution and geometry. For a cylindrical capacitor, the electric field between the cylinders can be found using Gauss's law. The charge $+Q$ on the inner cylinder creates a radial electric field. Applying Gauss's law to a cylindrical surface of radius $r$ and length $L$ : $E \cdot 2\pi rL = \frac{Q}{\epsilon_0}$ , so $E = \frac{Q}{2\pi\epsilon_0 Lr}$ . The energy density is therefore: $u = \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2}\epsilon_0 \left(\frac{Q}{2\pi\epsilon_0 Lr}\right)^2 = \frac{Q^2}{8\pi^2\epsilon_0 L^2 r^2}$ This matches answer choice B, which correctly shows the energy density is independent of the outer radius $b$ and proportional to $Q^2$ . Answer choice A incorrectly includes $\ln(b/a)$ in the numerator and has an extra factor of $\epsilon_0$ in the denominator. Answer choice C also incorrectly includes the logarithmic term and is missing a factor of $\epsilon_0$ in the denominator. Answer choice D has $\epsilon_0$ in the numerator instead of the denominator, which would give incorrect units for energy density. Remember that energy density depends only on the local electric field strength, not on the overall geometry of the capacitor. The logarithmic terms appear in capacitance calculations but not in local field-dependent quantities like energy density.
A spherical conductor of radius $R$ is given charge $Q$ and then connected by a thin wire to an initially uncharged spherical conductor of radius $2R$ . After equilibrium is established, what fraction of the original charge resides on the larger sphere?
Answer: $\frac{2}{3}$ of the charge because charge distributes proportionally to the respective radius values When conducting spheres are connected, you're dealing with electrostatic equilibrium. The key insight is that connected conductors must have the same electric potential, and for spherical conductors, the potential depends on both charge and radius. For a spherical conductor, the potential is $V = k\frac{Q}{R}$ , where $k$ is Coulomb's constant, $Q$ is the charge, and $R$ is the radius. When the spheres are connected, they reach the same potential: $V_1 = V_2$ . Let $Q_1$ be the final charge on the smaller sphere (radius $R$ ) and $Q_2$ be the final charge on the larger sphere (radius $2R$ ). Since potentials are equal: $k\frac{Q_1}{R} = k\frac{Q_2}{2R}$ This simplifies to $\frac{Q_1}{R} = \frac{Q_2}{2R}$ , so $Q_2 = 2Q_1$ . Since charge is conserved: $Q_1 + Q_2 = Q$ , we get $Q_1 + 2Q_1 = Q$ , so $Q_1 = \frac{Q}{3}$ and $Q_2 = \frac{2Q}{3}$ . Choice A incorrectly assumes charge distributes inversely with radius. Choice B wrongly applies surface area proportionality—this would give $\frac{4}{5}$ since the larger sphere has 4 times the surface area. Choice C also incorrectly uses inverse surface area relationship. Choice D correctly recognizes that charge distributes proportionally to radius values: the larger sphere gets twice the charge because its radius is twice as large. Remember: for connected conducting spheres, equal potentials mean charge distributes proportionally to the radii, not surface areas or their inverses.
Two identical capacitors, each with capacitance $C$ , are initially uncharged. One capacitor is connected to a battery of voltage $V$ and fully charged, then disconnected. This charged capacitor is then connected in parallel with the second uncharged capacitor. What is the final energy stored in the system compared to the initial energy?
Answer: The final energy is $\frac{1}{2}$ of the initial energy because the voltage across each capacitor becomes $V/2$ When capacitors are connected or disconnected from circuits, you need to track both charge conservation and energy changes carefully. Energy is not always conserved in these processes because connecting capacitors involves current flow and resistance losses. Let's trace through this step-by-step. Initially, one capacitor is charged to voltage $V$ with charge $Q = CV$ and energy $E_i = \frac{1}{2}CV^2$ . When you connect this charged capacitor in parallel with an identical uncharged capacitor, charge redistributes until both capacitors reach the same voltage. Since charge is conserved, the total charge $CV$ splits equally between the two identical capacitors: each gets $Q/2 = CV/2$ . The voltage across each capacitor becomes $V_{final} = \frac{Q/2}{C} = \frac{V}{2}$ . The total final energy is $E_f = 2 \times \frac{1}{2}C(V/2)^2 = \frac{1}{4}CV^2 = \frac{1}{2}E_i$ . Choice A incorrectly focuses on charge distribution but miscalculates the energy relationship. Choice C makes the common error of assuming energy conservation—this only applies in ideal circuits without resistance or current flow. When capacitors connect, the redistribution process always involves energy loss. Choice D incorrectly tries to use the increased total capacitance without properly accounting for the voltage change. The correct answer is B: the final energy is half the initial energy because the voltage drops to $V/2$ . Remember: when charged capacitors connect in parallel, energy is always lost during charge redistribution, even though charge itself is conserved.
A capacitor is charged to voltage $V_0$ and then disconnected from the charging source. A second, identical uncharged capacitor is then connected in parallel with the first. After the system reaches equilibrium, what fraction of the original energy remains in the system?
Answer: $\frac{1}{2}$ of the original energy because charge redistributes equally while maintaining energy conservation principles When capacitors are connected or disconnected while charged, you need to apply two key conservation principles: charge conservation and energy analysis. However, these situations often involve energy dissipation, so total energy is not always conserved. Initially, the charged capacitor has energy $E_0 = \frac{1}{2}CV_0^2$ . When you connect the identical uncharged capacitor in parallel, charge redistributes until both capacitors reach the same voltage. Since charge is conserved, the total charge $Q_0 = CV_0$ spreads equally between the two identical capacitors, giving each capacitor charge $Q_0/2$ . The final voltage across both capacitors becomes $V_f = \frac{Q_0/2}{C} = \frac{V_0}{2}$ . The final energy is $E_f = 2 \times \frac{1}{2}C(\frac{V_0}{2})^2 = \frac{1}{4}CV_0^2 = \frac{1}{2}E_0$ . Therefore, half the original energy remains. Option A incorrectly identifies the energy fraction - while energy does decrease due to losses during charge redistribution (typically as heat in connecting wires), the fraction is $\frac{1}{2}$ , not $\frac{1}{4}$ . Option B reflects the common misconception that energy is always conserved - but connecting charged capacitors involves current flow that dissipates energy. Option C gives an incorrect fraction and misunderstands the physical process. Remember: When charged capacitors are connected, charge conserves but energy typically dissipates. Always calculate the initial and final states separately using $E = \frac{1}{2}CV^2$ to find the energy difference.
A cylindrical capacitor consists of an inner conductor of radius $a$ and an outer conductor of radius $b$ (where $b > a$ ), with length $L \gg b$ . If the inner conductor carries charge $+Q$ and the outer conductor carries charge $-Q$ , the electric field at radius $r$ (where $a < r < b$ ) has magnitude closest to which expression?
Answer: $\frac{Q}{2\pi \epsilon_0 L r}$ because cylindrical symmetry gives inverse linear dependence on radius When you encounter cylindrical capacitors, the key is recognizing that the symmetry determines how the electric field varies with distance. Unlike point charges or infinite planes, cylindrical geometry creates a unique field pattern. To find the electric field, apply Gauss's law with a cylindrical Gaussian surface of radius $r$ and length $L$ . The electric field points radially outward and has constant magnitude on this surface. Since $\oint \vec{E} \cdot d\vec{A} = E \cdot 2\pi r L$ and the enclosed charge is $Q$ , Gauss's law gives: $E \cdot 2\pi r L = \frac{Q}{\epsilon_0}$ Solving for $E$ : $E = \frac{Q}{2\pi \epsilon_0 L r}$ This $1/r$ dependence is characteristic of cylindrical symmetry—the field decreases linearly with radius, not quadratically like point charges. Option A is wrong because it uses $1/r^2$ dependence, which applies to spherical symmetry (point charges), not cylindrical geometry. The reasoning incorrectly applies the inverse square law. Option B has the wrong coefficient. While it correctly identifies $1/r$ dependence, the factor of $4\pi$ in the denominator comes from spherical problems. The "factor of 2" reasoning is also incorrect. Option D doubles the correct answer by omitting the factor of 2 in the denominator. This suggests confusion about the cylindrical surface area calculation. Remember: the symmetry of the charge distribution determines the field's radial dependence. Spherical gives $1/r^2$ , cylindrical gives $1/r$ , and infinite planes give constant fields. Always match your Gaussian surface to the problem's symmetry.
Two parallel-plate capacitors with the same plate area $A$ but different separations $d_1$ and $d_2 = 2d_1$ are connected in parallel across a voltage source $V$ . A dielectric with dielectric constant $\kappa = 4$ completely fills the space in the second capacitor. What is the ratio of the energy stored in the first capacitor to the energy stored in the second capacitor?
Answer: $\frac{U_1}{U_2} = \frac{1}{2}$ , because the second capacitor has twice the separation but four times the dielectric constant When analyzing energy storage in capacitors connected in parallel, you need to consider both capacitance values and the fact that parallel capacitors share the same voltage. First, let's find each capacitance. For the first capacitor: $C_1 = \frac{\epsilon_0 A}{d_1}$ . For the second capacitor with dielectric: $C_2 = \frac{\kappa \epsilon_0 A}{d_2} = \frac{4\epsilon_0 A}{2d_1} = \frac{2\epsilon_0 A}{d_1}$ . So $C_2 = 2C_1$ . Since both capacitors are connected in parallel, they're at the same voltage $V$ . The energy stored in each capacitor is $U = \frac{1}{2}CV^2$ , so: - $U_1 = \frac{1}{2}C_1V^2$ - $U_2 = \frac{1}{2}C_2V^2 = \frac{1}{2}(2C_1)V^2 = C_1V^2$ Therefore: $\frac{U_1}{U_2} = \frac{\frac{1}{2}C_1V^2}{C_1V^2} = \frac{1}{2}$ Choice A correctly identifies this ratio and gives the right reasoning - the doubled separation reduces capacitance by half, but the 4× dielectric constant increases it by 4×, giving a net 2× increase in $C_2$ . Choice B incorrectly calculates the ratio as $\frac{1}{4}$ , missing that energy depends on $CV^2$ , not just $C$ . Choice C wrongly claims equal capacitances, ignoring the dielectric effect. Choice D gets both the ratio and reasoning backward, incorrectly stating that smaller separation means higher capacitance per unit area. Study tip: For parallel capacitors, always remember they share voltage, so energy ratios equal capacitance ratios. Calculate each capacitance carefully, accounting for both geometry and dielectric effects.
A parallel-plate capacitor with plate area $A$ and plate separation $d$ is connected to a battery and fully charged. The battery is then disconnected, and a dielectric material with dielectric constant $\kappa = 3$ is inserted to fill half the volume between the plates (covering the full area but only half the separation distance). What happens to the energy stored in the capacitor?
Answer: The energy decreases by a factor of 1.5 because the effective capacitance increases while charge remains constant When the battery is disconnected, charge Q remains constant. The dielectric fills half the separation, creating two capacitors in series: one with vacuum (C₁ = ε₀A/(d/2)) and one with dielectric (C₂ = κε₀A/(d/2) = 3ε₀A/(d/2)). The effective capacitance is 1/C_eff = 1/C₁ + 1/C₂ = (d/2)/(ε₀A) + (d/2)/(3ε₀A) = (d/2ε₀A)(1 + 1/3) = (2d/3ε₀A). So C_eff = 3ε₀A/2d = 1.5C₀. Since U = Q²/2C and Q is constant, U_new = Q²/(2 × 1.5C₀) = U₀/1.5.
A parallel-plate capacitor with plate area $A = 0.01\,\text{m}^2$ and separation $d = 2\,\text{mm}$ is connected to a 100 V battery. After reaching equilibrium, the battery is disconnected and the plates are slowly pulled apart until the separation becomes $d' = 6\,\text{mm}$ . What is the work done by the external agent pulling the plates apart?
Answer: $W = 4.42 \times 10^{-8}\,\text{J}$ , because the final energy is three times the initial energy at constant charge When a capacitor is disconnected from its battery, the charge remains constant while you change the plate separation. This creates a key energy relationship you need to recognize. Initially, the capacitance is $C_1 = \varepsilon_0 A/d = (8.85 \times 10^{-12})(0.01)/(0.002) = 4.42 \times 10^{-11}\,\text{F}$ . The stored energy is $U_1 = \frac{1}{2}CV^2 = \frac{1}{2}(4.42 \times 10^{-11})(100)^2 = 2.21 \times 10^{-8}\,\text{J}$ . After disconnection, the charge $Q = CV = 4.42 \times 10^{-9}\,\text{C}$ stays fixed. When separation triples to 6 mm, the new capacitance becomes $C_2 = C_1/3 = 1.47 \times 10^{-11}\,\text{F}$ . The final energy is $U_2 = \frac{Q^2}{2C_2} = \frac{Q^2}{2(C_1/3)} = 3 \times \frac{Q^2}{2C_1} = 3U_1 = 6.63 \times 10^{-8}\,\text{J}$ . The work done by the external agent equals the energy increase: $W = U_2 - U_1 = 6.63 \times 10^{-8} - 2.21 \times 10^{-8} = 4.42 \times 10^{-8}\,\text{J}$ . Choice A gives an incorrect energy difference calculation. Choice B represents only the initial energy, not the work done. Choice D gives the final total energy rather than the work performed. Study tip: For capacitor problems after battery disconnection, remember that charge stays constant, so energy scales inversely with capacitance. When plates separate, capacitance decreases and energy increases—the external work provides this energy increase.
A variable air-gap capacitor has plate area $A$ and adjustable separation distance $d$ . When connected to a constant voltage source $V$ , the capacitor plates experience an attractive force. If the plate separation is decreased by 10% while maintaining the voltage connection, the attractive force between the plates changes by approximately what factor?
Answer: Increases by a factor of 1.23 because force scales with the square of electric field strength The attractive force between capacitor plates is F = ½ε₀E²A, where E = V/d. So F = ½ε₀(V/d)²A = ½ε₀V²A/d². When d decreases to 0.9d, the new force is F' = ½ε₀V²A/(0.9d)² = ½ε₀V²A/(0.81d²) = F/0.81 ≈ 1.23F. The force increases because the electric field strength increases quadratically with decreasing separation.
Three capacitors with capacitances $C_1 = 2\,\mu\text{F}$ , $C_2 = 4\,\mu\text{F}$ , and $C_3 = 6\,\mu\text{F}$ are connected such that $C_1$ and $C_2$ are in parallel, and this combination is in series with $C_3$ . The entire network is connected across a 12 V battery. What is the charge stored on capacitor $C_3$ ?
Answer: $36\,\mu\text{C}$ , because the charge distributes according to the series voltage division rules When analyzing capacitor networks, you need to systematically find equivalent capacitances and apply the fundamental rules: capacitors in series share the same charge, while capacitors in parallel share the same voltage. First, find the equivalent capacitance of $C_1$ and $C_2$ in parallel: $C_{12} = C_1 + C_2 = 2 + 4 = 6\,\mu\text{F}$ . Now you have this $6\,\mu\text{F}$ combination in series with $C_3 = 6\,\mu\text{F}$ . For capacitors in series: $\frac{1}{C_{total}} = \frac{1}{C_{12}} + \frac{1}{C_3} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$ , so $C_{total} = 3\,\mu\text{F}$ . The total charge from the battery is $Q_{total} = C_{total} \times V = 3\,\mu\text{F} \times 12\,\text{V} = 36\,\mu\text{C}$ . Since $C_{12}$ and $C_3$ are in series, they must have the same charge flowing through them. Therefore, $C_3$ stores $36\,\mu\text{C}$ . Option A incorrectly assumes larger capacitance means more charge, ignoring the series connection rule. Option B mentions "series voltage division" but arrives at the wrong value—likely from a calculation error. Option C wrongly assumes $C_3$ sees the full 12V, but in series circuits, voltage divides between components. Remember this key principle: in series capacitor combinations, all capacitors store identical charge regardless of their individual capacitances. Focus on finding the total circuit charge first, then apply series rules.
A variable capacitor consists of two sets of parallel plates that can be rotated relative to each other to change the overlapping area. Initially, the plates fully overlap with area $A_0$ and separation $d$ , giving capacitance $C_0 = \epsilon_0 A_0/d$ . The capacitor is charged to voltage $V_0$ and disconnected from the battery. If the plates are then rotated so that only 25% of the area overlaps, what work must be done by an external agent?
Answer: $W = \frac{3\epsilon_0 A_0 V_0^2}{2d}$ , because the capacitance decreases and energy increases with constant charge When analyzing capacitor problems involving mechanical changes, always identify whether charge or voltage remains constant after disconnection from the battery. Here, the capacitor is disconnected, so charge $Q = C_0 V_0$ remains constant while capacitance changes. The initial energy is $U_i = \frac{1}{2}C_0 V_0^2 = \frac{\epsilon_0 A_0 V_0^2}{2d}$ . When the overlapping area reduces to 25%, the new capacitance becomes $C_f = \frac{\epsilon_0 (0.25A_0)}{d} = \frac{C_0}{4}$ . Since charge is constant, the final energy is $U_f = \frac{Q^2}{2C_f} = \frac{(C_0 V_0)^2}{2C_f} = \frac{C_0^2 V_0^2}{2(C_0/4)} = 4 \cdot \frac{C_0 V_0^2}{2} = 2\epsilon_0 A_0 V_0^2/d$ . The work done by an external agent equals the energy increase: $W = U_f - U_i = \frac{2\epsilon_0 A_0 V_0^2}{d} - \frac{\epsilon_0 A_0 V_0^2}{2d} = \frac{3\epsilon_0 A_0 V_0^2}{2d}$ . Answer A correctly identifies this result and the underlying physics. Answer B uses an incorrect proportionality assumption about area changes. Answer C makes an error in the energy difference calculation, likely forgetting the factor of 4 increase. Answer D incorrectly assumes voltage stays constant and that energy decreases—when capacitance decreases at constant charge, energy always increases because the electric field becomes stronger. Remember: when a capacitor is disconnected, decreasing capacitance at constant charge always increases stored energy, requiring positive work input.
Three capacitors with capacitances $C$ , $2C$ , and $3C$ are connected in a network where the $C$ and $2C$ capacitors are in series, and this combination is in parallel with the $3C$ capacitor. The entire network is connected across a voltage source $V$ . What fraction of the total stored energy is in the $3C$ capacitor?
Answer: $\frac{9}{11}$ , because the $3C$ capacitor experiences the full voltage while others share it When analyzing capacitor networks, you need to understand how capacitors combine and how energy distributes based on voltage and capacitance relationships. First, find the equivalent capacitance of the series combination ( $C$ and $2C$ ): $\frac{1}{C_{series}} = \frac{1}{C} + \frac{1}{2C} = \frac{3}{2C}$ , so $C_{series} = \frac{2C}{3}$ . This series combination is parallel to the $3C$ capacitor, giving total capacitance: $C_{total} = \frac{2C}{3} + 3C = \frac{11C}{3}$ . The key insight is voltage distribution. In parallel branches, voltage is the same across each branch. The $3C$ capacitor experiences the full voltage $V$ , while the series capacitors share this voltage. The energy stored in the $3C$ capacitor is $U_{3C} = \frac{1}{2}(3C)V^2 = \frac{3CV^2}{2}$ . Total energy stored is $U_{total} = \frac{1}{2}C_{total}V^2 = \frac{1}{2} \cdot \frac{11C}{3} \cdot V^2 = \frac{11CV^2}{6}$ . The fraction is: $\frac{U_{3C}}{U_{total}} = \frac{\frac{3CV^2}{2}}{\frac{11CV^2}{6}} = \frac{3CV^2}{2} \cdot \frac{6}{11CV^2} = \frac{9}{11}$ . Option A incorrectly assumes equal energy distribution regardless of capacitance values. Option B miscalculates by not considering the network's total energy properly. Option D correctly finds the equivalent capacitances but doesn't properly calculate the energy ratio. Remember: in parallel combinations, all elements see the same voltage, while in series combinations, elements share the voltage. Energy calculations depend critically on the actual voltage across each capacitor.
A parallel-plate capacitor is connected to a battery and allowed to fully charge. While still connected to the battery, a dielectric slab with dielectric constant $\kappa = 4$ is inserted between the plates, completely filling the space. Which of the following correctly describes the changes in the capacitor?
Answer: Voltage remains constant, charge increases by factor 4, and stored energy increases by factor 4 When analyzing capacitor problems involving dielectrics, the key factor is whether the capacitor remains connected to the battery. This determines which quantities stay constant and which change. Since the capacitor stays connected to the battery, the voltage across the plates must remain equal to the battery voltage - this is a fundamental constraint. The battery will supply or absorb charge as needed to maintain this constant voltage. When you insert a dielectric with $\kappa = 4$ , the capacitance increases by this same factor: $C_{new} = \kappa C_{original} = 4C$ . Since $Q = CV$ and voltage stays constant, the charge must increase proportionally with capacitance: $Q_{new} = C_{new}V = 4CV = 4Q_{original}$ . The stored energy $U = \frac{1}{2}CV^2$ also increases by factor 4, since capacitance quadruples while voltage remains constant. Choice A incorrectly assumes voltage increases - impossible when connected to a battery. Choice B makes the same voltage error and wrongly claims charge stays constant. Choice D correctly identifies constant voltage but incorrectly states that charge and energy decrease - they actually increase because the dielectric enhances the capacitor's ability to store charge at the same voltage. Choice C correctly describes all three changes: constant voltage (battery constraint), quadrupled charge (from increased capacitance), and quadrupled energy. Study tip: Always identify whether the capacitor is connected to or disconnected from the battery first. Connected = constant voltage; disconnected = constant charge. This immediately tells you which quantities can change.
A parallel-plate capacitor with initial capacitance $C_0$ is charged to voltage $V_0$ and then disconnected from the battery. A conducting slab of thickness $t$ and the same area as the plates is inserted halfway between the plates, which are separated by distance $d$ . What is the new voltage across the capacitor?
Answer: $V_0\left(1 - \frac{t}{d}\right)$ , because the effective separation is reduced by the thickness of the conducting slab When a capacitor is disconnected from its battery and then modified, you need to analyze how the geometry change affects both capacitance and voltage, keeping in mind that charge remains constant. Initially, the capacitor has charge $Q_0 = C_0 V_0$ . When you insert the conducting slab, it doesn't store charge itself—it simply redistributes the electric field. The key insight is that electric field cannot exist inside a conductor, so the field only exists in the air gaps on either side of the slab. The effective separation between the plates becomes $(d-t)$ because the conductor eliminates the electric field over thickness $t$ . Since $C_0 = \varepsilon_0 A/d$ , the new capacitance is $C_{new} = \varepsilon_0 A/(d-t)$ . With constant charge $Q_0$ , the new voltage is: $V_{new} = \frac{Q_0}{C_{new}} = \frac{C_0 V_0 d}{\varepsilon_0 A} \cdot \frac{(d-t)}{\varepsilon_0 A} = V_0 \frac{(d-t)}{d} = V_0\left(1-\frac{t}{d}\right)$ Option A incorrectly assumes the capacitance doesn't change—inserting a conductor definitely alters the field distribution. Option B treats this as two capacitors in series, but that's wrong because the conductor is at a single potential and doesn't create separate capacitors. Option D has the right physical reasoning about field gaps but arrives at an incorrect mathematical expression. Remember: when analyzing modified capacitors with constant charge, focus on how the geometry change affects the electric field distribution, then use $V = Q/C$ with the new capacitance.
A spherical capacitor consists of two concentric conducting spheres with inner radius $a$ and outer radius $b$ . The space between them is filled with two dielectric materials: the inner region (from $r = a$ to $r = c$ , where $a < c < b$ ) has dielectric constant $\kappa_1$ , and the outer region (from $r = c$ to $r = b$ ) has dielectric constant $\kappa_2$ . What is the capacitance of this system?
Answer: $C = 4\pi\epsilon_0\left[\frac{1}{\kappa_1}\left(\frac{1}{a} - \frac{1}{c}\right) + \frac{1}{\kappa_2}\left(\frac{1}{c} - \frac{1}{b}\right)\right]^{-1}$ , treating the regions as capacitors in series When you encounter a spherical capacitor with multiple dielectric layers, think of it as capacitors connected in series. The key insight is that electric field lines must pass through both dielectric regions sequentially, just like current flowing through resistors in series. To find the capacitance, start with the relationship between electric field and potential. In a spherical capacitor, the electric field at radius $r$ is $E = \frac{Q}{4\pi\epsilon r^2}$ , where $\epsilon$ depends on the dielectric material. The total voltage is the sum of voltage drops across both regions: $V = \int_a^c E_1 dr + \int_c^b E_2 dr = \frac{Q}{4\pi\epsilon_0\kappa_1}\left(\frac{1}{a} - \frac{1}{c}\right) + \frac{Q}{4\pi\epsilon_0\kappa_2}\left(\frac{1}{c} - \frac{1}{b}\right)$ Since $C = Q/V$ , we get answer B: the reciprocal of the bracketed expression multiplied by $4\pi\epsilon_0$ . Answer A incorrectly uses a simple average of dielectric constants, ignoring that the regions have different geometries and field strengths. Answer C attempts a weighted harmonic mean but uses incorrect weighting factors based on distances rather than the proper electrostatic analysis. Answer D simply multiplies both dielectric constants together, which has no physical basis for series-connected regions. Study tip: For any capacitor with multiple dielectric layers where field lines pass through each layer sequentially, always treat them as series capacitors. The total capacitance will involve adding the reciprocals of individual contributions, not averaging the dielectric constants.
Two identical parallel-plate capacitors are connected in series across a 12 V battery. Each capacitor has capacitance $C_0$ . If a dielectric material with dielectric constant $\kappa = 3$ is inserted between the plates of one capacitor only, what is the voltage across the capacitor without the dielectric?
Answer: 9.0 V, because the capacitor without dielectric has lower capacitance and higher voltage When capacitors are connected in series, the voltage divides between them based on their capacitances, but in an inverse relationship - the capacitor with smaller capacitance gets more voltage. Let's work through this step by step. Initially, both capacitors have capacitance $C_0$ . When the dielectric with $\kappa = 3$ is inserted into one capacitor, that capacitor's new capacitance becomes $C_1 = \kappa C_0 = 3C_0$ , while the other remains $C_2 = C_0$ . In series circuits, capacitors carry the same charge but divide the voltage inversely proportional to their capacitances. The voltage across each capacitor is: $V_1 = \frac{C_2}{C_1 + C_2} \times V_{total}$ and $V_2 = \frac{C_1}{C_1 + C_2} \times V_{total}$ For the capacitor without dielectric: $V_2 = \frac{3C_0}{3C_0 + C_0} \times 12V = \frac{3C_0}{4C_0} \times 12V = 9V$ Choice A incorrectly focuses only on the capacitance change without applying the voltage division rule. Choice B correctly identifies the inverse relationship but doesn't calculate the actual values - the voltage would be 3V for the capacitor with the dielectric. Choice C ignores that capacitance affects voltage division in series circuits - that's only true for parallel circuits or identical capacitors. Remember: in series capacitor circuits, smaller capacitance means larger voltage drop. When you see dielectric problems involving series capacitors, always recalculate the new capacitances first, then apply the inverse voltage division rule.

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