Difficulty: Intermediate | Prerequisites: Electric potential, Gauss's law, work-energy theorem
The potential energy of a system of point charges is the sum of kq_i q_j / r for every pair, and the work to bring a new charge in from infinity equals q times the potential at its destination. Capacitor circuits reduce to series and parallel combinations, where series capacitors share the same charge and parallel capacitors share the same voltage. Adding a dielectric increases a capacitor's capacitance and redistributes charge in the circuit. Electron acceleration problems are energy conservation: ½mv² = |q|ΔV.
Electric potential energy of a charge configuration (U)
The total energy stored in the arrangement, calculated by summing kq_i q_j / r over every unique pair. Think of it as the energy cost of assembling the charges from infinity.
Work done by the field to bring a charge from infinity
W = q × V_destination, where V is the potential at the destination created by all other charges already in place. Positive W means the field helps the charge arrive; negative W means external work is needed.
Capacitance (C)
The ratio of charge stored to voltage applied: C = Q / V. Measured in farads (F). In simple terms, it measures how much charge a device can hold per volt.
Series capacitors
Capacitors connected end to end so that the same charge flows onto each. The voltages add. The equivalent capacitance is smaller than any individual one: 1/C_eq = 1/C_1 + 1/C_2 + ...
Parallel capacitors
Capacitors connected across the same two nodes so they share the same voltage. The charges add. C_eq = C_1 + C_2 + ...
Dielectric constant (κ)
A dimensionless number (κ ≥ 1) describing how much an insulating material increases a capacitor's capacitance: C_new = κC_0. In simple terms, the dielectric weakens the internal field and lets the capacitor hold more charge at the same voltage.
X-ray tube
A device that accelerates electrons from a grounded cathode toward a high-voltage anode. The electrons gain kinetic energy equal to |q|ΔV, then produce X-rays when they slam into a target.
Cathode and anode
The cathode is the negative terminal (electron source). The anode is the positive terminal. Electrons travel from cathode to anode.
Parallel-plate capacitor
Two large parallel conducting plates separated by a gap. The field between them is approximately uniform: E = V/d = σ/ε0. The field outside (beyond both plates) is approximately zero if the charges are equal and opposite.
Potential energy of a pair of charges
U = kq₁q₂ / r. Positive U for like charges (repulsion, energy stored by pushing them together). Negative U for unlike charges (attraction, energy released when they come together).
Potential energy of multiple charges
Sum over all unique pairs: U_total = Σ kq_i q_j / r_ij.
For a square arrangement with charges +q, –q, –q, +q at the corners: there are four side pairs (two attractive, two repulsive) and two diagonal pairs (both attractive, since opposite charges sit diagonally). The diagonal distance is a√2. The net U comes out negative (U < 0) because the attractive-pair terms win.
Work done by the field to bring a charge from infinity
First compute V at the destination due to all charges already present: V = Σ kq_i / r_i.
W = q_new × V. If V is positive and q_new is positive, W > 0 (the field helps pull the charge in). If V is negative and q_new is positive, the field opposes the motion and W < 0.
Example: +q at (0, a) and –q at origin. Point A is at (a, 0). V_A = kq/r_1 + k(–q)/r_2 where r_1 and r_2 are the distances from each charge to A. If the distances are equal, V_A = 0 and W = 0.
Zero-potential points on an axis (dipole)
For a +q and –q pair on the y-axis, the potential on the y-axis is V(y) = kq/(|y – a|) – kq/|y|.
V = 0 wherever the contributions cancel. For charges of equal magnitude, there is exactly one such point between them (the midpoint if they are equal and opposite) where V = 0. Placing an additional +q there does not change U because U_new = q × V = q × 0 = 0.
On the y-axis outside both charges, V is dominated by the nearer charge and approaches zero only at infinity, so the number of finite zero-V points on the axis is typically one.
Identifying series and parallel
If two capacitors share a node with nothing else connected to it (no branch point), they are in series.
If two capacitors connect between the same two nodes, they are in parallel.
In the exam circuit: C₂ and C₃ are in series with each other, and that series combination is in parallel with C₁, all connected across a 9 V battery.
Solving the circuit (all capacitors C = 3 µF, battery = 9 V)
C₂ and C₃ in series: 1/C_23 = 1/3 + 1/3 = 2/3, so C_23 = 1.5 µF.
C_23 is in parallel with C₁: both see the full 9 V.
Q₁ = C₁ × V = 3 × 9 = 27 µC.
Q₂ = Q₃ = C_23 × V = 1.5 × 9 = 13.5 µC (series capacitors share the same charge).
Effect of inserting a dielectric into C₃ (κ = 3)
C₃ becomes κC = 9 µF.
New series combination: 1/C_23' = 1/3 + 1/9 = 4/9, so C_23' = 2.25 µF.
C₁ is still in parallel across the same 9 V battery, so Q₁ does not change. The voltage across C₁ is still 9 V, giving Q₁ = 27 µC.
Q₂ = Q₃ = C_23' × 9 = 20.25 µC. Q₂ increases because the series combination's capacitance increased.
Speed of an accelerated electron
Energy conservation: ½mv² = |q|ΔV.
Solve for v: v = √(2|q|ΔV / m).
For ΔV = 11 kV, m = 9.1 × 10⁻³¹ kg, |q| = 1.6 × 10⁻¹⁹ C: v ≈ 6.2 × 10⁷ m/s.
Surface charge density on the plates
The field between the plates is E = V/d = 11,000 / 0.10 = 1.1 × 10⁵ N/C.
For equal and opposite plates: E = σ/ε0, so σ = ε0 E = (8.85 × 10⁻¹²)(1.1 × 10⁵) ≈ 0.97 µC/m².
Field outside the parallel plates
With equal and opposite charge densities, the fields from the two plates cancel outside. E = 0 to the left of the cathode and to the right of the anode.
This is the hallmark of the parallel-plate capacitor: the field is confined between the plates.
Potential energy of a pair of point charges
U = \frac{k q_1 q_2}{r}Potential at a point due to a collection of point charges
V = \sum_i \frac{k q_i}{r_i}Work done by the field bringing charge q from infinity to a point with potential V
W = q VCapacitors in series
\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdotsSeries capacitors carry the same charge Q.
Capacitors in parallel
C_{\text{eq}} = C_1 + C_2 + \cdotsParallel capacitors share the same voltage V.
Capacitance with a dielectric
C = \kappa C_0Kinetic energy of a charge accelerated through ΔV
\frac{1}{2} m v^2 = |q| \, |\Delta V| \quad \Rightarrow \quad v = \sqrt{\frac{2|q|\,|\Delta V|}{m}}Surface charge density from field between parallel plates
\sigma = \varepsilon_0 E = \varepsilon_0 \frac{V}{d}Students often think capacitors in series share the same voltage. They do not. Series capacitors share the same charge; their voltages add up to the source voltage.
Students often assume that adding a dielectric to one capacitor in a series combination changes the voltage across a parallel capacitor. It does not if the parallel capacitor is directly across the battery. The battery fixes its voltage.
Students often forget to sum over all pairs when computing U for a multi-charge system. For four charges, there are six pairs, not four.
Students often mix up the work done by the field (W = qV) with the work done by an external agent (W_ext = –qV). The signs are opposite. Read the question carefully to see which one it asks for.
⚠️ "What is the potential energy of this arrangement?" requires summing all pairs, including diagonals. For a square of alternating charges, U < 0.
⚠️ "At how many points on the y-axis can a charge be placed without changing U?" translates to: where is V = 0 on the y-axis? For a ±q dipole, there is typically one finite point.
⚠️ Capacitor questions test whether you can identify series vs parallel, compute Q on each, and then predict what changes when a dielectric is inserted. The parallel capacitor across the battery is the trap: its voltage (and therefore its charge) does not change.
⚠️ Electron speed calculations: plug into v = √(2|q|ΔV/m). Watch the units (convert kV to V). A common wrong answer is 4.1 × 10⁸ m/s, which exceeds the speed of light, so a quick sanity check catches it.
⚠️ For parallel plates with equal and opposite charge, the field outside is zero. This is a standard three-answer question.
True or false: Capacitors in series have the same voltage across them. (False. They have the same charge. Their voltages add to the total.)
Fill in the blank: For two 3 µF capacitors in series across a 9 V battery, the charge on each is ______ . (13.5 µC. C_eq = 1.5 µF, Q = 1.5 × 9 = 13.5 µC.)
True or false: Inserting a dielectric into C₃ (which is in series with C₂) changes the charge on C₁ (which is in parallel across the battery). (False. C₁ still sees 9 V, so Q₁ stays the same.)
Fill in the blank: The speed of an electron accelerated through 11 kV is approximately ______ m/s. (6.2 × 10⁷ m/s.)
True or false: The potential energy of a square arrangement with alternating +q and –q charges at the corners is positive. (False. It is negative because attractive pairs dominate.)
Q: A +3 µC charge sits at (0, 5 cm) and a −3 µC charge at the origin. What is the work done by the field to bring a +3 µC charge from infinity to point A = (5 cm, 0)?
A: Compute V at A: V_A = k(3 × 10⁻⁶)/r₁ + k(−3 × 10⁻⁶)/r₂. The distances from A to each charge are r₁ = √(0.05² + 0.05²) = 0.05√2 m and r₂ = 0.05 m. V_A = k(3 × 10⁻⁶)(1/0.0707 – 1/0.05) ≈ –157 kV (net negative, because the negative charge is closer). W = qV_A = (3 × 10⁻⁶)(–157,000) ≈ –0.47 J. The field opposes bringing another positive charge to a negative-potential location. (Closest answer: W = 0.47 J if the question asks magnitude, or check sign conventions.)
Q: At how many points on the y-axis can a +q charge be placed without changing the system's potential energy?
A: One point. V = 0 at the midpoint y = 2.5 cm (for equal and opposite charges ±q). Placing +q where V = 0 adds zero energy.
Q: Three identical capacitors C = 3 µF are connected with C₂ and C₃ in series, and that combination in parallel with C₁, across 9 V. What is Q₁?
A: C₁ sees the full 9 V (it is in parallel with the battery). Q₁ = 3 × 9 = 27 µC.
Q: In the same circuit, a dielectric (κ = 3) is inserted into C₃. Does Q₂ increase, decrease, or stay the same?
A: Q₂ increases. C₃ becomes 9 µF, the series combination rises from 1.5 to 2.25 µF, and Q₂ = Q₃ = 2.25 × 9 = 20.25 µC (up from 13.5 µC).
Q: An electron is accelerated from rest through 11 kV. What is its final speed?
A: v = √(2 × 1.6 × 10⁻¹⁹ × 11,000 / 9.1 × 10⁻³¹) ≈ 6.2 × 10⁷ m/s.
Q: For a parallel-plate capacitor with equal and opposite charge on the plates, what is the electric field outside the plates?
A: Zero. The fields from the two plates cancel outside.
Capacitor circuits are the bridge between electrostatics and DC circuits. The same series/parallel logic returns with resistors (though the formulas swap: series resistors add, parallel resistors use the reciprocal rule). Dielectrics connect to the broader topic of polarisation and bound charge, which appears later in the course.
The X-ray tube problem ties energy conservation to electrostatics. The same v = √(2qΔV/m) calculation shows up in cathode ray tubes, mass spectrometers, and particle accelerators.
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