A parallel-plate capacitor is charged so that it holds charge Q at voltage V. If you then double the charge Q stored on it (by connecting a stronger battery), what happens to the capacitance C?
AC doubles — since C = Q/V and Q increased, C must increase
BC stays the same — capacitance depends only on geometry and material, not on Q or V
CC halves — the stronger field from more charge reduces the effective capacitance
DC increases, and V also doubles, so the ratio C = Q/V stays the same only by coincidence
Capacitance C = Q/V is a geometric constant — it characterizes the device, not the charge state. When you double Q, the voltage V also doubles proportionally (since V = Q/C), keeping the ratio C = Q/V unchanged. C depends only on plate area A, separation d, and dielectric constant κᵣ: C = ε₀κᵣA/d. This is the central insight: Q and V are conjugate variables that both change together, while C remains fixed by geometry.
Question 2 Multiple Choice
A cylindrical capacitor has inner radius a, outer radius b, and length ℓ. If you double the length ℓ while keeping a and b the same, what happens to capacitance?
ACapacitance doubles — C_cylinder = 2πε₀ℓ/ln(b/a), so doubling ℓ doubles C
BCapacitance increases by a factor of ln(2) — the logarithm picks up the extra length
CCapacitance is halved — a longer cylinder distributes charge over more surface area, reducing efficiency
DCapacitance is unchanged — it depends only on the ratio b/a, not the absolute length
The formula C_cylinder = 2πε₀κᵣℓ/ln(b/a) shows that capacitance scales linearly with ℓ. Doubling ℓ doubles C. Physically, a longer cylinder provides twice as much plate area (the cylindrical surfaces), just as increasing plate area A doubles a parallel-plate capacitor's capacitance. The log term ln(b/a) captures the radial geometry and is fixed when a and b don't change.
Question 3 True / False
Inserting a dielectric material with relative permittivity κᵣ > 1 between the plates of a charged capacitor increases its capacitance because the dielectric polarizes, reducing the effective electric field and lowering the voltage for the same stored charge.
TTrue
FFalse
Answer: True
This is correct. The dielectric's molecules align with the applied field, creating bound surface charges that partially oppose the free charges on the plates. This reduces the net electric field E, and since V = Ed (for a parallel-plate capacitor), V decreases for the same Q. Lower V with the same Q means C = Q/V increases. The dielectric effectively 'helps' the capacitor store charge more efficiently. All three geometry formulas gain a factor of κᵣ: C = ε₀κᵣA/d, etc.
Question 4 True / False
A parallel-plate capacitor with larger plate separation d stores more charge at a given voltage, so increasing d increases capacitance.
TTrue
FFalse
Answer: False
This is backwards. C = ε₀A/d — capacitance is inversely proportional to separation d. A larger gap means the electric field E = V/d is weaker for the same voltage, so fewer charges are induced on the plates (Q = CV decreases). Intuitively, moving the plates farther apart weakens the attractive interaction between opposite charges, reducing the device's ability to store charge per volt. To increase capacitance, you want a smaller d, not larger.
Question 5 Short Answer
Why does capacitance C = Q/V remain constant as Q and V change — what physical property does this represent?
Think about your answer, then reveal below.
Model answer: Capacitance measures a geometric property of the device: how much charge it can hold per unit of voltage difference. When you increase Q by adding more charge, the electric field between the conductors grows proportionally, and so does the voltage V = ∫E·dr. Since both Q and V scale together, their ratio C = Q/V stays fixed. This means C is a property of the geometry (plate area, separation, shape) and the material (dielectric constant) — not of how much charge happens to be stored at the moment. It characterizes the device's intrinsic ability to store charge, independent of its current state.
This independence from Q and V is what makes capacitance a useful circuit parameter: you can specify a capacitor by its capacitance value C, and then calculate V from Q (or vice versa) using V = Q/C for any charge state. The same reasoning explains why the derivation always cancels Q: finding E from Gauss's law gives a field proportional to Q, integrating to get V gives a result proportional to Q, and dividing Q by V eliminates Q entirely, leaving only geometric factors.