A parallel-plate capacitor is charged by a battery to 12V, then disconnected. The plate separation is then doubled. What happens to the capacitance?
ACapacitance doubles because the electric field spreads over a larger region
BCapacitance halves because C = ε₀A/d and d doubled
CCapacitance stays the same because the voltage across the plates didn't change
DCapacitance stays the same because the charge on the plates didn't change
Capacitance C = ε₀A/d depends only on geometry — plate area A, separation d, and permittivity ε₀. Doubling d halves C, regardless of charge or voltage. Options C and D reflect the common misconception that capacitance depends on operating conditions. It doesn't: C is a property of the physical geometry, like resistance is a property of a resistor's material and dimensions. After disconnecting from the battery, Q is fixed, but C still changes — meaning V must change to satisfy Q = CV.
Question 2 Multiple Choice
Two identical capacitors, each with capacitance C, are connected in series. What is the total capacitance of the combination?
A2C — the capacitances add together
BC — series combination equals one capacitor
CC/2 — series combination is smaller than either individual capacitor
D4C — charge builds up across both gaps
Series capacitors combine as 1/C_total = 1/C₁ + 1/C₂. For two identical capacitors C: 1/C_total = 2/C, so C_total = C/2. Series combination is always smaller than the smallest individual capacitor. This surprises students who expect series to work like series resistors (which add). The physical reason: in series, the same charge Q sits on each capacitor, but the voltages add — you need more voltage per unit charge, which means less capacitance. Parallel is where capacitances add: C_parallel = C₁ + C₂.
Question 3 True / False
The energy stored in a charged capacitor resides on the charged plates themselves.
TTrue
FFalse
Answer: False
The energy is stored in the electric field between the plates, not on the charges. The energy density in an electric field is u = ½ε₀E², and integrating over the volume between the plates gives exactly U = ½CV². This is more than semantic: it anticipates a deeper principle in electrodynamics that fields are real physical entities carrying energy and momentum. The charges are the source of the field, but the energy belongs to the field.
Question 4 True / False
Increasing the plate area of a parallel-plate capacitor while keeping the separation fixed will increase its capacitance.
TTrue
FFalse
Answer: True
From C = ε₀A/d, capacitance is directly proportional to plate area A. Larger plates can accumulate more charge for the same voltage because a larger surface area holds more charge while maintaining the same surface charge density σ = Q/A. More facing area means more 'storage space' for charge without increasing the electric field strength, so more charge can be stored per volt applied.
Question 5 Short Answer
Why does the formula C = ε₀A/d show that capacitance is a geometric property, and what does this mean for how capacitance changes when voltage is applied?
Think about your answer, then reveal below.
Model answer: C = ε₀A/d contains only geometric quantities (plate area A, separation d) and a material constant (ε₀). Voltage and charge do not appear because capacitance is defined as the ratio C = Q/V — it is constant for a given capacitor regardless of how much charge you put on it or what voltage you apply. Changing the voltage changes Q proportionally so that Q/V stays fixed. Capacitance only changes if you physically alter the geometry: resize the plates, change the separation, or insert a dielectric material.
This geometric nature of capacitance is analogous to resistance being a property of a resistor's material and dimensions, not of the current flowing through it. Just as R = ρL/A doesn't contain current, C = ε₀A/d doesn't contain charge or voltage. This is why the common misconceptions — that capacitance depends on charge or voltage — are so fundamental: they confuse a fixed property of the device with the variable state of the circuit.