Capacitance with a dielectric and smaller gap
The plate separation of a parallel-plate capacitor is halved, and a dielectric with κ = 3 is inserted. By what factor does its capacitance change?
Answer C: explanation for every choice
A
This assumes halving the separation halves C. C is inversely proportional to d.
B
This accounts for the dielectric but ignores the smaller gap.
C
Correct. C = κε₀A/d. The dielectric multiplies C by 3, and halving d doubles it: 3 × 2 = 6.
D
This treats C as proportional to 1/d².
TakeawayC = κε₀A/d.
Equation usedCapacitance and stored energy
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