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A parallel-plate capacitor has area $A$, gap $d$, capacitance $C_0$. If the gap is halved, the new capacitance is:
A$2C_0$, since $C \propto 1/d$ by $C = \varepsilon_0 A/d$
B$C_0/2$, the inverse direction of scaling here
C$C_0$, since neither area nor dielectric changed
D$4C_0$, the area-and-gap factor combined together
Answer & Solution
Correct answer: A. $2C_0$, since $C \propto 1/d$ by $C = \varepsilon_0 A/d$
$C = \varepsilon_0 A/d$; halving $d$ doubles $C$.
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