A parallel-plate air capacitor has capacitance $C_0$. A dielectric slab of dielectric constant $k$ is inserted between the plates, completely filling the gap. The new capacitance is: 
A$C_0$ (unchanged)
B$C_0 / k$
C$k C_0$
D$C_0 / k^2$
Answer & Solution
Correct answer: C. $k C_0$
A dielectric reduces the effective field in the gap by a factor $k$, hence the voltage by $k$ for the same charge. Since $C = Q/V$, the capacitance increases by a factor $k$: $C = k C_0 = k A \varepsilon_0/d$.
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