Q.How capacitance changes if the distance between the plates of a parallel plate capacitor is halved?

Capacitance of a parallel plate capacitor is inversely proportional to the plate separation.

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Doubling both $A$ and $d$ leaves $C$ (and hence $Q$, $V$) unchanged, but the field $E=V/d$ halves, so only the energy density changes.

Working

Capacitance of a parallel plate capacitor: $C = \dfrac{\varepsilon_0 A}{d}$

If both $A \to 2A$ and $d \to 2d$:

$C' = \dfrac{\varepsilon_0(2A)}{2d} = \dfrac{\varepsilon_0 A}{d} = C$ — unchanged.

Since capacitance is unchanged and the capacitor is isolated (charge $Q$ fixed), the voltage $V=Q/C$ is also unchanged, and so is $Q$.

However, the field between the plates $E = V/d$. With $V$ unchanged but $d$ doubled:

$E' = \dfrac{V}{2d} = \dfrac{E}{2}$

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For a parallel plate capacitor, $C = \dfrac{\varepsilon_0 A}{d}$, so $C \propto 1/d$; halving $d$ doubles $C$.

The capacitance of a parallel plate capacitor (plate area $A$, separation $d$, vacuum/air between plates) is

$C = \frac{\varepsilon_0 A}{d}$

With $A$ unchanged, $C$ depends only on $1/d$. If the separation is halved, $d \to d/2$, so …

Capacitance depends only on the capacitor's geometry, not on the applied voltage, so it is unaffected when $V\to2V$; the charge $Q=CV$, however, doubles.

Working

For a parallel plate capacitor,

$C=\dfrac{\varepsilon_0A}{d}$

depends only on the plate area $A$, separation $d$, and the dielectric — none of which change when the applied voltage is changed. So the capacitance $C$ stays exactly the same.

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C is inversely proportional to plate separation, so doubling d gives half the capacitance, C/2.

The capacitance of a parallel-plate capacitor is

C = epsilon_0 A / d

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Parallel plate capacitor capacitance C = ε₀A/d, option (vi).

For a parallel plate capacitor with plate area A, plate separation d, and vacuum (or air) between the plates, the capacitance is

C = ε₀A/d,

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