An air-filled parallel-plate capacitor is charged from a source of emf - AQA - A-Level Physics - Question 23 - 2018 - Paper 2

When the separation between the plates of a capacitor is doubled, with the capacitor disconnected from the source of emf, the charge on the plates remains constant. The electric field $E$ between the plates is defined by the formula:

$E = \frac{V}{d}$

where $V$ is the voltage and $d$ is the distance between the plates. Since the voltage increases as the distance doubles while the charge remains constant, the new electric field $E_f$ is given by:

$E_f = \frac{V'}{d'}$

With $V'$ as the new voltage which will approximately double if the capacitor's capacitance remains the same, and $d'$ as $2d$. This results in:

$E_f = \frac{2V}{2d} = \frac{V}{d} = E$

Thus, the final electric field remains the same, $E$.