Answer:
D) Q/2
Explanation:
The relationship between charge Q, capacitance C and voltage drop V across a capacitor is
[tex]Q=CV[/tex] (1)
In the first part of the problem, we have that the charge stored on the capacitor is Q, when the voltage supplied is V. The capacitance of the parallel-plate capacitor is given by
[tex]C=\frac{\epsilon_0 A}{d}[/tex]
where [tex]\epsilon_0[/tex] is the vacuum permittivity, A is the area of the plates, d is the separation between the plates.
Later, the voltage of the battery is kept constant, V, while the separation between the plates of the capacitor is doubled: [tex]d'=2d[/tex]. The capacitance becomes
[tex]C'=\frac{\epsilon_0 A}{d'}=\frac{\epsilon_0 A}{2d}=\frac{C}{2}[/tex]
And therefore, the new charge stored on the capacitor will be
[tex]Q'=C'V=\frac{C}{2}V=\frac{Q}{2}[/tex]
