A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then:
A. The energy stored in the capacitor decreases
B. The capacitance of the capacitor increases
C. The charge on the capacitor decreases
D. The voltage across the capacitor increases

Hint:First let us see what a parallel plate capacitor is: An arrangement of electrodes and insulating material or dielectric are created by parallel plate capacitors. The plates are charged and an electrical field is formed between them when two parallel plates are attached by a battery, and this configuration is known as the parallel plate capacitor.

Complete answer:
A parallel plate, which consists of two metal plates with a distance between them is the simplest configuration for a capacitor. Electrons are deposited on one plate while an equal number of electrons are withdrawn from the other plate. This very specific property of a parallel plate capacitor makes it ideal for extracting harmonics from the supply of AC. For electronic circuits for different purposes, a parallel plate capacitor is used. It can be used in transducers also.

In a parallel plate capacitor, the electric field $E$ is directly proportional to the charge $Q$ as there will be more field lines if there is more charge. So,
$E\alpha Q$ ...... (i)
Also, voltage across the parallel plates is given by:
$V = Ed$
Thus,
$V\alpha E$ ...... (ii)
From equation (i) and (ii), we get:
$Q\alpha CV$

Different capacitors hold different quantities of charge for the same applied voltage, based on their physical characteristics. We define their capacitance $C$ to be such that the load $Q$ contained in the capacitor is proportional to the load. The charge deposited in the capacitor is given by:
$
Q = CV \\
\Rightarrow C = \dfrac{Q}{V} \\
$
At the point when a voltage $V$ is applied to the capacitor, it stores a charge $Q$, as appeared. We can perceive how its capacitance relies upon surface area $A$ and distance d by considering the qualities of the Coulomb power. We realize that like charges repulse, not at all like charges pull in, and the force between charges falls down because of separation.

So it appears to be very sensible that the greater the plates are, the more charge they can store—on the grounds that the charges can spread out additional. Hence $C$ ought to be more prominent for bigger $A$ . Also, the closer the plates are together, the more noteworthy the attraction of the opposite charges on them. So $C$ should be greater for smaller $d$ .

Hence, the generalised equation for a parallel plate capacitor is given as:
$C = {\varepsilon _ \circ }\left( {\dfrac{A}{d}} \right)$
where ${\varepsilon _ \circ }$ represents the absolute permittivity of the dielectric material being used. The distance between the plates $d$ grows as the plates are separated and hence the capacitance decreases.

Therefore the voltage through the capacitor increases as $Q = CV$ to maintain the charge constant.Hence, option D is correct.

Note:Electrical current cannot directly pass through the capacitor, as the resistor or inductor does, due to the insulating properties of the dielectric layer between the two plates. However, the charging and unloading of the two plates has the consequence that the current is flowing.