If the distance between the plates of a capacitor is halved and the area of the plates is doubled,what will be the new capacitance?

Two parallel plates separated by $d \ mm$ are kept at a potential difference of $V \ V$. $A$ particle of mass $m$ and charge $q$ enters the region between them with some velocity. The acceleration of the particle will be

$A$ parallel plate capacitor has circular plates of $10\, cm$ radius separated by an air-gap of $1\, mm$. It is charged by connecting the plates to a $100\, V$ battery. The change in energy stored in the capacitor when the plates are moved to a distance of $1\, cm$ while remaining connected to the battery is:

Four metallic plates,each with a surface area of one side $A$,are placed at a distance $d$ from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between $a$ and $b$ is

$A$ metal slab of thickness $b = d/2$ is inserted between the plates of a parallel plate capacitor of plate separation $d$. What is the ratio of the new capacitance to the original capacitance?

$A$ parallel plate capacitor with plates of area $1\,m^2$ each,are at a separation of $0.1\,m.$ If the electric field between the plates is $100\,N/C,$ what is the magnitude of the charge on each plate?