Charges of $2Q$ and $-Q$ are placed on two plates of a parallel plate capacitor. If the capacitance of the capacitor is $C$,the potential difference between the plates is:

$A$ parallel-plate capacitor consists of a fixed plate and a movable plate that is allowed to slide in the direction parallel to the plates. Let $x$ be the distance of overlap,as shown in the figure. The separation between the plates is fixed. Assume that the plates are electrically isolated,so that their charges $\pm Q$ are constant. Force on the movable plate is proportional to

Write the capacitance of a parallel plate capacitor with a medium of dielectric constant $K$.

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?

In a parallel plate capacitor,the capacity can be increased by decreasing

Two identical metal plates are given charges $q_1$ and $q_2$ $(q_2 < q_1)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $C$,the potential difference $V$ between the plates is