$A$ parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge,potential,and capacitance respectively are:

$A$ parallel plate capacitor with area $200\,cm^2$ and separation between the plates $1.5\,cm$,is connected across a battery of $emf$ $V$. If the force of attraction between the plates is $25\times10^{-6}\,N$,the value of $V$ is approximately........$V$ $\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}\,\frac{{{C^2}}}{{N{m^2}}}} \right)$

$A$ 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

Two parallel plate capacitors have their plate areas $100 \,cm^2$ and $500 \,cm^2$ respectively. If they have the same charge and potential,and the distance between the plates of the first capacitor is $0.5 \,mm$,then the distance between the plates of the second capacitor is ........ $cm$.

Two circular plates each of radius 'r' are kept parallel to each other distance 'd' apart. The capacitance of the capacitor formed is '$C_1$'. If the radius of each of the plates is increased to $\sqrt{3}$ times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes '$C_2$'. The ratio $C_1 : C_2$ is

If the charge on each plate of a parallel plate capacitor is $Q$ and the magnitude of the electric field between the plates is $E$,then the force on each plate of the parallel plate capacitor will be: