The capacity of a parallel plate capacitor is $10\,\mu F$,when the distance between its plates is $8\,cm$. If the distance between the plates is reduced to $4\,cm$,then the capacity of this parallel plate capacitor will be.........$\mu F$.

$A$ 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?

The capacitance of a parallel plate capacitor is $12\,\mu F$. If the distance between the plates is doubled and the area is halved,then the new capacitance will be.........$\mu F$.

One plate of a parallel plate capacitor is connected to a spring as shown in the figure. The area of each plate of the capacitor is $A$ and the distance between the plates is $d$,when the battery is not connected and the spring is unstretched. After connecting the battery,in the steady state the distance between the plates is $0.75 d$,then the force constant of the spring is

$A$ parallel plate capacitor is made of two plates with a diameter of $4 \ cm$. What should be the distance between the two plates so that its capacitance is equal to the capacitance of a sphere with a diameter of $20 \ cm$?

Two protons $A$ and $B$ are placed in the space between the plates of a parallel plate capacitor charged up to $V$ volts (see figure). If the forces on the protons are $F_A$ and $F_B$ respectively,then: