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.........$\mu F$.

As shown in the figure,two parallel plate capacitors having equal plate area of $200 \, cm^2$ are joined in such a way that $a \neq b$. The equivalent capacitance of the combination is $x \varepsilon_0 \, F$. The value of $x$ 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 $20 \ cm$?

What is a parallel plate capacitor? Obtain the formula for the capacitance of such a parallel plate capacitor and state the factors on which the value of capacitance depends.

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

$A$ 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