In a parallel plate capacitor,the capacity increases if

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.

$A$ parallel plate capacitor consisting of two circular plates of radius $10 \ cm$ is being charged by a constant current of $0.15 \ A$. If the rate of change of potential difference between the plates is $7 \times 10^8 \ V/s$,then the integer value of the distance between the parallel plates is $—$ (Take $\epsilon_0 = 9 \times 10^{-12} \ F/m, \pi = 22/7$) . . . . . . $\mu m$.

Calculate the capacitance of two plates of equal area of $1 \ m^2$ separated by a distance of $1 \ mm$.

Two parallel plates separated by a distance of $5\,mm$ are kept at a potential difference of $50\,V.$ $A$ particle of mass $10^{-15}\,kg$ and charge $10^{-11}\,C$ enters in it with a velocity $10^7\,m/s.$ The acceleration of the particle will be

$A$ parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles,then