question_answer

A parallel plate capacitor with air between the plates has a capacitance of 8 pF. The separation between the plates is now reduced by its half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of the capacitance of the capacitor in the second case.

Or

Two metal plates form a parallel plate capacitor. The distance between the plates is d. A metal sheet of thickness d/2 and of the same area is introduced between the plates. What is the ratio of the capacitances in the two cases?

Answer:

80 pF    Or         1 : 2 Capacitance \[C=8pF\] We know that, \[C={{e}_{0}}A\,/\,d\] Where, \[A=\] area of plates \[d=\] Separation between the plates If plate separation is reduced to half i.e. \[\frac{d}{2}\] and space between the plates is filled with medium of dielectric constant 5, the new capacitance is \[C'=\frac{5{{\varepsilon }_{0}}A}{d\,/\,2}=\frac{10{{\varepsilon }_{0}}A}{d}=10C=10\times 8=80pF\] Or Let initial capacitance be C, \[C={{\varepsilon }_{0}}A\,/\,d\] When a metal sheet of thickness \[\frac{d}{2}\] is introduced between the plates the new capacitance is \[C'=\frac{{{\varepsilon }_{0}}A}{d-\frac{d}{2}}\] \[=\frac{{{\varepsilon }_{0}}A}{d\,/\,2}=\frac{2{{\varepsilon }_{0}}A}{d}=2C\] Ratio of capacitances \[\frac{C}{C'}=\frac{1}{2}\]