A parallel plate capacitor with air between the plates has a capacitance of 8 pF.What will be the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 5?

A Parallel plate capacitor with air between plates has a capacitance of 8pF(1pF=10^-12F) . What will be the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric Constant=6? (Given varepsilon_0=8.85xx10^-12C^2N^-1m^-2)

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^-12 F) . What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

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

A parallel plate capacitor with air between the plates has capacitance of 9pF. The separation between its plates is 'd'. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k_1=3 and thickness d/3 while the other one has dielectric constant k_2=6 and thickness (2d)/(3) . Capacitance of the capacitor is now

A parallel plate capacitor with oil between the plates has a capacitance C(dielectric constant of oil is 2). If the oil is removed, then capacitance of the capacitor becomes