What will be the change in the capacitance of a capacitor, if the separation between the plates is doubled?

What will be the change in the capacitance of a parallel plate capacitor, if the surface area of its plates is doubled and separation between plates doubled?

To reduce the capacitance of parallel plate capacitor, the space between the plate is

A parallel plate air capacitor has a capacity of 2 pF. If the separation between its plates is doubled and a mica sheet is introduced between its plates, its capacity becomes 6 pF. What is the dielectric constant of mica?

The capacitance of an air filled parallel plate capacitor is 10 muF . The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of 40 xx 10^(-12) farads. The dielectirc consatnt of wax is

A parallel plate capacitor with air inbetween the paltes 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 5. Calculate the new value of capacitance.

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

A parallel plate capacitor with air between the plates has a capacitance C. If the distance between the plates is doubled and the space between the plates is filled with a dielectric of dielectric constant 6, then the capacitance will become.

AN air filled parallel plate capacitor has a capacitance of 2.1 pF. The separation of the plates is doubled, and wax is inserted between them . The new capacitance is 2.6 p F Find the dielectric constant of the wax.