A parallel plate air capacitor has a capacitance of 2pF. Now, the separation between the plates is doubled, and the space is filled with wax. If the capacitance rises to 6pF, what is the dielectric constant of wax?

Explanation

Initial separation between the plates = d, area of each plate =`alpha`.
Capacitance in the 1st case,
`C_1= (in_0alpha)/(d)`
Final separation between the plates =2d, dielectric constant of wax = k.
Capacitance in the 2nd case,
`C_2 = (k in_0alpha)/(2d)`
`(C_1)/(C_2)=(2)/(k)" or" k= 2xx(C_2)/(C_1)= 2xx(6pF)/(2pF)=6`.