A parallel plate capacitor with air between the plates has a capacitance of 8pF (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?
Step 1: Initial capacitance Let the initial distance between the plates be d. Let A be the area of each plate. Capacitance of the parallel plate capacitor is given by: Ci=dA∈0 ....(1) Step 2: Capacitance after inserting dielectric Let k be the dielectric constant. Cf=d′kA∈0 As the distance between plates is reduced to half, new distance is d′=2d ∴Cf=d2kA∈0=2kCi=2×6×8pF=96pF
