In a parallel plate capacitor, the separation between the plates is 3mm with air between them. Now a 1mm thick layer of a material of dielectric constant 2 is introduced between the plates due to which the capacity increases. In order to bring its capacity of the original value, the separation between the plates must be made-

Options

A. 4.5 mm
B. 3.5 mm
C. 2.5 mm
D. 1.5 mm

Answer

B

Explanation

`C_(1)=(epsi_(0)A)/(d) and C_(2)=(epsi_(0)A)/((d-t)+(t)/(K))`
In the second case, `d'=(d-t)+(t)/(K)`
and t=1 mm and K=2
`therefore d'=(3-1)+(1)/(2)+(5)/(2)mm`
thus the effective distance is decreased by `(1)/(2)` mm and the capacitance is increased. hence to restore the capacity to its original value, the original plate separation should be increased by `1/2` mm or it should be made
`3+(1)/(2)=3.5mm`.