Capacitance of a parallel plate capacitor becomes `4//3` times its original value if a dielectric slab of thickness `t=d/2` is inserted between the plates [d is the separation between the plates]. The dielectric constant of the slab is

Options

A. 4
B. 8
C. 2
D. 6

Answer

C

Explanation

Thickness of dielectric slab `t=(d)/(2)`. Let its dielectric constant be K. then,
`C.=(epsi_(0)A)/(d-t+(t)/(K))=(epsi_(0)A)/(d-(d)/(2)+(d)/(2K))`
`=(epsi_(0)A)/((d)/(2)(1+(1)/(K)))=(4)/(3)*(epsi_(0)A)/(d)` (given)
`impliesK=2`