A slab of material of dielectric constant K has the same area as the plates of a parallel capacitor, but has a thickness `((3)/(4) d)`,
where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates

Options

A. `(3K)/(K+4)`
B. `(3)/(4)K`
C. `(4K)/(K+3)`
D. `(4)/(3)K`

Answer

C

Explanation

The capacitance of a parallel plate capacitor in the absence of the dielectric is
`C_(0) = (epsilon_(0)A)/(d)` …(i)
where A is the area of each plate and d is the distance between them.
The capacitance of a parallel plate capacitor in the presence of dielectric slab of thickness t and dielectric constant K, is
`C = (epsilon_(0)A)/((d-t)+((t)/(K))) = (epsilon_(0)A)/((d-(3)/(4)d)+((3d)/(4K)))`
`C = (epsilon_(0)A)/((d)/(4) + (3d)/(4K)) = (4K epsilon_(0)A)/(d(K+3))` ...(ii)
Divide (ii) by (i), we get
`(C )/(C_(0)) = (4 K epsilon_(0)A)/(d(K+3)) xx (d)/(epsilon_(0)A) = (4K)/(K + 3)`