Capacitance of a capacitor becomes `7/6` times of its original value if a dielectric slab of thickess `t=2/4d` is introduced in between the plates, `d` is the separation between the plates. The dielectric constant of the dielectric slab is

Options

A. `14/11`
B. `11/14`
C. `7/11`
D. `11/7`

Answer

A

Explanation

The capacitance of a capacitor is given by
`C_(1)=(epsilon_(0)A)/d`……..i
If dielectric slab of constant `K` and thickness `t=2/3 d` is introduced, then capacitance becomes
`C_(2)=(epsilon_(0)A)/(d-t(1-1/K))`
`=(epsilon_(0)A)/(d-2/3d(1-1/K))` .........ii
Given `C_(2)=7/6C_(1)`
Then eq. (i) becomes,
`C_(2)=(epsilon_(0)A)/(d[(1-2/3)+2/(3K)])`
`impliesC_(2)=(C_(1))/((1/3+2/(3K)))`
`implies 7/6C_(1)=(C_(1))/(1/3+2/(3K))`
`=1/3+2/(3K)=6/7`
`implies2/(3K)=6/7-1/3=11/21`
or `33K=21xx2`
`K=42/33=14/11`