A parallel plate capacitor is made of two square plates of side 'a' , separated by a distance d much less than a. The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this Capacitor is :
a) `(Kin_(0)a^(2))/(2d(K+1))` b) `(Kin_(0)a^(2))/(d(K-1)) In K` c) `(Kin_(0)a^(2))/(d)In K` d) `(1)/(2)(Kin_(0)a^(2))/(d)`

Options

A. `1/2 (K epsilon_(0)a^(2))/d`
B. `( epsilon_(0)a^(2))/d "In" K`
C. `(K epsilon_(0)a^(2))/(d(K-1))"In" K`
D. `(K epsilon_(0)a^(2))/(2d(K+1))`

Answer

C

Explanation

For vacuum `k_(1)=1`
For dielectric `k_(2)=k`
Consier two elemental slabs. (one in vacum and other in dielectric) having capacitances `dC_(1)` and `dC_(2).dC_(1)` and `dC_(2)` are in series.
`1/(dC)=1/(dC_(1))+1/(dC_(2))=y/(epsilon_(0)k_(1)adx)+(d-y)/(epsilon_(0)k_(2)adx),dC=` Capitances of composite elemental slabs.
`implies1/(dC)=(k_(2)y+(d-y)k_(1))/(epsilon_(0)k_(1)k_(2)a dx)=(ky+d-y)/(epsilon_(0)kadx)`
All such dC's are in parallel.
`:.` Total capacitance `C=int dC=epsilon_(0)ka int(dx)/(ky+d-y)`....i
Now `(d-y)/x=d/aimpliesx=(d-y)a/d`.......ii
From equation i and ii
`C=epsilon_(0)ka[a/d] int_(0)^(d)(dy)/((k-1)y+d)=(epsilon_(0)ka^(2))/d(In[(k-1)y+d]_(0)^(d))/((k-1))`
`C=(epsilon_(0)ka^(2))/d(Ink)/((k-1))`