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)(Kin_(0)a^(2))/(d)`
B. `(Kin_(0)a^(2))/(d)lnK`
C. `(Kin_(0)a^(2))/(d(K-1))lnK`
D. `(Kin_(0)a^(4))/(2d(K+1))`

Answer

C

Explanation

For vacuum `k_(1)=1`
For dielectric `k_(2)=k`
Consider two elemental slabs (one in vacuum and other in dielectric) having capacitances `dC_(1) & dC_(2),dC_(1) and dC_(2)` are in series.
`(1)/(dC)=(1)/(dC_(1))+(1)/(dC_(2))=(y)/(in_(0)k_(1)adx)+(d-y)/(in_(0)k_(2)adx),dC`=Capacitance of composite elemental slabs.
`implies (1)/(dC)=(k_(2)+y+(d-y)k_(1))/(in_(0)k_(1)k_(2)adx)=(ky+d-y)/(in_(0)kadx)`
All such dC's are in parallel `therefore`Total capacitance, `C=intdC=in_(0)kaint(dx)/(ky+d-y)` . . (i)
Now, `(d-y)/(x)=(d)/(a) implies x=(d-y)(a)/(d) ` . . (ii) , from equation (i) & (ii)
`C=in_(0)ka[a/d]int_(0)^(-d)(dy)/((k-1)y+d)=(in_(0)ka^(2))/(d)(ln[(k-1)y+d]_(0)^(d))/((k-1)),C=(in_(0)ka^(2))/(d)(lnk)/((k-1))` i.e., option (C).