A capacitor is formed by two square metal plates of edge a, separated by a distance d, Dielectrics of dielectric constants #K_1# and #K_2# are filled in the gap as shown in figure. Find the capacitance.?

Consider an infinitesimal strip of of capacitor of thickness #dx# at a distance #x# as shown. Keeping in view the position of terminlas we see that the infinitesimal capacitors so formed are connected in series.
Plate separation for lower part is #xtanalpha# whereas for the upper part it is #(d-xtanalpha)#.
We know that equivalent capacitance of capacitors connected in series is given by the expression

#1/(dC_(eq))=1/(dC_(K_1))+1/(dC_(K_2))#

Inserting value for Parallel-plate capacitor of plate area #A# and plate separation #t# as #C=epsilon_0A/t#, we get

#1/(dC_(eq))=(d-xtanalpha)/(K_1epsilon_0(adx))+(xtanalpha)/(K_2epsilon_0(adx))#
#=>1/(dC_(eq))=1/(epsilon_0adx)[(d-xtanalpha)/(K_1)+(xtanalpha)/(K_2)]#
#=>dC_(eq)=(epsilon_0adx)/[(d-xtanalpha)/(K_1)+(xtanalpha)/(K_2)]#
#=>dC_(eq)=(K_1K_2epsilon_0adx)/[K_2(d-xtanalpha)+K_1(xtanalpha)]#
#=>dC_(eq)=(K_1K_2epsilon_0adx)/[K_2d+(K_1-K_2)(tanalpha)x]#

Integrating both sides with respect to respective variables from #C=0toC_(eq) and x=0toa# we get

#C_(eq)=(K_1K_2epsilon_0a)int_0^a(dx)/[K_2d+(K_1-K_2)(tanalpha)x]#

Using the following integral we get

#int(dx)/(b*x+c)=ln(abs(b*x+c))/b#

#C_(eq)=(K_1K_2epsilon_0a)/((K_1-K_2)(tanalpha))[ln(|K_2d+(K_1-K_2)(tanalpha)x|)]_0^a#

#=>C_(eq)=(K_1K_2epsilon_0a)/((K_1-K_2)(tanalpha))[ln(|K_2d+(K_1-K_2)(tanalpha)a|)-(ln(|K_2d|)]#

Inserting #tanalpha=d/a#, we get
#C_(eq)=(K_1K_2epsilon_0a^2)/((K_1-K_2)d)[ln(|K_2d+(K_1-K_2)d|)-ln(|K_2d|)]#

Simplifying

#C_(eq)=(K_1K_2epsilon_0a^2)/((K_1-K_2)d)[ln((K_1d)-(ln(K_2d)]#
#=>C_(eq)=(K_1K_2epsilon_0a^2)/((K_1-K_2)d)ln(K_1/K_2)#