The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K_1 and K_2 with thickness 3 8 d and d 2 , respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. (If K_1=1.25 K_2, the value of K_1 is:
A. 2.66
C_1= _0 A / d aligned & C _2= _0 ~A 3 ~d 8 ~K _1 + d 2 ~K _2 + ( d - 3 ~d 8 - d 2 ) \\ & C _2= _0 ~A / d 3 8 ~K _1 + 1 2 ~K _2 + 1 8 \\ & = _0 ~A / d 3 8 ~K _1 + 1 2 ( 4 5 ~K _1 ) + 1 8 \\ & C _2=2 C _1 \\ & _0 ~A / d 3 8 ~K _1 + 5 8 ~K _1 + 1 8 =2 _0 ~A / d \\ & 1 1 ~K _1 + 1 8 =2 1 ~K _1 + 1 8 = 1 2 ~K _1= 8 3 =2.66 aligned
Related: Physics — Capacitance · All PYQ Banks