In the figure a capacitor is filled with dielectrics `K_1=2, K_2=3,K_3=6' and 'epsilon_0 A/d=C`. The resultant capacitance between P and Q is

In the figure a capacitor is filled with dielectrics K_(1) , K_(2) and K_(3) . The resultant capacitance is

Two parallel plate capacitors, each of capacitance 40 muF, are connected is series. The space between the plates of one capacitor is filled with a dielectric material of dielectric constant K =4. Find the equivalent capacitacne of the system.

Two identical parallel plate (air) capacitors c_(1) and c_(2) have capacitances C each. The space between their plates is now filled with dielectrics as shown. If two capacitors still have equal capacitance, obtain the relation betweenn dielectric constants k,k_(1) and k_(2) .

A capacitor of plate area A and separation d is filled with two dielectrics of dielectric constant K_(1) = 6 and K_(2) = 4 . New capacitance will be

Space between the plates of a parallel plate capacitor is filled with a dielectric whose dielectric constant varies with distance as per the relation: K(X)=K_(0)+lambdaX , ( lambda= constant, K_(0)= constant, X is perpendicular distance from one plate to a point inside dielectric). The capacitance C_(1) of this capacitor, would be related to its vacuum capacitance C_(0) per the relation (d = plate separation):

Air filled capacitor of capacitance 2 muF is filled with three dielectric material of dielectric constant K_(1) = 4, K _(2) = 4 and K_(3) = 6 as shown in the figure . The new capacitance of the capacitor is

Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K_(1),K_(2) and K_(3) . The first capacitor is filled as shown in fig. I, and the second one is filled as shown in fig. II. If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be ( E_(1) refers to capacitor (I) and E_(2) to capacitor (II)):

Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K_(1), K_(2) and K_(3) . The first capacitor is filled as shown in fig. I, and the second one is filled as shown in fig. II. If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be ( E_(1) refers to capacitor (I) and E_(2) to capacitor (II)):

A parallel plate capacitor of plate area A and separation d filled with three dielectric materials as show in figure. The dielectric constants are K_1 , K_2 and K_3 respectively. The capacitance across AB will be :

The capacitance of a parallel plate capacitor with plate area A and separation d is C . The space between the plates in filled with two wedges of dielectric constants K_(1) and K_(2) respectively. Find the capacitance of resulting capacitor.