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 capaciitor is…

Question

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 capaciitor is filled as shown in figure

I,

and the second one is filled as shown in figure

I I .

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

(I I)

):

Aicharya · Accepted Answer

Energy stored in a capacitor E = 1 2 C V 2 &#8658; E 1 E 2 = C 1 C 2 C 1 = d 3 &#8201; &#949; 0 k 1 A + d 3 &#949; 0 A k 2 + d 3 &#949; 0 A k 3 - 1 = d 3 A &#949; 0 - 1 1 k 1 + 1 k 2 + 1 k 3 - 1 = d 3 A &#949; 0 - 1 k 1 k 2 + k 2 k 3 + k 3 k 1 k 1 k 2 k 3 - 1 C 1 = 3 A &#949; 0 d k 1 k 2 k 3 k 1 k 2 + k 2 k 3 + k 3 k 1 C 2 = &#949; 0 A 3 k 1 d + &#949; 0 A 3 k 2 d + &#949; 0 A 3 k 3 d &#8756; C 1 C 2 = E 1 E 2 = 9 &#8201; k 1 k 2 k 3 k 1 + k 2 + k 3 k 1 k 2 + k 2 k 3 + k 3 k 1

Two identical parallel plate capacitors, of capacitance C each, have plates of area A , separated by a…

Solution