Question Number 40732 by Tinkutara last updated on 26/Jul/18
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Commented by Tinkutara last updated on 26/Jul/18
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Commented by Tinkutara last updated on 26/Jul/18
Please help in comparing energy of C2 initially and finally.
Answered by ajfour last updated on 27/Jul/18
![Before dielectric is inserted C_(eq) =((C_1 C_2 )/(C_1 +C_2 )) q_1 =q_2 =C_(eq) V_b (V_b being emf of B) =((C_1 C_2 V_b )/(C_1 +C_2 )) V_1 =(q_1 /C_1 ) =((C_2 V_b )/(C_1 +C_2 )) ; V_2 =((C_1 V_b )/(C_1 +C_2 )) U_1 =(q_1 ^2 /(2C_1 )) = (((C_1 C_2 )/(C_1 +C_2 )))^2 (V_b ^( 2) /(2C_1 )) = ((C_1 C_2 ^( 2) V_b ^( 2) )/(2(C_1 +C_2 )^2 )) U_2 =(q_2 ^2 /(2C_2 )) = ((C_1 ^( 2) C_2 ^ V_b ^( 2) )/(2(C_1 +C_2 )^2 )) let dielectric constant of slab be K. After it is slipped in C_2 ′=KC_2 C_(eq) ′=((KC_1 C_2 )/(C_1 +KC_2 )) > ((C_1 C_2 )/(C_1 +C_2 )) q_1 ′=q_2 ′=C_(eq) ′ V_b =((KC_1 C_2 V_b )/(C_1 +KC_2 )) > q_1 (=q_2 ) V_1 ′=((q_1 ′)/C_1 ) =((KC_2 V_b )/(C_1 +KC_2 )) > V_1 V_2 ′=((q_2 ′)/(KC_2 ))=((C_1 V_b )/(C_1 +KC_2 )) < V_2 U_1 ′=(((q_1 ′)^2 )/(2C_1 ))=(1/(2C_1 ))(((KC_1 C_2 V_b )/(C_1 +KC_2 )))^2 > U_1 U_2 ′=(((q_2 ′)^2 )/(2KC_2 ))=(1/(2KC_2 ))(((KC_1 C_2 V_b )/(C_1 +KC_2 )))^2 ((U_2 ′)/U_2 )=(((1/(2KC_2 ))(((KC_1 C_2 V_b )/(C_1 +KC_2 )))^2 )/((C_1 ^( 2) C_2 ^ V_b ^( 2) )/(2(C_1 +C_2 )^2 ))) =((K(C_1 +C_2 )^2 )/((C_1 +KC_2 )^2 )) = K[((1+(C_1 /C_2 ))/((C_1 /C_2 )+K))]^2 = K(((1+r)/(K+r)))^2 where r=(C_1 /C_2 ) So i believe it depends on r &K.](https://www.tinkutara.com/question/Q40739.png)
Commented by Tinkutara last updated on 27/Jul/18
Thanks Sir! I also believe it should be.
Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18
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