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Question-40732




Question Number 40732 by Tinkutara last updated on 26/Jul/18
Commented by Tinkutara last updated on 26/Jul/18
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.
BeforedielectricisinsertedCeq=C1C2C1+C2q1=q2=CeqVb(VbbeingemfofB)=C1C2VbC1+C2V1=q1C1=C2VbC1+C2;V2=C1VbC1+C2U1=q122C1=(C1C2C1+C2)2Vb22C1=C1C22Vb22(C1+C2)2U2=q222C2=C12C2Vb22(C1+C2)2letdielectricconstantofslabbeK.AfteritisslippedinC2=KC2Ceq=KC1C2C1+KC2>C1C2C1+C2q1=q2=CeqVb=KC1C2VbC1+KC2>q1(=q2)V1=q1C1=KC2VbC1+KC2>V1V2=q2KC2=C1VbC1+KC2<V2U1=(q1)22C1=12C1(KC1C2VbC1+KC2)2>U1U2=(q2)22KC2=12KC2(KC1C2VbC1+KC2)2U2U2=12KC2(KC1C2VbC1+KC2)2C12C2Vb22(C1+C2)2=K(C1+C2)2(C1+KC2)2=K[1+C1C2C1C2+K]2=K(1+rK+r)2wherer=C1C2Soibelieveitdependsonr&K.
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
excellent....
excellent.

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