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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.](Q40739.png)
$${Before}\:{dielectric}\:{is}\:{inserted} \\ $$$${C}_{{eq}} =\frac{{C}_{\mathrm{1}} {C}_{\mathrm{2}} }{{C}_{\mathrm{1}} +{C}_{\mathrm{2}} } \\ $$$${q}_{\mathrm{1}} ={q}_{\mathrm{2}} ={C}_{{eq}} {V}_{{b}} \:\:\:\:\:\left({V}_{{b}} \:{being}\:{emf}\:{of}\:{B}\right) \\ $$$$\:\:\:\:=\frac{{C}_{\mathrm{1}} {C}_{\mathrm{2}} {V}_{{b}} }{{C}_{\mathrm{1}} +{C}_{\mathrm{2}} } \\ $$$${V}_{\mathrm{1}} =\frac{{q}_{\mathrm{1}} }{{C}_{\mathrm{1}} }\:=\frac{{C}_{\mathrm{2}} {V}_{{b}} }{{C}_{\mathrm{1}} +{C}_{\mathrm{2}} }\:;\:\:{V}_{\mathrm{2}} =\frac{{C}_{\mathrm{1}} {V}_{{b}} }{{C}_{\mathrm{1}} +{C}_{\mathrm{2}} } \\ $$$${U}_{\mathrm{1}} =\frac{{q}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}{C}_{\mathrm{1}} }\:=\:\left(\frac{{C}_{\mathrm{1}} {C}_{\mathrm{2}} }{{C}_{\mathrm{1}} +{C}_{\mathrm{2}} }\right)^{\mathrm{2}} \frac{{V}_{{b}} ^{\:\:\mathrm{2}} }{\mathrm{2}{C}_{\mathrm{1}} } \\ $$$$\:\:\:\:=\:\frac{{C}_{\mathrm{1}} {C}_{\mathrm{2}} ^{\:\mathrm{2}} {V}_{{b}} ^{\:\:\mathrm{2}} }{\mathrm{2}\left({C}_{\mathrm{1}} +{C}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${U}_{\mathrm{2}} =\frac{{q}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}{C}_{\mathrm{2}} }\:=\:\frac{{C}_{\mathrm{1}} ^{\:\mathrm{2}} {C}_{\mathrm{2}} ^{\:} {V}_{{b}} ^{\:\:\mathrm{2}} }{\mathrm{2}\left({C}_{\mathrm{1}} +{C}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${let}\:{dielectric}\:{constant}\:{of}\:{slab}\:{be}\:{K}. \\ $$$${After}\:{it}\:{is}\:{slipped}\:{in} \\ $$$${C}_{\mathrm{2}} '={KC}_{\mathrm{2}} \\ $$$${C}_{{eq}} '=\frac{{KC}_{\mathrm{1}} {C}_{\mathrm{2}} }{{C}_{\mathrm{1}} +{KC}_{\mathrm{2}} }\:>\:\frac{{C}_{\mathrm{1}} {C}_{\mathrm{2}} }{{C}_{\mathrm{1}} +{C}_{\mathrm{2}} } \\ $$$${q}_{\mathrm{1}} '={q}_{\mathrm{2}} '={C}_{{eq}} '\:{V}_{{b}} =\frac{{KC}_{\mathrm{1}} {C}_{\mathrm{2}} {V}_{{b}} }{{C}_{\mathrm{1}} +{KC}_{\mathrm{2}} }\:>\:{q}_{\mathrm{1}} \left(={q}_{\mathrm{2}} \right) \\ $$$${V}_{\mathrm{1}} '=\frac{{q}_{\mathrm{1}} '}{{C}_{\mathrm{1}} }\:=\frac{{KC}_{\mathrm{2}} {V}_{{b}} }{{C}_{\mathrm{1}} +{KC}_{\mathrm{2}} }\:>\:{V}_{\mathrm{1}} \\ $$$${V}_{\mathrm{2}} '=\frac{{q}_{\mathrm{2}} '}{{KC}_{\mathrm{2}} }=\frac{{C}_{\mathrm{1}} {V}_{{b}} }{{C}_{\mathrm{1}} +{KC}_{\mathrm{2}} }\:<\:{V}_{\mathrm{2}} \\ $$$${U}_{\mathrm{1}} '=\frac{\left({q}_{\mathrm{1}} '\right)^{\mathrm{2}} }{\mathrm{2}{C}_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{2}{C}_{\mathrm{1}} }\left(\frac{{KC}_{\mathrm{1}} {C}_{\mathrm{2}} {V}_{{b}} }{{C}_{\mathrm{1}} +{KC}_{\mathrm{2}} }\right)^{\mathrm{2}} \:>\:{U}_{\mathrm{1}} \\ $$$${U}_{\mathrm{2}} '=\frac{\left({q}_{\mathrm{2}} '\right)^{\mathrm{2}} }{\mathrm{2}{KC}_{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}{KC}_{\mathrm{2}} }\left(\frac{{KC}_{\mathrm{1}} {C}_{\mathrm{2}} {V}_{{b}} }{{C}_{\mathrm{1}} +{KC}_{\mathrm{2}} }\right)^{\mathrm{2}} \: \\ $$$$\:\:\:\:\:\frac{{U}_{\mathrm{2}} '}{{U}_{\mathrm{2}} }=\frac{\frac{\mathrm{1}}{\mathrm{2}{KC}_{\mathrm{2}} }\left(\frac{{KC}_{\mathrm{1}} {C}_{\mathrm{2}} {V}_{{b}} }{{C}_{\mathrm{1}} +{KC}_{\mathrm{2}} }\right)^{\mathrm{2}} }{\frac{{C}_{\mathrm{1}} ^{\:\mathrm{2}} {C}_{\mathrm{2}} ^{\:} {V}_{{b}} ^{\:\:\mathrm{2}} }{\mathrm{2}\left({C}_{\mathrm{1}} +{C}_{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:=\frac{{K}\left({C}_{\mathrm{1}} +{C}_{\mathrm{2}} \right)^{\mathrm{2}} }{\left({C}_{\mathrm{1}} +{KC}_{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:{K}\left[\frac{\mathrm{1}+\frac{{C}_{\mathrm{1}} }{{C}_{\mathrm{2}} }}{\frac{{C}_{\mathrm{1}} }{{C}_{\mathrm{2}} }+{K}}\right]^{\mathrm{2}} \\ $$$$\:\:\:=\:{K}\left(\frac{\mathrm{1}+{r}}{{K}+{r}}\right)^{\mathrm{2}} \:\:\:{where}\:{r}=\frac{{C}_{\mathrm{1}} }{{C}_{\mathrm{2}} } \\ $$$$\:\:{So}\:{i}\:{believe}\:{it}\:{depends}\:{on}\:{r}\:\&{K}. \\ $$
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
$${excellent}.... \\ $$