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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

	$B e f o r e d i e l e c t r i c i s i n s e r t e d$ $C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}}$ $q_{1} = q_{2} = C_{e q} V_{b} (V_{b} b e i n g e m f o f B)$ $= \frac{C_{1} C_{2} V_{b}}{C_{1} + C_{2}}$ $V_{1} = \frac{q_{1}}{C_{1}} = \frac{C_{2} V_{b}}{C_{1} + C_{2}}; V_{2} = \frac{C_{1} V_{b}}{C_{1} + C_{2}}$ $U_{1} = \frac{q_{1}^{2}}{2 C_{1}} = {(\frac{C_{1} C_{2}}{C_{1} + C_{2}})}^{2} \frac{V_{b}^{2}}{2 C_{1}}$ $= \frac{C_{1} C_{2}^{2} V_{b}^{2}}{2 {(C_{1} + C_{2})}^{2}}$ $U_{2} = \frac{q_{2}^{2}}{2 C_{2}} = \frac{C_{1}^{2} C_{2}^{} V_{b}^{2}}{2 {(C_{1} + C_{2})}^{2}}$ $l e t d i e l e c t r i c c o n s t a n t o f s l a b b e K .$ $A f t e r i t i s s l i p p e d i n$ $C_{2}^{'} = {K C}_{2}$ $C_{e q}^{'} = \frac{{K C}_{1} C_{2}}{C_{1} + {K C}_{2}} > \frac{C_{1} C_{2}}{C_{1} + C_{2}}$ $q_{1}^{'} = q_{2}^{'} = C_{e q}^{'} V_{b} = \frac{{K C}_{1} C_{2} V_{b}}{C_{1} + {K C}_{2}} > q_{1} (= q_{2})$ $V_{1}^{'} = \frac{q_{1}^{'}}{C_{1}} = \frac{{K C}_{2} V_{b}}{C_{1} + {K C}_{2}} > V_{1}$ $V_{2}^{'} = \frac{q_{2}^{'}}{{K C}_{2}} = \frac{C_{1} V_{b}}{C_{1} + {K C}_{2}} < V_{2}$ $U_{1}^{'} = \frac{{(q_{1}^{'})}^{2}}{2 C_{1}} = \frac{1}{2 C_{1}} {(\frac{{K C}_{1} C_{2} V_{b}}{C_{1} + {K C}_{2}})}^{2} > U_{1}$ $U_{2}^{'} = \frac{{(q_{2}^{'})}^{2}}{2 {K C}_{2}} = \frac{1}{2 {K C}_{2}} {(\frac{{K C}_{1} C_{2} V_{b}}{C_{1} + {K C}_{2}})}^{2}$ $\frac{U_{2}^{'}}{U_{2}} = \frac{\frac{1}{2 {K C}_{2}} {(\frac{{K C}_{1} C_{2} V_{b}}{C_{1} + {K C}_{2}})}^{2}}{\frac{C_{1}^{2} C_{2}^{} V_{b}^{2}}{2 {(C_{1} + C_{2})}^{2}}}$ $= \frac{K {(C_{1} + C_{2})}^{2}}{{(C_{1} + {K C}_{2})}^{2}} = K {[\frac{1 + \frac{C_{1}}{C_{2}}}{\frac{C_{1}}{C_{2}} + K}]}^{2}$ $= K {(\frac{1 + r}{K + r})}^{2} w h e r e r = \frac{C_{1}}{C_{2}}$ $S o i b e l i e v e i t d e p e n d s o n r & K .$
	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

	$e x c e l l e n t . . . .$
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