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Question-36722




Question Number 36722 by Tinkutara last updated on 04/Jun/18
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18
the eletric field is E^→ =Ei^�   let assume  a positive charge Q is situated  at point p(−l,0) which produce Electric field  now potential at point A is V_A   V_A =(Q/(4Πε_0 (l+a)))=(Q/(4Πε_0 ((√(l^2 +a^2 +2la)) )))  V_B =(Q/(4Πε_0 ((√(l^2 +a^2 )) )))=(Q/(4Πε_0 ((√(l^2 +a^2 )))))  V_c =(Q/(4Πε_0 (l−a)))=(Q/(4Πε_0 (√(l^2 +a^2 −2la))))  V_D =(Q/(4Πε_0 ((√(l^2 +a^2 )) )))=(Q/(4Πε_0 (√(l^2 +a^2 ))))   so V_A  is minimum  V_c  is maximum  V_B =V_D
$${the}\:{eletric}\:{field}\:{is}\:\overset{\rightarrow} {{E}}={E}\hat {{i}} \\ $$$${let}\:{assume}\:\:{a}\:{positive}\:{charge}\:{Q}\:{is}\:{situated} \\ $$$${at}\:{point}\:{p}\left(−{l},\mathrm{0}\right)\:{which}\:{produce}\:{Electric}\:{field} \\ $$$${now}\:{potential}\:{at}\:{point}\:{A}\:{is}\:{V}_{{A}} \\ $$$${V}_{{A}} =\frac{{Q}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} \left({l}+{a}\right)}=\frac{{Q}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} \left(\sqrt{{l}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{la}}\:\right)} \\ $$$${V}_{{B}} =\frac{{Q}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} \left(\sqrt{{l}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\right)}=\frac{{Q}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} \left(\sqrt{\left.{l}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}\right.} \\ $$$${V}_{{c}} =\frac{{Q}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} \left({l}−{a}\right)}=\frac{{Q}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} \sqrt{{l}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{la}}} \\ $$$${V}_{{D}} =\frac{{Q}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} \left(\sqrt{{l}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\right)}=\frac{{Q}}{\mathrm{4}\Pi\epsilon_{\mathrm{0}} \sqrt{{l}^{\mathrm{2}} +{a}^{\mathrm{2}} }}\: \\ $$$${so}\:{V}_{{A}} \:{is}\:{minimum} \\ $$$${V}_{{c}} \:{is}\:{maximum} \\ $$$${V}_{{B}} ={V}_{{D}} \\ $$
Commented by Tinkutara last updated on 06/Jun/18
Thank you very much Sir! I got the answer. ��������

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